YES

We show the termination of the TRS R:

  a__zeros() -> cons(|0|(),zeros())
  a__U11(tt(),L) -> s(a__length(mark(L)))
  a__and(tt(),X) -> mark(X)
  a__isNat(|0|()) -> tt()
  a__isNat(length(V1)) -> a__isNatList(V1)
  a__isNat(s(V1)) -> a__isNat(V1)
  a__isNatIList(V) -> a__isNatList(V)
  a__isNatIList(zeros()) -> tt()
  a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
  a__isNatList(nil()) -> tt()
  a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
  a__length(nil()) -> |0|()
  a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
  mark(zeros()) -> a__zeros()
  mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
  mark(length(X)) -> a__length(mark(X))
  mark(and(X1,X2)) -> a__and(mark(X1),X2)
  mark(isNat(X)) -> a__isNat(X)
  mark(isNatList(X)) -> a__isNatList(X)
  mark(isNatIList(X)) -> a__isNatIList(X)
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(|0|()) -> |0|()
  mark(tt()) -> tt()
  mark(s(X)) -> s(mark(X))
  mark(nil()) -> nil()
  a__zeros() -> zeros()
  a__U11(X1,X2) -> U11(X1,X2)
  a__length(X) -> length(X)
  a__and(X1,X2) -> and(X1,X2)
  a__isNat(X) -> isNat(X)
  a__isNatList(X) -> isNatList(X)
  a__isNatIList(X) -> isNatIList(X)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__U11#(tt(),L) -> mark#(L)
p3: a__and#(tt(),X) -> mark#(X)
p4: a__isNat#(length(V1)) -> a__isNatList#(V1)
p5: a__isNat#(s(V1)) -> a__isNat#(V1)
p6: a__isNatIList#(V) -> a__isNatList#(V)
p7: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p8: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1)
p9: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p10: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p11: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p12: a__length#(cons(N,L)) -> a__and#(a__isNatList(L),isNat(N))
p13: a__length#(cons(N,L)) -> a__isNatList#(L)
p14: mark#(zeros()) -> a__zeros#()
p15: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p16: mark#(U11(X1,X2)) -> mark#(X1)
p17: mark#(length(X)) -> a__length#(mark(X))
p18: mark#(length(X)) -> mark#(X)
p19: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p20: mark#(and(X1,X2)) -> mark#(X1)
p21: mark#(isNat(X)) -> a__isNat#(X)
p22: mark#(isNatList(X)) -> a__isNatList#(X)
p23: mark#(isNatIList(X)) -> a__isNatIList#(X)
p24: mark#(cons(X1,X2)) -> mark#(X1)
p25: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p15, p16, p17, p18, p19, p20, p21, p22, p23, p24, p25}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__isNatList#(L)
p3: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p4: a__isNat#(s(V1)) -> a__isNat#(V1)
p5: a__isNat#(length(V1)) -> a__isNatList#(V1)
p6: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p7: a__and#(tt(),X) -> mark#(X)
p8: mark#(s(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(isNatIList(X)) -> a__isNatIList#(X)
p11: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1)
p12: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p13: a__isNatIList#(V) -> a__isNatList#(V)
p14: mark#(isNatList(X)) -> a__isNatList#(X)
p15: mark#(isNat(X)) -> a__isNat#(X)
p16: mark#(and(X1,X2)) -> mark#(X1)
p17: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p18: mark#(length(X)) -> mark#(X)
p19: mark#(length(X)) -> a__length#(mark(X))
p20: a__length#(cons(N,L)) -> a__and#(a__isNatList(L),isNat(N))
p21: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p22: a__U11#(tt(),L) -> mark#(L)
p23: mark#(U11(X1,X2)) -> mark#(X1)
p24: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__U11#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (9,3)
          tt_A() = (2,2)
          a__length#_A(x1) = ((0,1),(0,0)) x1 + (9,3)
          mark_A(x1) = ((1,1),(0,1)) x1 + (6,2)
          cons_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(1,1)) x2 + (4,2)
          a__isNatList#_A(x1) = (7,3)
          a__isNat#_A(x1) = (7,3)
          s_A(x1) = ((0,1),(0,1)) x1 + (5,0)
          length_A(x1) = ((0,1),(0,1)) x1 + (14,5)
          a__and#_A(x1,x2) = ((0,1),(0,0)) x2 + (7,3)
          a__isNat_A(x1) = ((0,1),(0,0)) x1 + (3,2)
          isNatList_A(x1) = (1,0)
          mark#_A(x1) = ((0,1),(0,0)) x1 + (7,3)
          isNatIList_A(x1) = ((0,1),(0,0)) x1 + (3,0)
          a__isNatIList#_A(x1) = (7,3)
          isNat_A(x1) = ((0,1),(0,0)) x1 + (1,0)
          and_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(0,1)) x2
          a__isNatList_A(x1) = (7,2)
          a__and_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(0,1)) x2 + (4,0)
          U11_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (12,5)
          a__zeros_A() = (5,2)
          |0|_A() = (5,0)
          zeros_A() = (0,0)
          a__U11_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(0,1)) x2 + (23,5)
          a__length_A(x1) = ((0,1),(0,1)) x1 + (22,5)
          a__isNatIList_A(x1) = ((0,1),(0,0)) x1 + (8,2)
          nil_A() = (6,2)
  
    precedence:
    
      |0| > nil > tt = mark = s = length = a__isNat = isNatIList = isNat = a__isNatList = a__and = U11 = a__U11 = a__length = a__isNatIList > a__zeros > cons > a__U11# = a__length# = a__isNatList# = a__isNat# = a__and# = mark# = a__isNatIList# > and > zeros > isNatList
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = []
      pi(cons) = []
      pi(a__isNatList#) = []
      pi(a__isNat#) = []
      pi(s) = []
      pi(length) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(mark#) = []
      pi(isNatIList) = []
      pi(a__isNatIList#) = []
      pi(isNat) = []
      pi(and) = []
      pi(a__isNatList) = []
      pi(a__and) = []
      pi(U11) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p20

