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: lexicographic order
        interpretations:
          a__U11#_A(x1,x2) = ((0,0),(1,0)) x2 + (5,23)
          tt_A() = (6,14)
          a__length#_A(x1) = ((0,0),(1,0)) x1 + (5,17)
          mark_A(x1) = x1 + (6,17)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (6,21)
          a__isNatList#_A(x1) = (5,10)
          a__isNat#_A(x1) = (5,10)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,15)
          length_A(x1) = ((1,0),(1,0)) x1 + (14,9)
          a__and#_A(x1,x2) = ((0,0),(1,0)) x2 + (5,10)
          a__isNat_A(x1) = ((0,0),(1,0)) x1 + (6,20)
          isNatList_A(x1) = ((0,0),(1,0)) x1 + (0,1)
          mark#_A(x1) = ((0,0),(1,0)) x1 + (5,10)
          isNatIList_A(x1) = ((1,0),(0,0)) x1 + (10,14)
          a__isNatIList#_A(x1) = ((0,0),(1,0)) x1 + (5,15)
          isNat_A(x1) = ((0,0),(1,0)) x1 + (0,4)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (0,18)
          a__isNatList_A(x1) = ((0,0),(1,0)) x1 + (6,14)
          a__and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (0,18)
          U11_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (14,2)
          a__zeros_A() = (12,22)
          |0|_A() = (0,23)
          zeros_A() = (6,22)
          a__U11_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (14,2)
          a__length_A(x1) = ((1,0),(1,0)) x1 + (14,19)
          a__isNatIList_A(x1) = ((1,0),(1,1)) x1 + (10,15)
          nil_A() = (0,24)
  
    precedence:
    
      a__isNat = isNat > mark = a__isNatList > a__zeros > zeros > isNatIList > and = a__and = a__isNatIList > a__isNatIList# > cons > a__U11# = a__length# > a__isNatList# = a__isNat# = a__and# = mark# > a__length = nil > length > |0| > tt = U11 = a__U11 > isNatList > s
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = [1]
      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) = [1]
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = [1]
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p22

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__and#(a__isNatList(L),isNat(N))
p21: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),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__length#(cons(N,L)) -> a__and#(a__isNatList(L),isNat(N))
p4: a__and#(tt(),X) -> mark#(X)
p5: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p6: mark#(U11(X1,X2)) -> mark#(X1)
p7: mark#(length(X)) -> a__length#(mark(X))
p8: a__length#(cons(N,L)) -> a__isNatList#(L)
p9: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p10: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p11: a__isNat#(length(V1)) -> a__isNatList#(V1)
p12: a__isNat#(s(V1)) -> a__isNat#(V1)
p13: mark#(length(X)) -> mark#(X)
p14: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p15: mark#(and(X1,X2)) -> mark#(X1)
p16: mark#(isNat(X)) -> a__isNat#(X)
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: lexicographic order
        interpretations:
          a__U11#_A(x1,x2) = (5,17)
          tt_A() = (6,4)
          a__length#_A(x1) = (5,17)
          mark_A(x1) = x1 + (6,5)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (6,18)
          a__and_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (0,6)
          a__isNatList_A(x1) = ((0,0),(1,0)) x1 + (6,11)
          isNat_A(x1) = ((0,0),(1,0)) x1 + (0,10)
          a__and#_A(x1,x2) = ((0,0),(1,0)) x2 + (5,6)
          mark#_A(x1) = ((0,0),(1,0)) x1 + (5,6)
          U11_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (12,4)
          length_A(x1) = ((1,0),(0,0)) x1 + (12,7)
          a__isNatList#_A(x1) = (5,6)
          a__isNat_A(x1) = ((0,0),(1,0)) x1 + (6,14)
          isNatList_A(x1) = ((0,0),(1,0)) x1 + (0,10)
          a__isNat#_A(x1) = (5,6)
          s_A(x1) = x1
          and_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (0,2)
          isNatIList_A(x1) = ((1,0),(1,0)) x1 + (4,6)
          a__isNatIList#_A(x1) = ((0,0),(1,0)) x1 + (5,7)
          a__zeros_A() = (13,26)
          |0|_A() = (0,9)
          zeros_A() = (7,22)
          a__U11_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (12,8)
          a__length_A(x1) = ((1,0),(0,0)) x1 + (12,8)
          a__isNatIList_A(x1) = ((1,0),(1,0)) x1 + (7,11)
          nil_A() = (6,10)
  