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__isNatList#(L)
p3: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p4: a__isNat#(s(V1)) -> a__isNat#(V1)
p5: a__isNat#(length(V1)) -> a__isNatList#(V1)
p6: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p7: a__and#(tt(),X) -> mark#(X)
p8: mark#(s(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(isNatIList(X)) -> a__isNatIList#(X)
p11: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1)
p12: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p13: a__isNatIList#(V) -> a__isNatList#(V)
p14: mark#(isNatList(X)) -> a__isNatList#(X)
p15: mark#(isNat(X)) -> a__isNat#(X)
p16: mark#(and(X1,X2)) -> mark#(X1)
p17: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p18: mark#(length(X)) -> mark#(X)
p19: mark#(length(X)) -> a__length#(mark(X))
p20: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p21: a__U11#(tt(),L) -> mark#(L)
p22: mark#(U11(X1,X2)) -> mark#(X1)
p23: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21, p22, p23}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p3: a__U11#(tt(),L) -> mark#(L)
p4: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p5: mark#(U11(X1,X2)) -> mark#(X1)
p6: mark#(length(X)) -> a__length#(mark(X))
p7: a__length#(cons(N,L)) -> a__isNatList#(L)
p8: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p9: a__and#(tt(),X) -> mark#(X)
p10: mark#(length(X)) -> mark#(X)
p11: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p12: mark#(and(X1,X2)) -> mark#(X1)
p13: mark#(isNat(X)) -> a__isNat#(X)
p14: a__isNat#(length(V1)) -> a__isNatList#(V1)
p15: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p16: a__isNat#(s(V1)) -> a__isNat#(V1)
p17: mark#(isNatList(X)) -> a__isNatList#(X)
p18: mark#(isNatIList(X)) -> a__isNatIList#(X)
p19: a__isNatIList#(V) -> a__isNatList#(V)
p20: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p21: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1)
p22: mark#(cons(X1,X2)) -> mark#(X1)
p23: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__U11#_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (2,0)
          tt_A() = (0,0)
          a__length#_A(x1) = ((1,0),(1,1)) x1 + (2,0)
          mark_A(x1) = x1
          cons_A(x1,x2) = x1 + ((1,1),(1,1)) x2
          a__and_A(x1,x2) = x1 + x2
          a__isNatList_A(x1) = ((0,1),(1,1)) x1
          isNat_A(x1) = ((0,1),(1,1)) x1
          mark#_A(x1) = ((1,1),(0,1)) x1
          U11_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
          length_A(x1) = ((1,1),(1,1)) x1 + (1,1)
          a__isNatList#_A(x1) = ((1,1),(1,0)) x1
          a__and#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(0,1)) x2
          a__isNat_A(x1) = ((0,1),(1,1)) x1
          isNatList_A(x1) = ((0,1),(1,1)) x1
          and_A(x1,x2) = x1 + x2
          a__isNat#_A(x1) = ((0,1),(1,0)) x1
          s_A(x1) = x1
          isNatIList_A(x1) = ((0,1),(1,1)) x1 + (1,1)
          a__isNatIList#_A(x1) = ((1,1),(1,0)) x1 + (2,1)
          a__zeros_A() = (0,0)
          |0|_A() = (0,0)
          zeros_A() = (0,0)
          a__U11_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
          a__length_A(x1) = ((1,1),(1,1)) x1 + (1,1)
          a__isNatIList_A(x1) = ((0,1),(1,1)) x1 + (1,1)
          nil_A() = (1,1)
  
    precedence:
    
      mark = a__and = a__isNatList = and = isNatIList = a__zeros = a__isNatIList > a__isNat > tt > isNatList > cons = mark# = a__isNatList# = a__and# = a__isNat# = a__isNatIList# > a__U11# = a__length# = s = a__U11 = a__length > |0| > U11 > isNat > length > zeros = nil
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = []
      pi(cons) = []
      pi(a__and) = []
      pi(a__isNatList) = []
      pi(isNat) = []
      pi(mark#) = []
      pi(U11) = [2]
      pi(length) = []
      pi(a__isNatList#) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(and) = []
      pi(a__isNat#) = []
      pi(s) = []
      pi(isNatIList) = []
      pi(a__isNatIList#) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p3: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p4: mark#(U11(X1,X2)) -> mark#(X1)
p5: mark#(length(X)) -> a__length#(mark(X))
p6: a__length#(cons(N,L)) -> a__isNatList#(L)
p7: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p8: a__and#(tt(),X) -> mark#(X)
p9: mark#(length(X)) -> mark#(X)
p10: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p11: mark#(and(X1,X2)) -> mark#(X1)
p12: mark#(isNat(X)) -> a__isNat#(X)
p13: a__isNat#(length(V1)) -> a__isNatList#(V1)
p14: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p15: a__isNat#(s(V1)) -> a__isNat#(V1)
p16: mark#(isNatList(X)) -> a__isNatList#(X)
p17: mark#(isNatIList(X)) -> a__isNatIList#(X)
p18: a__isNatIList#(V) -> a__isNatList#(V)
p19: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p20: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1)
p21: mark#(cons(X1,X2)) -> mark#(X1)
p22: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21, p22}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__isNatList#(L)
p3: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p4: a__isNat#(s(V1)) -> a__isNat#(V1)
p5: a__isNat#(length(V1)) -> a__isNatList#(V1)
p6: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p7: a__and#(tt(),X) -> mark#(X)
p8: mark#(s(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(isNatIList(X)) -> a__isNatIList#(X)
p11: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1)
p12: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p13: a__isNatIList#(V) -> a__isNatList#(V)
p14: mark#(isNatList(X)) -> a__isNatList#(X)
p15: mark#(isNat(X)) -> a__isNat#(X)
p16: mark#(and(X1,X2)) -> mark#(X1)
p17: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p18: mark#(length(X)) -> mark#(X)
p19: mark#(length(X)) -> a__length#(mark(X))
p20: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p21: mark#(U11(X1,X2)) -> mark#(X1)
p22: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__U11#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (10,2)
          tt_A() = (3,3)
          a__length#_A(x1) = ((0,1),(0,0)) x1 + (10,2)
          mark_A(x1) = ((0,1),(0,1)) x1 + (3,3)
          cons_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (3,3)
          a__isNatList#_A(x1) = (8,2)
          a__isNat#_A(x1) = (8,2)
          s_A(x1) = x1
          length_A(x1) = ((0,0),(0,1)) x1 + (8,5)
          a__and#_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,1),(0,0)) x2 + (5,2)
          a__isNat_A(x1) = (3,3)
          isNatList_A(x1) = (2,0)
          mark#_A(x1) = ((0,1),(0,0)) x1 + (8,2)
          isNatIList_A(x1) = ((1,1),(1,1)) x1 + (4,8)
          a__isNatIList#_A(x1) = ((0,1),(0,0)) x1 + (14,2)
          isNat_A(x1) = (2,0)
          and_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(0,1)) x2 + (1,0)
          a__and_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(0,1)) x2 + (3,0)
          a__isNatList_A(x1) = (3,3)
          U11_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (4,5)
          a__zeros_A() = (4,5)
          |0|_A() = (0,0)
          zeros_A() = (0,2)
          a__U11_A(x1,x2) = x1 + ((0,0),(0,1)) x2 + (5,5)
          a__length_A(x1) = ((0,0),(0,1)) x1 + (8,5)
          a__isNatIList_A(x1) = ((1,1),(1,1)) x1 + (5,8)
          nil_A() = (0,0)
  
    precedence:
    
      a__isNatList# = a__isNat# = a__and# = mark# = a__isNatIList# > a__U11# = a__length# > tt = mark = cons = s = length = a__isNat = isNatList = isNatIList = isNat = and = a__and = a__isNatList = U11 = a__zeros = |0| = zeros = a__U11 = a__length = a__isNatIList = nil
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = []
      pi(cons) = []
      pi(a__isNatList#) = []
      pi(a__isNat#) = []
      pi(s) = []
      pi(length) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(mark#) = []
      pi(isNatIList) = []
      pi(a__isNatIList#) = []
      pi(isNat) = []
      pi(and) = []
      pi(a__and) = []
      pi(a__isNatList) = []
      pi(U11) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p19

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__isNatList#(L)
p3: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p4: a__isNat#(s(V1)) -> a__isNat#(V1)
p5: a__isNat#(length(V1)) -> a__isNatList#(V1)
p6: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p7: a__and#(tt(),X) -> mark#(X)
p8: mark#(s(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(isNatIList(X)) -> a__isNatIList#(X)
p11: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1)
p12: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p13: a__isNatIList#(V) -> a__isNatList#(V)
p14: mark#(isNatList(X)) -> a__isNatList#(X)
p15: mark#(isNat(X)) -> a__isNat#(X)
p16: mark#(and(X1,X2)) -> mark#(X1)
p17: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p18: mark#(length(X)) -> mark#(X)
p19: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p20: mark#(U11(X1,X2)) -> mark#(X1)
p21: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p3: a__length#(cons(N,L)) -> a__isNatList#(L)
p4: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p5: a__and#(tt(),X) -> mark#(X)
p6: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p7: mark#(U11(X1,X2)) -> mark#(X1)
p8: mark#(length(X)) -> mark#(X)
p9: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p10: mark#(and(X1,X2)) -> mark#(X1)
p11: mark#(isNat(X)) -> a__isNat#(X)
p12: a__isNat#(length(V1)) -> a__isNatList#(V1)
p13: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p14: a__isNat#(s(V1)) -> a__isNat#(V1)
p15: mark#(isNatList(X)) -> a__isNatList#(X)
p16: mark#(isNatIList(X)) -> a__isNatIList#(X)
p17: a__isNatIList#(V) -> a__isNatList#(V)
p18: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p19: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1)
p20: mark#(cons(X1,X2)) -> mark#(X1)
p21: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__U11#_A(x1,x2) = ((1,0),(0,0)) x1 + (2,4)
          tt_A() = (7,1)
          a__length#_A(x1) = (9,4)
          mark_A(x1) = ((1,0),(1,1)) x1 + (7,3)
          cons_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (7,0)
          a__and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (0,3)
          a__isNatList_A(x1) = (7,4)
          isNat_A(x1) = (0,4)
          a__isNatList#_A(x1) = (8,4)
          a__and#_A(x1,x2) = ((1,0),(0,0)) x2 + (8,4)
          a__isNat_A(x1) = (7,4)
          isNatList_A(x1) = (0,1)
          mark#_A(x1) = ((1,0),(0,0)) x1 + (8,4)
          U11_A(x1,x2) = x1 + x2 + (6,1)
          length_A(x1) = x1 + (6,2)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (0,1)
          a__isNat#_A(x1) = (8,4)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,2)
          isNatIList_A(x1) = ((1,1),(0,0)) x1 + (8,1)
          a__isNatIList#_A(x1) = ((1,1),(0,0)) x1 + (10,4)
          a__zeros_A() = (15,9)
          |0|_A() = (0,10)
          zeros_A() = (8,1)
          a__U11_A(x1,x2) = x1 + x2 + (6,1)
          a__length_A(x1) = x1 + (6,8)
          a__isNatIList_A(x1) = ((1,1),(0,1)) x1 + (8,12)
          nil_A() = (7,2)
  
    precedence:
    
      mark = cons = a__and = a__isNatList = a__isNat = a__zeros = a__isNatIList > tt > isNatIList > zeros > a__length > s = a__U11 > |0| > length > a__U11# = a__length# = isNat = a__isNatList# = a__and# = isNatList = mark# = U11 = and = a__isNat# = a__isNatIList# = nil
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = []
      pi(cons) = []
      pi(a__and) = []
      pi(a__isNatList) = []
      pi(isNat) = []
      pi(a__isNatList#) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(mark#) = []
      pi(U11) = []
      pi(length) = []
      pi(and) = []
      pi(a__isNat#) = []
      pi(s) = []
      pi(isNatIList) = []
      pi(a__isNatIList#) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = [1]
      pi(a__length) = [1]
      pi(a__isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p19