    precedence:
    
      isNatList = zeros > U11 > mark = cons = a__zeros = a__length = a__isNatIList > length > tt = a__isNatList = isNat = a__isNat > a__U11 > s > a__isNatIList# > a__U11# = a__length# = a__and# = mark# = a__isNatList# = a__isNat# > isNatIList > |0| > a__and = and > 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__and#) = []
      pi(mark#) = []
      pi(U11) = []
      pi(length) = []
      pi(a__isNatList#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(a__isNat#) = []
      pi(s) = []
      pi(and) = [2]
      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:

  p18

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__and#(a__isNatList(L),isNat(N))
p4: a__and#(tt(),X) -> mark#(X)
p5: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p6: mark#(U11(X1,X2)) -> mark#(X1)
p7: mark#(length(X)) -> a__length#(mark(X))
p8: a__length#(cons(N,L)) -> a__isNatList#(L)
p9: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p10: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p11: a__isNat#(length(V1)) -> a__isNatList#(V1)
p12: a__isNat#(s(V1)) -> a__isNat#(V1)
p13: mark#(length(X)) -> mark#(X)
p14: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p15: mark#(and(X1,X2)) -> mark#(X1)
p16: mark#(isNat(X)) -> a__isNat#(X)
p17: mark#(isNatList(X)) -> a__isNatList#(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, 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#(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#(length(X)) -> a__length#(mark(X))
p16: a__length#(cons(N,L)) -> a__and#(a__isNatList(L),isNat(N))
p17: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p18: mark#(U11(X1,X2)) -> mark#(X1)
p19: 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: lexicographic order
        interpretations:
          a__U11#_A(x1,x2) = ((1,0),(1,1)) x2 + (4,10)
          tt_A() = (4,10)
          a__length#_A(x1) = x1
          mark_A(x1) = ((1,0),(1,1)) x1 + (4,10)
          cons_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (4,10)
          a__isNatList#_A(x1) = (2,5)
          a__isNat#_A(x1) = (2,5)
          s_A(x1) = x1
          length_A(x1) = x1 + (3,1)
          a__and#_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,1)
          a__isNat_A(x1) = (4,10)
          isNatList_A(x1) = (0,0)
          mark#_A(x1) = ((1,0),(1,1)) x1 + (2,5)
          isNat_A(x1) = (0,0)
          and_A(x1,x2) = x1 + x2
          a__isNatList_A(x1) = (4,10)
          a__and_A(x1,x2) = x1 + ((1,0),(1,1)) x2
          U11_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (3,3)
          a__zeros_A() = (5,12)
          |0|_A() = (0,0)
          zeros_A() = (1,1)
          a__U11_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (3,13)
          a__length_A(x1) = x1 + (3,3)
          a__isNatIList_A(x1) = ((1,0),(1,1)) x1 + (4,21)
          isNatIList_A(x1) = ((1,0),(1,1)) x1 + (0,11)
          nil_A() = (0,9)
  
    precedence:
    
      nil > isNat > a__U11# = a__length# = zeros > a__isNatList# = a__isNat# = a__and# = mark# > tt = mark = a__isNat = and = a__isNatList = a__and = a__isNatIList > a__zeros > cons > a__length > length > s = isNatList = U11 = a__U11 > isNatIList > |0|
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = []
      pi(cons) = [2]
      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(a__isNatList) = []
      pi(a__and) = []
      pi(U11) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = [2]
      pi(a__length) = [1]
      pi(a__isNatIList) = []
      pi(isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p18