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p3: a__length#(cons(N,L)) -> a__isNatList#(L)
p4: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p5: a__and#(tt(),X) -> mark#(X)
p6: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p7: mark#(U11(X1,X2)) -> mark#(X1)
p8: mark#(length(X)) -> mark#(X)
p9: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p10: mark#(and(X1,X2)) -> mark#(X1)
p11: mark#(isNat(X)) -> a__isNat#(X)
p12: a__isNat#(length(V1)) -> a__isNatList#(V1)
p13: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p14: a__isNat#(s(V1)) -> a__isNat#(V1)
p15: mark#(isNatList(X)) -> a__isNatList#(X)
p16: mark#(isNatIList(X)) -> a__isNatIList#(X)
p17: a__isNatIList#(V) -> a__isNatList#(V)
p18: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p19: mark#(cons(X1,X2)) -> mark#(X1)
p20: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__isNatList#(L)
p3: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p4: a__isNat#(s(V1)) -> a__isNat#(V1)
p5: a__isNat#(length(V1)) -> a__isNatList#(V1)
p6: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p7: a__and#(tt(),X) -> mark#(X)
p8: mark#(s(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(isNatIList(X)) -> a__isNatIList#(X)
p11: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p12: a__isNatIList#(V) -> a__isNatList#(V)
p13: mark#(isNatList(X)) -> a__isNatList#(X)
p14: mark#(isNat(X)) -> a__isNat#(X)
p15: mark#(and(X1,X2)) -> mark#(X1)
p16: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p17: mark#(length(X)) -> mark#(X)
p18: mark#(U11(X1,X2)) -> mark#(X1)
p19: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p20: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__U11#_A(x1,x2) = (3,3)
          tt_A() = (0,2)
          a__length#_A(x1) = (3,3)
          mark_A(x1) = ((0,0),(1,1)) x1 + (0,2)
          cons_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(1,1)) x2 + (0,2)
          a__isNatList#_A(x1) = (3,3)
          a__isNat#_A(x1) = (3,3)
          s_A(x1) = ((0,0),(1,1)) x1
          length_A(x1) = ((0,0),(1,1)) x1
          a__and#_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(0,1)) x2 + (1,1)
          a__isNat_A(x1) = (0,2)
          isNatList_A(x1) = (0,0)
          mark#_A(x1) = ((0,1),(0,1)) x1 + (3,3)
          isNatIList_A(x1) = ((0,0),(1,1)) x1 + (0,2)
          a__isNatIList#_A(x1) = ((0,1),(0,1)) x1 + (5,3)
          isNat_A(x1) = (0,0)
          and_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(1,1)) x2
          U11_A(x1,x2) = x1 + ((0,0),(1,1)) x2
          a__and_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(1,1)) x2
          a__isNatList_A(x1) = (0,2)
          a__zeros_A() = (0,3)
          |0|_A() = (0,0)
          zeros_A() = (0,1)
          a__U11_A(x1,x2) = x1 + ((0,0),(1,1)) x2
          a__length_A(x1) = ((0,0),(1,1)) x1
          a__isNatIList_A(x1) = ((0,0),(1,1)) x1 + (0,3)
          nil_A() = (0,1)
  
    precedence:
    
      a__U11# = a__length# = a__isNatList# = a__isNat# = a__and# = mark# = a__isNatIList# > tt = mark = cons = a__isNat = a__and = a__isNatList = a__zeros = a__isNatIList > isNat > isNatIList > and > s = length = U11 = a__U11 = a__length > |0| = zeros = nil > isNatList
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = []
      pi(cons) = []
      pi(a__isNatList#) = []
      pi(a__isNat#) = []
      pi(s) = []
      pi(length) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(mark#) = []
      pi(isNatIList) = []
      pi(a__isNatIList#) = []
      pi(isNat) = []
      pi(and) = []
      pi(U11) = []
      pi(a__and) = []
      pi(a__isNatList) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p11

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__isNatList#(L)
p3: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p4: a__isNat#(s(V1)) -> a__isNat#(V1)
p5: a__isNat#(length(V1)) -> a__isNatList#(V1)
p6: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p7: a__and#(tt(),X) -> mark#(X)
p8: mark#(s(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(isNatIList(X)) -> a__isNatIList#(X)
p11: a__isNatIList#(V) -> a__isNatList#(V)
p12: mark#(isNatList(X)) -> a__isNatList#(X)
p13: mark#(isNat(X)) -> a__isNat#(X)
p14: mark#(and(X1,X2)) -> mark#(X1)
p15: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p16: mark#(length(X)) -> mark#(X)
p17: mark#(U11(X1,X2)) -> mark#(X1)
p18: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p19: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p3: a__length#(cons(N,L)) -> a__isNatList#(L)
p4: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p5: a__and#(tt(),X) -> mark#(X)
p6: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p7: mark#(U11(X1,X2)) -> mark#(X1)
p8: mark#(length(X)) -> mark#(X)
p9: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p10: mark#(and(X1,X2)) -> mark#(X1)
p11: mark#(isNat(X)) -> a__isNat#(X)
p12: a__isNat#(length(V1)) -> a__isNatList#(V1)
p13: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p14: a__isNat#(s(V1)) -> a__isNat#(V1)
p15: mark#(isNatList(X)) -> a__isNatList#(X)
p16: mark#(isNatIList(X)) -> a__isNatIList#(X)
p17: a__isNatIList#(V) -> a__isNatList#(V)
p18: mark#(cons(X1,X2)) -> mark#(X1)
p19: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__U11#_A(x1,x2) = (0,1)
          tt_A() = (0,0)
          a__length#_A(x1) = (0,1)
          mark_A(x1) = ((1,1),(1,1)) x1
          cons_A(x1,x2) = x1 + ((1,1),(1,1)) x2
          a__and_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(1,1)) x2
          a__isNatList_A(x1) = (0,0)
          isNat_A(x1) = (0,0)
          a__isNatList#_A(x1) = (0,1)
          a__and#_A(x1,x2) = ((1,1),(1,1)) x2 + (0,1)
          a__isNat_A(x1) = (0,0)
          isNatList_A(x1) = (0,0)
          mark#_A(x1) = ((1,1),(1,1)) x1 + (0,1)
          U11_A(x1,x2) = x1 + ((1,1),(1,1)) x2
          length_A(x1) = ((0,1),(1,0)) x1
          and_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(1,1)) x2
          a__isNat#_A(x1) = (0,1)
          s_A(x1) = x1
          isNatIList_A(x1) = ((0,1),(1,1)) x1 + (1,2)
          a__isNatIList#_A(x1) = ((0,0),(0,1)) x1 + (2,3)
          a__zeros_A() = (2,2)
          |0|_A() = (0,0)
          zeros_A() = (1,1)
          a__U11_A(x1,x2) = x1 + ((1,1),(1,1)) x2
          a__length_A(x1) = ((0,1),(1,0)) x1
          a__isNatIList_A(x1) = ((1,1),(1,1)) x1 + (3,3)
          nil_A() = (1,1)
  
    precedence:
    
      mark = a__and = a__isNatList = isNat = a__isNat = U11 = and = s = a__isNatIList# = a__zeros = a__U11 = a__length = a__isNatIList = nil > a__U11# = a__length# = a__isNatList# = a__and# = mark# = a__isNat# > tt = cons = isNatList = zeros > |0| > isNatIList > length
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = []
      pi(cons) = []
      pi(a__and) = []
      pi(a__isNatList) = []
      pi(isNat) = []
      pi(a__isNatList#) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(mark#) = []
      pi(U11) = []
      pi(length) = []
      pi(and) = []
      pi(a__isNat#) = []
      pi(s) = []
      pi(isNatIList) = []
      pi(a__isNatIList#) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p16