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#(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#(length(X)) -> a__length#(mark(X))
p16: a__length#(cons(N,L)) -> a__and#(a__isNatList(L),isNat(N))
p17: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p18: 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}


-- 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__and#(a__isNatList(L),isNat(N))
p4: a__and#(tt(),X) -> mark#(X)
p5: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
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__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p10: a__isNat#(length(V1)) -> a__isNatList#(V1)
p11: a__isNat#(s(V1)) -> a__isNat#(V1)
p12: mark#(length(X)) -> mark#(X)
p13: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p14: mark#(and(X1,X2)) -> mark#(X1)
p15: mark#(isNat(X)) -> a__isNat#(X)
p16: mark#(isNatList(X)) -> a__isNatList#(X)
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 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: lexicographic order
        interpretations:
          a__U11#_A(x1,x2) = ((1,0),(0,0)) x2 + (39,103)
          tt_A() = (24,105)
          a__length#_A(x1) = ((1,0),(0,0)) x1 + (18,103)
          mark_A(x1) = ((1,0),(1,1)) x1 + (21,104)
          cons_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (21,31)
          a__and_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (4,2)
          a__isNatList_A(x1) = ((1,0),(1,0)) x1 + (12,7)
          isNat_A(x1) = ((1,0),(1,1)) x1 + (4,54)
          a__and#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,29)
          mark#_A(x1) = x1 + (19,106)
          U11_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (20,29)
          length_A(x1) = ((1,0),(1,0)) x1 + (20,2)
          a__isNatList#_A(x1) = ((1,0),(1,1)) x1 + (22,1)
          a__isNat_A(x1) = ((1,0),(1,1)) x1 + (25,0)
          isNatList_A(x1) = ((1,0),(1,1)) x1 + (3,0)
          a__isNat#_A(x1) = ((1,0),(0,0)) x1 + (3,53)
          s_A(x1) = x1
          and_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (4,1)
          a__zeros_A() = (22,107)
          |0|_A() = (0,106)
          zeros_A() = (1,105)
          a__U11_A(x1,x2) = ((1,0),(1,0)) x2 + (41,30)
          a__length_A(x1) = ((1,0),(1,0)) x1 + (20,9)
          a__isNatIList_A(x1) = ((1,0),(1,0)) x1 + (24,105)
          isNatIList_A(x1) = ((1,0),(1,1)) x1 + (4,32)
          nil_A() = (98,107)
  
    precedence:
    
      mark = a__and = s = a__U11 = a__length > length = a__isNat = zeros > a__zeros > mark# > nil > and = a__isNatIList = isNatIList > a__isNat# > a__isNatList# > a__U11# = a__length# = a__and# > a__isNatList > isNat > cons > isNatList > U11 > |0| > tt
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = []
      pi(cons) = []
      pi(a__and) = []
      pi(a__isNatList) = []
      pi(isNat) = [1]
      pi(a__and#) = []
      pi(mark#) = [1]
      pi(U11) = []
      pi(length) = []
      pi(a__isNatList#) = []
      pi(a__isNat) = []
      pi(isNatList) = [1]
      pi(a__isNat#) = []
      pi(s) = []
      pi(and) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p15

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__and#(a__isNatList(L),isNat(N))
p4: a__and#(tt(),X) -> mark#(X)
p5: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
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__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p10: a__isNat#(length(V1)) -> a__isNatList#(V1)
p11: a__isNat#(s(V1)) -> a__isNat#(V1)
p12: mark#(length(X)) -> mark#(X)
p13: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p14: mark#(and(X1,X2)) -> mark#(X1)
p15: mark#(isNatList(X)) -> a__isNatList#(X)
p16: mark#(cons(X1,X2)) -> mark#(X1)
p17: 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}