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p3: a__length#(cons(N,L)) -> a__isNatList#(L)
p4: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p5: a__and#(tt(),X) -> mark#(X)
p6: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p7: mark#(U11(X1,X2)) -> mark#(X1)
p8: mark#(length(X)) -> mark#(X)
p9: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p10: mark#(and(X1,X2)) -> mark#(X1)
p11: mark#(isNat(X)) -> a__isNat#(X)
p12: a__isNat#(length(V1)) -> a__isNatList#(V1)
p13: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p14: a__isNat#(s(V1)) -> a__isNat#(V1)
p15: mark#(isNatList(X)) -> a__isNatList#(X)
p16: a__isNatIList#(V) -> a__isNatList#(V)
p17: mark#(cons(X1,X2)) -> mark#(X1)
p18: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p17, p18}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__isNatList#(L)
p3: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p4: a__isNat#(s(V1)) -> a__isNat#(V1)
p5: a__isNat#(length(V1)) -> a__isNatList#(V1)
p6: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p7: a__and#(tt(),X) -> mark#(X)
p8: mark#(s(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(isNatList(X)) -> a__isNatList#(X)
p11: mark#(isNat(X)) -> a__isNat#(X)
p12: mark#(and(X1,X2)) -> mark#(X1)
p13: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p14: mark#(length(X)) -> mark#(X)
p15: mark#(U11(X1,X2)) -> mark#(X1)
p16: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p17: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__U11#_A(x1,x2) = ((0,1),(0,0)) x2 + (1,0)
          tt_A() = (0,0)
          a__length#_A(x1) = ((0,1),(0,0)) x1 + (1,0)
          mark_A(x1) = ((1,1),(0,1)) x1
          cons_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(1,1)) x2
          a__isNatList#_A(x1) = (0,0)
          a__isNat#_A(x1) = (0,0)
          s_A(x1) = x1
          length_A(x1) = ((0,1),(0,1)) x1 + (2,2)
          a__and#_A(x1,x2) = ((0,1),(0,0)) x2
          a__isNat_A(x1) = ((1,0),(0,0)) x1
          isNatList_A(x1) = ((0,1),(0,0)) x1
          mark#_A(x1) = ((0,1),(0,0)) x1
          isNat_A(x1) = ((1,0),(0,0)) x1
          and_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(0,1)) x2
          U11_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,0),(1,1)) x2 + (2,2)
          a__and_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(0,1)) x2
          a__isNatList_A(x1) = ((0,1),(0,0)) x1
          a__zeros_A() = (1,1)
          |0|_A() = (2,0)
          zeros_A() = (0,1)
          a__U11_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          a__length_A(x1) = ((0,1),(0,1)) x1 + (3,2)
          a__isNatIList_A(x1) = ((1,1),(0,0)) x1 + (1,0)
          isNatIList_A(x1) = ((1,1),(0,0)) x1 + (1,0)
          nil_A() = (3,1)
  
    precedence:
    
      mark = a__isNat = a__and = a__isNatList = a__zeros = zeros = a__isNatIList = isNatIList > isNatList > a__isNatList# = a__isNat# = a__and# = mark# > length > cons = and = a__length > a__U11 > s = isNat > a__U11# = tt = a__length# = U11 = |0| = nil
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = []
      pi(cons) = []
      pi(a__isNatList#) = []
      pi(a__isNat#) = []
      pi(s) = []
      pi(length) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(mark#) = []
      pi(isNat) = []
      pi(and) = []
      pi(U11) = [2]
      pi(a__and) = []
      pi(a__isNatList) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = [2]
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p3: a__isNat#(s(V1)) -> a__isNat#(V1)
p4: a__isNat#(length(V1)) -> a__isNatList#(V1)
p5: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p6: a__and#(tt(),X) -> mark#(X)
p7: mark#(s(X)) -> mark#(X)
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(isNatList(X)) -> a__isNatList#(X)
p10: mark#(isNat(X)) -> a__isNat#(X)
p11: mark#(and(X1,X2)) -> mark#(X1)
p12: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p13: mark#(length(X)) -> mark#(X)
p14: mark#(U11(X1,X2)) -> mark#(X1)
p15: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p16: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14}
  {p1, p16}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(U11(X1,X2)) -> mark#(X1)
p2: mark#(length(X)) -> mark#(X)
p3: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p4: a__and#(tt(),X) -> mark#(X)
p5: mark#(and(X1,X2)) -> mark#(X1)
p6: mark#(isNat(X)) -> a__isNat#(X)
p7: a__isNat#(length(V1)) -> a__isNatList#(V1)
p8: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p9: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p10: a__isNat#(s(V1)) -> a__isNat#(V1)
p11: mark#(isNatList(X)) -> a__isNatList#(X)
p12: mark#(cons(X1,X2)) -> mark#(X1)
p13: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,1),(1,1)) x1 + (3,0)
          U11_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
          length_A(x1) = ((1,1),(1,1)) x1 + (1,1)
          and_A(x1,x2) = x1 + x2
          a__and#_A(x1,x2) = ((1,1),(1,1)) x2 + (3,0)
          mark_A(x1) = x1
          tt_A() = (0,0)
          isNat_A(x1) = ((0,0),(1,1)) x1
          a__isNat#_A(x1) = ((1,1),(1,1)) x1 + (1,0)
          a__isNatList#_A(x1) = ((0,1),(1,1)) x1 + (3,0)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,1),(1,1)) x2
          a__isNat_A(x1) = ((0,0),(1,1)) x1
          isNatList_A(x1) = ((0,0),(1,1)) x1
          s_A(x1) = ((0,1),(1,0)) x1
          a__zeros_A() = (0,0)
          |0|_A() = (0,0)
          zeros_A() = (0,0)
          a__U11_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
          a__length_A(x1) = ((1,1),(1,1)) x1 + (1,1)
          a__and_A(x1,x2) = x1 + x2
          a__isNatIList_A(x1) = ((0,1),(1,1)) x1 + (1,1)
          a__isNatList_A(x1) = ((0,0),(1,1)) x1
          isNatIList_A(x1) = ((0,1),(1,1)) x1 + (1,1)
          nil_A() = (1,1)
  
    precedence:
    
      a__isNat# > mark# = U11 = length = and = a__and# = mark = tt = isNat = a__isNatList# = cons = a__isNat = isNatList = s = a__zeros = |0| = zeros = a__U11 = a__length = a__and = a__isNatIList = a__isNatList = isNatIList = nil
    
    partial status:
  
      pi(mark#) = []
      pi(U11) = []
      pi(length) = []
      pi(and) = []
      pi(a__and#) = []
      pi(mark) = []
      pi(tt) = []
      pi(isNat) = []
      pi(a__isNat#) = []
      pi(a__isNatList#) = []
      pi(cons) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(s) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__and) = []
      pi(a__isNatIList) = []
      pi(a__isNatList) = []
      pi(isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(U11(X1,X2)) -> mark#(X1)
p2: mark#(length(X)) -> mark#(X)
p3: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p4: a__and#(tt(),X) -> mark#(X)
p5: mark#(and(X1,X2)) -> mark#(X1)
p6: a__isNat#(length(V1)) -> a__isNatList#(V1)
p7: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p8: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p9: a__isNat#(s(V1)) -> a__isNat#(V1)
p10: mark#(isNatList(X)) -> a__isNatList#(X)
p11: mark#(cons(X1,X2)) -> mark#(X1)
p12: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(U11(X1,X2)) -> mark#(X1)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(isNatList(X)) -> a__isNatList#(X)
p5: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p6: a__isNat#(s(V1)) -> a__isNat#(V1)
p7: a__isNat#(length(V1)) -> a__isNatList#(V1)
p8: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p9: a__and#(tt(),X) -> mark#(X)
p10: mark#(and(X1,X2)) -> mark#(X1)
p11: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p12: mark#(length(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = x1
          U11_A(x1,x2) = x1 + ((0,0),(1,1)) x2 + (3,2)
          s_A(x1) = x1
          cons_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (2,5)
          isNatList_A(x1) = (0,0)
          a__isNatList#_A(x1) = (0,0)
          a__isNat#_A(x1) = (0,0)
          length_A(x1) = x1 + (1,4)
          a__and#_A(x1,x2) = x2
          a__isNat_A(x1) = (2,5)
          tt_A() = (2,5)
          and_A(x1,x2) = x1 + ((1,1),(0,1)) x2
          mark_A(x1) = ((1,1),(1,1)) x1 + (2,5)
          a__zeros_A() = (6,9)
          |0|_A() = (0,0)
          zeros_A() = (3,1)
          a__U11_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (4,4)
          a__length_A(x1) = x1 + (4,4)
          a__and_A(x1,x2) = x1 + ((1,1),(1,1)) x2
          a__isNatIList_A(x1) = ((0,1),(1,1)) x1 + (9,7)
          a__isNatList_A(x1) = (2,5)
          isNatIList_A(x1) = ((0,1),(1,0)) x1 + (1,6)
          nil_A() = (1,1)
          isNat_A(x1) = (0,0)
  
    precedence:
    
      isNatList = a__isNat = and = mark = a__and = a__isNatIList = a__isNatList > length = a__length > a__zeros = zeros = a__U11 > tt > cons > |0| > isNatIList > mark# = U11 = s = a__isNatList# = a__isNat# = a__and# = nil = isNat
    
    partial status:
  
      pi(mark#) = []
      pi(U11) = [1]
      pi(s) = []
      pi(cons) = [2]
      pi(isNatList) = []
      pi(a__isNatList#) = []
      pi(a__isNat#) = []
      pi(length) = [1]
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(tt) = []
      pi(and) = []
      pi(mark) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = [2]
      pi(a__length) = [1]
      pi(a__and) = []
      pi(a__isNatIList) = []
      pi(a__isNatList) = []
      pi(isNatIList) = []
      pi(nil) = []
      pi(isNat) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(U11(X1,X2)) -> mark#(X1)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(isNatList(X)) -> a__isNatList#(X)
p4: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p5: a__isNat#(s(V1)) -> a__isNat#(V1)
p6: a__isNat#(length(V1)) -> a__isNatList#(V1)
p7: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p8: a__and#(tt(),X) -> mark#(X)
p9: mark#(and(X1,X2)) -> mark#(X1)
p10: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p11: mark#(length(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(U11(X1,X2)) -> mark#(X1)
p2: mark#(length(X)) -> mark#(X)
p3: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p4: a__and#(tt(),X) -> mark#(X)
p5: mark#(and(X1,X2)) -> mark#(X1)
p6: mark#(isNatList(X)) -> a__isNatList#(X)
p7: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p8: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p9: a__isNat#(length(V1)) -> a__isNatList#(V1)
p10: a__isNat#(s(V1)) -> a__isNat#(V1)
p11: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,0),(1,0)) x1 + (4,5)
          U11_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(0,0)) x2 + (3,0)
          length_A(x1) = ((1,0),(0,0)) x1 + (3,1)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (0,5)
          a__and#_A(x1,x2) = ((1,0),(1,0)) x2 + (4,5)
          mark_A(x1) = ((1,0),(1,0)) x1 + (5,6)
          tt_A() = (5,3)
          isNatList_A(x1) = (0,1)
          a__isNatList#_A(x1) = (4,5)
          cons_A(x1,x2) = ((1,1),(0,0)) x2 + (5,1)
          a__isNat_A(x1) = (5,6)
          a__isNat#_A(x1) = (4,5)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,4)
          a__zeros_A() = (11,1)
          |0|_A() = (6,2)
          zeros_A() = (6,0)
          a__U11_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(0,0)) x2 + (3,4)
          a__length_A(x1) = x1 + (3,3)
          a__and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (0,6)
          a__isNatIList_A(x1) = ((1,1),(1,1)) x1 + (6,10)
          a__isNatList_A(x1) = (5,6)
          isNatIList_A(x1) = ((1,1),(1,0)) x1 + (4,1)
          nil_A() = (4,2)
          isNat_A(x1) = (0,1)
  
    precedence:
    
      and = mark = isNatList = a__isNat = a__length = a__and = a__isNatIList = a__isNatList > U11 = a__U11 > tt = a__zeros > mark# = length = a__and# = a__isNatList# = cons = a__isNat# = nil > s > zeros > |0| = isNatIList = isNat
    
    partial status:
  