-- 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#(and(X1,X2)) -> mark#(X1)
p12: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p13: mark#(length(X)) -> mark#(X)
p14: mark#(length(X)) -> a__length#(mark(X))
p15: a__length#(cons(N,L)) -> a__and#(a__isNatList(L),isNat(N))
p16: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p17: 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: lexicographic order
        interpretations:
          a__U11#_A(x1,x2) = (2,9)
          tt_A() = (3,10)
          a__length#_A(x1) = (2,9)
          mark_A(x1) = x1 + (3,17)
          cons_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (3,23)
          a__isNatList#_A(x1) = (2,7)
          a__isNat#_A(x1) = (2,7)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,6)
          length_A(x1) = x1 + (11,18)
          a__and#_A(x1,x2) = ((0,0),(1,0)) x2 + (2,7)
          a__isNat_A(x1) = ((0,0),(1,0)) x1 + (3,11)
          isNatList_A(x1) = ((0,0),(1,0)) x1 + (0,3)
          mark#_A(x1) = ((0,0),(1,0)) x1 + (2,7)
          and_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (0,8)
          a__isNatList_A(x1) = ((0,0),(1,0)) x1 + (3,20)
          isNat_A(x1) = (1,8)
          a__and_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (0,8)
          U11_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (12,37)
          a__zeros_A() = (7,25)
          |0|_A() = (0,1)
          zeros_A() = (4,9)
          a__U11_A(x1,x2) = ((1,0),(0,0)) x2 + (14,36)
          a__length_A(x1) = x1 + (11,18)
          a__isNatIList_A(x1) = x1 + (5,19)
          isNatIList_A(x1) = x1 + (4,18)
          nil_A() = (4,2)
  
    precedence:
    
      a__isNat = isNat > a__isNatIList = isNatIList > mark = cons = isNatList = a__isNatList = a__zeros > and = a__and > nil > a__U11 > a__U11# = tt = a__length# = a__isNatList# = a__isNat# = a__and# = mark# = U11 > |0| > a__length > s = length > zeros
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = []
      pi(cons) = []
      pi(a__isNatList#) = []
      pi(a__isNat#) = []
      pi(s) = []
      pi(length) = [1]
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(mark#) = []
      pi(and) = []
      pi(a__isNatList) = []
      pi(isNat) = []
      pi(a__and) = [2]
      pi(U11) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p15

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#(isNatList(X)) -> a__isNatList#(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#(length(X)) -> a__length#(mark(X))
p15: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p16: 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}


-- 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#(length(X)) -> a__length#(mark(X))
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#(isNatList(X)) -> a__isNatList#(X)
p12: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p13: a__isNat#(length(V1)) -> a__isNatList#(V1)
p14: a__isNat#(s(V1)) -> a__isNat#(V1)
p15: mark#(cons(X1,X2)) -> mark#(X1)
p16: 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: lexicographic order
        interpretations:
          a__U11#_A(x1,x2) = ((1,0),(1,1)) x2 + (10,19)
          tt_A() = (9,24)
          a__length#_A(x1) = ((1,0),(1,1)) x1 + (2,1)
          mark_A(x1) = x1 + (8,10)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (8,10)
          a__and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (0,9)
          a__isNatList_A(x1) = ((1,0),(1,1)) x1 + (4,22)
          isNat_A(x1) = ((1,0),(1,1)) x1 + (3,20)
          a__isNatList#_A(x1) = ((1,0),(0,0)) x1 + (6,15)
          a__and#_A(x1,x2) = ((1,0),(0,0)) x2 + (6,2)
          a__isNat_A(x1) = ((1,0),(0,0)) x1 + (10,26)
          isNatList_A(x1) = x1 + (1,11)
          mark#_A(x1) = ((1,0),(0,0)) x1 + (6,2)
          U11_A(x1,x2) = ((1,0),(1,0)) x2 + (11,14)
          length_A(x1) = ((1,0),(1,0)) x1 + (9,14)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (0,1)
          a__isNat#_A(x1) = ((1,0),(1,1)) x1 + (0,1)
          s_A(x1) = x1
          a__zeros_A() = (8,27)
          |0|_A() = (0,26)
          zeros_A() = (0,17)
          a__U11_A(x1,x2) = ((1,0),(1,0)) x2 + (17,23)
          a__length_A(x1) = ((1,0),(1,0)) x1 + (9,15)
          a__isNatIList_A(x1) = ((1,0),(0,0)) x1 + (10,25)
          isNatIList_A(x1) = ((1,0),(0,0)) x1 + (3,16)
          nil_A() = (12,25)
  
    precedence:
    
      a__length > mark = s = a__U11 > a__zeros > cons = isNatList > a__U11# = a__length# > a__isNatList# > a__and# = mark# = a__isNat# > length > a__isNatIList > a__isNat > a__isNatList > a__and = U11 = and > tt > isNatIList > |0| > zeros = nil > isNat
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = [1]
      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(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(isNatIList) = []
      pi(nil) = []

The next rules are strictly ordered:

  p13

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#(length(X)) -> a__length#(mark(X))
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#(isNatList(X)) -> a__isNatList#(X)
p12: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p13: a__isNat#(s(V1)) -> a__isNat#(V1)
p14: mark#(cons(X1,X2)) -> mark#(X1)
p15: 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, p14, p15}
  {p13}


-- 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__and#(a__isNat(V1),isNatList(V2))
p4: a__and#(tt(),X) -> mark#(X)
p5: mark#(s(X)) -> mark#(X)
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: mark#(isNatList(X)) -> a__isNatList#(X)
p8: mark#(and(X1,X2)) -> mark#(X1)
p9: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p10: mark#(length(X)) -> mark#(X)
p11: mark#(length(X)) -> a__length#(mark(X))
p12: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L)
p13: 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: lexicographic order
        interpretations:
          a__U11#_A(x1,x2) = ((0,0),(1,0)) x2 + (24,26)
          tt_A() = (0,2)
          a__length#_A(x1) = ((0,0),(1,0)) x1 + (24,26)
          mark_A(x1) = x1 + (0,11)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (0,5)
          a__isNatList#_A(x1) = ((0,0),(1,0)) x1 + (24,4)
          a__and#_A(x1,x2) = ((0,0),(1,0)) x2 + (24,4)
          a__isNat_A(x1) = x1
          isNatList_A(x1) = x1 + (0,3)
          mark#_A(x1) = ((0,0),(1,0)) x1 + (24,4)
          s_A(x1) = x1
          and_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (0,12)
          length_A(x1) = ((1,0),(0,0)) x1 + (23,14)
          a__and_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (0,12)
          a__isNatList_A(x1) = x1 + (0,13)
          isNat_A(x1) = x1
          U11_A(x1,x2) = ((1,0),(0,0)) x2 + (23,5)
          a__zeros_A() = (1,13)
          |0|_A() = (0,3)
          zeros_A() = (1,3)
          a__U11_A(x1,x2) = ((1,0),(0,0)) x2 + (23,15)
          a__length_A(x1) = ((1,0),(0,0)) x1 + (23,15)
          a__isNatIList_A(x1) = x1 + (1,12)
          isNatIList_A(x1) = x1 + (1,1)
          nil_A() = (1,1)
  
    precedence:
    
      mark > a__isNat = zeros > a__and = a__isNatList > isNat > s = a__U11 = a__length > tt > length = a__isNatIList > isNatList = a__zeros > U11 = isNatIList > a__U11# = a__length# = a__isNatList# = a__and# = mark# > and > cons = nil > |0|
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = []
      pi(mark) = [1]
      pi(cons) = [2]
      pi(a__isNatList#) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = [1]
      pi(mark#) = []
      pi(s) = []
      pi(and) = []
      pi(length) = []
      pi(a__and) = [2]
      pi(a__isNatList) = [1]
      pi(isNat) = []
      pi(U11) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__isNatIList) = [1]
      pi(isNatIList) = [1]
      pi(nil) = []