      pi(mark#) = []
      pi(U11) = []
      pi(length) = []
      pi(and) = []
      pi(a__and#) = []
      pi(mark) = []
      pi(tt) = []
      pi(isNatList) = []
      pi(a__isNatList#) = []
      pi(cons) = []
      pi(a__isNat) = []
      pi(a__isNat#) = []
      pi(s) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__and) = []
      pi(a__isNatIList) = []
      pi(a__isNatList) = []
      pi(isNatIList) = []
      pi(nil) = []
      pi(isNat) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(length(X)) -> mark#(X)
p2: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p3: a__and#(tt(),X) -> mark#(X)
p4: mark#(and(X1,X2)) -> mark#(X1)
p5: mark#(isNatList(X)) -> a__isNatList#(X)
p6: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p7: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p8: a__isNat#(length(V1)) -> a__isNatList#(V1)
p9: a__isNat#(s(V1)) -> a__isNat#(V1)
p10: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(length(X)) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(isNatList(X)) -> a__isNatList#(X)
p4: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p5: a__isNat#(s(V1)) -> a__isNat#(V1)
p6: a__isNat#(length(V1)) -> a__isNatList#(V1)
p7: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p8: a__and#(tt(),X) -> mark#(X)
p9: mark#(and(X1,X2)) -> mark#(X1)
p10: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((0,1),(0,0)) x1 + (1,6)
          length_A(x1) = ((0,0),(0,1)) x1 + (6,6)
          s_A(x1) = x1
          isNatList_A(x1) = ((0,0),(0,1)) x1 + (2,9)
          a__isNatList#_A(x1) = ((0,1),(0,0)) x1 + (10,6)
          cons_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(1,1)) x2 + (1,5)
          a__isNat#_A(x1) = ((0,1),(0,0)) x1 + (5,6)
          a__and#_A(x1,x2) = ((0,1),(0,0)) x2 + (1,6)
          a__isNat_A(x1) = ((0,1),(0,1)) x1 + (8,8)
          tt_A() = (7,8)
          and_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(1,1)) x2
          mark_A(x1) = ((0,1),(0,1)) x1 + (0,5)
          a__zeros_A() = (2,8)
          |0|_A() = (0,0)
          zeros_A() = (0,3)
          a__U11_A(x1,x2) = ((0,0),(1,1)) x2 + (6,11)
          a__length_A(x1) = ((0,0),(0,1)) x1 + (6,6)
          a__and_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(1,1)) x2
          a__isNatIList_A(x1) = ((0,1),(0,1)) x1 + (9,18)
          a__isNatList_A(x1) = ((0,1),(0,1)) x1 + (8,14)
          isNatIList_A(x1) = ((0,0),(0,1)) x1 + (1,14)
          nil_A() = (0,0)
          isNat_A(x1) = ((0,0),(0,1)) x1 + (1,8)
          U11_A(x1,x2) = ((0,0),(1,1)) x2 + (5,6)
  
    precedence:
    
      a__zeros > zeros > mark = a__and = a__isNatIList = isNatIList > |0| > length = a__U11 = a__length > a__isNat = nil = isNat > a__isNatList > mark# = s = isNatList = a__and# = tt > a__isNatList# = cons = a__isNat# = and = U11
    
    partial status:
  
      pi(mark#) = []
      pi(length) = []
      pi(s) = []
      pi(isNatList) = []
      pi(a__isNatList#) = []
      pi(cons) = []
      pi(a__isNat#) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(tt) = []
      pi(and) = []
      pi(mark) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__and) = []
      pi(a__isNatIList) = []
      pi(a__isNatList) = []
      pi(isNatIList) = []
      pi(nil) = []
      pi(isNat) = []
      pi(U11) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(length(X)) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p4: a__isNat#(s(V1)) -> a__isNat#(V1)
p5: a__isNat#(length(V1)) -> a__isNatList#(V1)
p6: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p7: a__and#(tt(),X) -> mark#(X)
p8: mark#(and(X1,X2)) -> mark#(X1)
p9: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5}
  {p1, p2, p7, p8, p9}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__isNat#(length(V1)) -> a__isNatList#(V1)
p2: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p3: a__isNat#(s(V1)) -> a__isNat#(V1)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__isNat#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          length_A(x1) = ((0,0),(1,1)) x1 + (1,1)
          a__isNatList#_A(x1) = ((1,1),(0,0)) x1 + (2,2)
          cons_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(1,1)) x2 + (1,1)
          s_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      a__isNat# = length = cons = s > a__isNatList#
    
    partial status:
  
      pi(a__isNat#) = []
      pi(length) = []
      pi(a__isNatList#) = []
      pi(cons) = [2]
      pi(s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p2: a__isNat#(s(V1)) -> a__isNat#(V1)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__isNat#(s(V1)) -> a__isNat#(V1)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__isNat#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      a__isNat# = s
    
    partial status:
  
      pi(a__isNat#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(length(X)) -> mark#(X)
p2: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p3: a__and#(tt(),X) -> mark#(X)
p4: mark#(and(X1,X2)) -> mark#(X1)
p5: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,0),(1,0)) x1 + (0,1)
          length_A(x1) = ((1,0),(0,0)) x1 + (3,1)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (0,1)
          a__and#_A(x1,x2) = ((1,0),(1,0)) x2 + (0,1)
          mark_A(x1) = ((1,0),(1,1)) x1 + (2,8)
          tt_A() = (2,6)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,7)
          a__zeros_A() = (2,4)
          cons_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (2,1)
          |0|_A() = (3,5)
          zeros_A() = (0,1)
          a__U11_A(x1,x2) = ((1,0),(1,0)) x2 + (5,7)
          a__length_A(x1) = ((1,0),(1,0)) x1 + (3,6)
          a__and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (0,8)
          a__isNat_A(x1) = ((0,0),(1,0)) x1 + (2,14)
          a__isNatList_A(x1) = ((0,0),(1,0)) x1 + (2,8)
          a__isNatIList_A(x1) = ((1,0),(1,0)) x1 + (3,11)
          isNatIList_A(x1) = ((1,0),(0,0)) x1 + (3,2)
          nil_A() = (4,5)
          isNatList_A(x1) = ((0,0),(1,0)) x1 + (0,2)
          isNat_A(x1) = ((0,0),(1,0)) x1 + (1,9)
          U11_A(x1,x2) = ((1,0),(1,0)) x2 + (4,1)
  
    precedence:
    
      and = mark = tt = a__zeros = |0| = a__and = a__isNat = a__isNatList = a__isNatIList = nil = isNat > zeros > mark# = length = a__and# = cons = a__length = isNatIList = isNatList > a__U11 > s > U11
    
    partial status:
  