The next rules are strictly ordered:

  p13

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


-- 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#(length(X)) -> a__length#(mark(X))
p7: mark#(length(X)) -> mark#(X)
p8: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p9: mark#(and(X1,X2)) -> mark#(X1)
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 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: lexicographic order
        interpretations:
          a__U11#_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (22,11)
          tt_A() = (23,49)
          a__length#_A(x1) = x1 + (0,1)
          mark_A(x1) = x1 + (22,32)
          cons_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (22,29)
          a__and_A(x1,x2) = x1 + x2 + (3,47)
          a__isNatList_A(x1) = ((1,0),(1,1)) x1 + (11,28)
          isNat_A(x1) = ((1,0),(1,1)) x1 + (3,2)
          a__isNatList#_A(x1) = ((1,0),(1,1)) x1 + (21,79)
          a__and#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,72)
          a__isNat_A(x1) = ((1,0),(1,1)) x1 + (24,1)
          isNatList_A(x1) = ((1,0),(1,1)) x1 + (5,30)
          mark#_A(x1) = x1 + (21,48)
          length_A(x1) = ((1,0),(0,0)) x1 + (23,3)
          and_A(x1,x2) = x1 + x2 + (3,47)
          s_A(x1) = x1
          a__zeros_A() = (23,81)
          |0|_A() = (0,2)
          zeros_A() = (1,50)
          a__U11_A(x1,x2) = ((1,0),(0,0)) x2 + (45,3)
          a__length_A(x1) = ((1,0),(0,0)) x1 + (23,3)
          a__isNatIList_A(x1) = ((1,0),(1,1)) x1 + (23,29)
          isNatIList_A(x1) = ((1,0),(1,1)) x1 + (2,32)
          nil_A() = (13,1)
          U11_A(x1,x2) = ((1,0),(0,0)) x2 + (45,1)
  
    precedence:
    
      a__isNatIList > tt = nil > mark = length = s = a__U11 = a__length > cons = a__zeros > a__isNatList = isNatList > mark# > a__U11# = a__length# = a__and = a__isNatList# = a__and# = and > isNat = a__isNat = isNatIList > |0| > zeros = U11
    
    partial status:
  
      pi(a__U11#) = []
      pi(tt) = []
      pi(a__length#) = [1]
      pi(mark) = [1]
      pi(cons) = []
      pi(a__and) = []
      pi(a__isNatList) = [1]
      pi(isNat) = []
      pi(a__isNatList#) = [1]
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = [1]
      pi(mark#) = []
      pi(length) = []
      pi(and) = []
      pi(s) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__isNatIList) = []
      pi(isNatIList) = []
      pi(nil) = []
      pi(U11) = []

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__length#(cons(N,L)) -> a__isNatList#(L)
p3: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p4: a__and#(tt(),X) -> mark#(X)
p5: mark#(length(X)) -> a__length#(mark(X))
p6: mark#(length(X)) -> mark#(X)
p7: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p8: mark#(and(X1,X2)) -> mark#(X1)
p9: mark#(isNatList(X)) -> a__isNatList#(X)
p10: mark#(cons(X1,X2)) -> mark#(X1)
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 estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: a__length#(cons(N,L)) -> a__isNatList#(L)
p2: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p3: a__and#(tt(),X) -> mark#(X)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(isNatList(X)) -> a__isNatList#(X)
p7: mark#(and(X1,X2)) -> mark#(X1)
p8: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p9: mark#(length(X)) -> mark#(X)
p10: mark#(length(X)) -> a__length#(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: lexicographic order
        interpretations:
          a__length#_A(x1) = x1 + (12,1)
          cons_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (15,27)
          a__isNatList#_A(x1) = ((1,0),(0,0)) x1 + (14,24)
          a__and#_A(x1,x2) = ((1,0),(1,0)) x2 + (28,24)
          a__isNat_A(x1) = ((1,0),(0,0)) x1 + (2,46)
          isNatList_A(x1) = ((1,0),(0,0)) x1
          tt_A() = (2,1)
          mark#_A(x1) = ((1,0),(1,0)) x1 + (27,23)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,3)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (26,25)
          mark_A(x1) = x1 + (15,48)
          length_A(x1) = ((1,0),(1,0)) x1 + (26,0)
          a__zeros_A() = (16,28)
          |0|_A() = (0,29)
          zeros_A() = (1,2)
          a__U11_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (41,2)
          a__length_A(x1) = ((1,0),(1,0)) x1 + (26,29)
          a__and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (26,26)
          a__isNatIList_A(x1) = ((1,0),(0,0)) x1 + (15,47)
          a__isNatList_A(x1) = ((1,0),(0,0)) x1 + (15,45)
          isNatIList_A(x1) = ((1,0),(0,0)) x1 + (1,0)
          nil_A() = (0,0)
          isNat_A(x1) = ((1,0),(0,0)) x1 + (1,0)
          U11_A(x1,x2) = ((1,0),(0,0)) x2 + (27,1)
  
    precedence:
    
      a__isNatIList = a__isNatList > a__isNat > tt > mark > a__zeros > cons = a__and > length = |0| = a__U11 = a__length = U11 > and > a__length# = zeros = isNatIList > a__isNatList# = a__and# = isNatList = mark# = nil > s = isNat
    
    partial status:
  
      pi(a__length#) = [1]
      pi(cons) = []
      pi(a__isNatList#) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(tt) = []
      pi(mark#) = []
      pi(s) = []
      pi(and) = []
      pi(mark) = []
      pi(length) = []
      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:

  p10

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

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


-- Reduction pair.

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

p1: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p2: a__and#(tt(),X) -> mark#(X)
p3: mark#(length(X)) -> mark#(X)
p4: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p5: mark#(and(X1,X2)) -> mark#(X1)
p6: mark#(isNatList(X)) -> a__isNatList#(X)
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: 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: lexicographic order
        interpretations:
          a__isNatList#_A(x1) = ((1,0),(1,1)) x1 + (1,2)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (20,28)
          a__and#_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
          a__isNat_A(x1) = ((1,0),(0,0)) x1 + (2,26)
          isNatList_A(x1) = ((1,0),(1,1)) x1 + (2,1)
          tt_A() = (0,4)
          mark#_A(x1) = ((1,0),(1,1)) x1 + (0,6)
          length_A(x1) = ((1,0),(0,0)) x1 + (21,7)
          and_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (34,7)
          mark_A(x1) = ((1,0),(1,0)) x1 + (20,8)
          s_A(x1) = x1
          a__zeros_A() = (42,29)
          |0|_A() = (0,30)
          zeros_A() = (22,5)
          a__U11_A(x1,x2) = ((1,0),(0,0)) x2 + (41,29)
          a__length_A(x1) = ((1,0),(0,0)) x1 + (21,29)
          a__and_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (34,33)
          a__isNatIList_A(x1) = ((1,0),(1,1)) x1 + (20,10)
          a__isNatList_A(x1) = ((1,0),(1,0)) x1 + (19,2)
          isNatIList_A(x1) = ((1,0),(0,0)) x1 + (1,29)
          nil_A() = (1,3)
          isNat_A(x1) = ((1,0),(0,0)) x1 + (1,25)
          U11_A(x1,x2) = ((1,0),(0,0)) x2 + (22,7)
  
    precedence:
    