      pi(mark#) = []
      pi(length) = []
      pi(and) = []
      pi(a__and#) = []
      pi(mark) = []
      pi(tt) = []
      pi(s) = []
      pi(a__zeros) = []
      pi(cons) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__and) = []
      pi(a__isNat) = []
      pi(a__isNatList) = []
      pi(a__isNatIList) = []
      pi(isNatIList) = []
      pi(nil) = []
      pi(isNatList) = []
      pi(isNat) = []
      pi(U11) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p2: a__and#(tt(),X) -> mark#(X)
p3: mark#(and(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p2: a__and#(tt(),X) -> mark#(X)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(and(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,1),(0,0)) x1 + (2,2)
          and_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (2,3)
          a__and#_A(x1,x2) = ((1,1),(0,0)) x2 + (3,2)
          mark_A(x1) = ((1,1),(1,1)) x1 + (6,0)
          tt_A() = (3,2)
          s_A(x1) = x1
          a__zeros_A() = (9,3)
          cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (6,0)
          |0|_A() = (0,0)
          zeros_A() = (1,2)
          a__U11_A(x1,x2) = ((1,1),(1,1)) x2 + (21,14)
          a__length_A(x1) = x1 + (15,14)
          a__and_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (3,4)
          a__isNat_A(x1) = ((1,1),(1,1)) x1 + (3,2)
          length_A(x1) = x1 + (9,7)
          a__isNatList_A(x1) = ((0,1),(1,1)) x1 + (14,8)
          a__isNatIList_A(x1) = ((0,1),(1,1)) x1 + (14,8)
          isNatIList_A(x1) = ((0,1),(1,0)) x1 + (7,1)
          nil_A() = (2,2)
          isNatList_A(x1) = ((0,0),(1,1)) x1 + (7,1)
          isNat_A(x1) = ((1,1),(1,0)) x1 + (2,1)
          U11_A(x1,x2) = ((1,0),(1,1)) x2 + (21,1)
  
    precedence:
    
      mark = tt = a__length = a__and = a__isNat = length = a__isNatList = a__isNatIList > mark# > s = a__U11 > and > a__zeros = |0| = nil > a__and# = zeros > cons = isNatIList = isNatList = isNat = U11
    
    partial status:
  
      pi(mark#) = []
      pi(and) = []
      pi(a__and#) = []
      pi(mark) = []
      pi(tt) = []
      pi(s) = []
      pi(a__zeros) = []
      pi(cons) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__and) = []
      pi(a__isNat) = []
      pi(length) = []
      pi(a__isNatList) = []
      pi(a__isNatIList) = []
      pi(isNatIList) = []
      pi(nil) = []
      pi(isNatList) = []
      pi(isNat) = []
      pi(U11) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__and#(tt(),X) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(and(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(s(X)) -> mark#(X)
p2: mark#(and(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          s_A(x1) = x1 + (1,1)
          and_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
  
    precedence:
    
      mark# = s = and
    
    partial status:
  
      pi(mark#) = []
      pi(s) = [1]
      pi(and) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(and(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(and(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = x1 + (1,1)
          and_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (2,1)
  
    precedence:
    
      and > mark#
    
    partial status:
  
      pi(mark#) = [1]
      pi(and) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__U11#(tt(),L) -> a__length#(mark(L))
p2: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__U11#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,0)
          tt_A() = (22,12)
          a__length#_A(x1) = ((1,0),(1,1)) x1 + (8,6)
          mark_A(x1) = x1 + (13,0)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(0,1)) x2
          a__and_A(x1,x2) = ((0,1),(0,0)) x1 + x2 + (2,0)
          a__isNatList_A(x1) = x1 + (15,3)
          isNat_A(x1) = ((1,0),(1,0)) x1 + (1,3)
          a__zeros_A() = (19,1)
          |0|_A() = (10,13)
          zeros_A() = (7,1)
          a__U11_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(0,1)) x2 + (7,1)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          a__length_A(x1) = ((1,1),(0,1)) x1 + (14,2)
          a__isNat_A(x1) = ((1,0),(1,0)) x1 + (13,3)
          length_A(x1) = ((1,1),(0,1)) x1 + (14,2)
          a__isNatIList_A(x1) = x1 + (16,11)
          isNatIList_A(x1) = x1 + (4,11)
          nil_A() = (23,12)
          U11_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(0,1)) x2 + (7,1)
          isNatList_A(x1) = x1 + (10,3)
          and_A(x1,x2) = ((0,1),(0,0)) x1 + x2 + (1,0)
  
    precedence:
    
      mark = a__isNatList = a__zeros = a__isNat = a__isNatIList > zeros > a__U11 = a__length > isNatList > nil = U11 > a__U11# = |0| > cons > a__length# = a__and = isNat > tt > isNatIList = and > s = length
    
    partial status:
  
      pi(a__U11#) = [2]
      pi(tt) = []
      pi(a__length#) = [1]
      pi(mark) = [1]
      pi(cons) = [2]
      pi(a__and) = []
      pi(a__isNatList) = []
      pi(isNat) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(s) = []
      pi(a__length) = []
      pi(a__isNat) = []
      pi(length) = []
      pi(a__isNatIList) = []
      pi(isNatIList) = []
      pi(nil) = []
      pi(U11) = []
      pi(isNatList) = []
      pi(and) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)

and R consists of:

r1: a__zeros() -> cons(|0|(),zeros())
r2: a__U11(tt(),L) -> s(a__length(mark(L)))
r3: a__and(tt(),X) -> mark(X)
r4: a__isNat(|0|()) -> tt()
r5: a__isNat(length(V1)) -> a__isNatList(V1)
r6: a__isNat(s(V1)) -> a__isNat(V1)
r7: a__isNatIList(V) -> a__isNatList(V)
r8: a__isNatIList(zeros()) -> tt()
r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2))
r10: a__isNatList(nil()) -> tt()
r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2))
r12: a__length(nil()) -> |0|()
r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L)
r14: mark(zeros()) -> a__zeros()
r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r16: mark(length(X)) -> a__length(mark(X))
r17: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r18: mark(isNat(X)) -> a__isNat(X)
r19: mark(isNatList(X)) -> a__isNatList(X)
r20: mark(isNatIList(X)) -> a__isNatIList(X)
r21: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r22: mark(|0|()) -> |0|()
r23: mark(tt()) -> tt()
r24: mark(s(X)) -> s(mark(X))
r25: mark(nil()) -> nil()
r26: a__zeros() -> zeros()
r27: a__U11(X1,X2) -> U11(X1,X2)
r28: a__length(X) -> length(X)
r29: a__and(X1,X2) -> and(X1,X2)
r30: a__isNat(X) -> isNat(X)
r31: a__isNatList(X) -> isNatList(X)
r32: a__isNatIList(X) -> isNatIList(X)

The estimated dependency graph contains the following SCCs:

  (no SCCs)