      mark = s = |0| = a__U11 = a__length > isNatList = nil > a__and = a__isNatIList = a__isNatList > and > a__isNatList# = a__and# = mark# = isNatIList > a__isNat > tt > cons = a__zeros > length > zeros > U11 > isNat
    
    partial status:
  
      pi(a__isNatList#) = [1]
      pi(cons) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(tt) = []
      pi(mark#) = [1]
      pi(length) = []
      pi(and) = [1]
      pi(mark) = []
      pi(s) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(a__and) = []
      pi(a__isNatIList) = [1]
      pi(a__isNatList) = []
      pi(isNatIList) = []
      pi(nil) = []
      pi(isNat) = []
      pi(U11) = []

The next rules are strictly ordered:

  p4

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


-- Reduction pair.

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

p1: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p2: a__and#(tt(),X) -> mark#(X)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(isNatList(X)) -> a__isNatList#(X)
p6: mark#(and(X1,X2)) -> mark#(X1)
p7: 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: lexicographic order
        interpretations:
          a__isNatList#_A(x1) = ((0,0),(1,0)) x1 + (2,3)
          cons_A(x1,x2) = x1 + x2 + (8,0)
          a__and#_A(x1,x2) = ((0,0),(1,0)) x2 + (2,4)
          a__isNat_A(x1) = ((1,0),(1,0)) x1 + (0,12)
          isNatList_A(x1) = ((1,0),(0,0)) x1 + (1,2)
          tt_A() = (0,2)
          mark#_A(x1) = ((0,0),(1,0)) x1 + (2,3)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,1)
          and_A(x1,x2) = x1 + x2 + (9,1)
          length_A(x1) = ((1,0),(0,0)) x1 + (5,8)
          a__zeros_A() = (14,12)
          |0|_A() = (0,11)
          zeros_A() = (6,0)
          a__U11_A(x1,x2) = ((1,0),(0,0)) x2 + (13,2)
          a__length_A(x1) = ((1,0),(0,0)) x1 + (5,10)
          mark_A(x1) = ((1,0),(1,1)) x1 + (8,7)
          a__isNatIList_A(x1) = x1 + (4,1)
          a__isNatList_A(x1) = ((1,0),(1,0)) x1 + (3,9)
          a__and_A(x1,x2) = x1 + x2 + (9,2)
          isNatIList_A(x1) = x1
          nil_A() = (0,12)
          isNat_A(x1) = ((1,0),(1,0)) x1 + (0,11)
          U11_A(x1,x2) = ((1,0),(0,0)) x2 + (5,1)
  
    precedence:
    
      |0| > s > cons = a__isNat = tt = a__zeros = mark = a__isNatList = a__and > a__U11 > length = a__length > a__isNatList# = a__and# = mark# > isNatList > and = zeros = U11 > isNat > a__isNatIList = isNatIList = nil
    
    partial status:
  
      pi(a__isNatList#) = []
      pi(cons) = []
      pi(a__and#) = []
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(tt) = []
      pi(mark#) = []
      pi(s) = []
      pi(and) = [2]
      pi(length) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = []
      pi(mark) = []
      pi(a__isNatIList) = [1]
      pi(a__isNatList) = []
      pi(a__and) = []
      pi(isNatIList) = []
      pi(nil) = []
      pi(isNat) = []
      pi(U11) = []

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: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p2: a__and#(tt(),X) -> mark#(X)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(isNatList(X)) -> a__isNatList#(X)
p6: 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}


-- Reduction pair.

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

p1: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p2: a__and#(tt(),X) -> mark#(X)
p3: mark#(length(X)) -> mark#(X)
p4: mark#(isNatList(X)) -> a__isNatList#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: 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: lexicographic order
        interpretations:
          a__isNatList#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          cons_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (8,8)
          a__and#_A(x1,x2) = x2 + (0,1)
          a__isNat_A(x1) = ((1,0),(0,0)) x1 + (5,5)
          isNatList_A(x1) = ((1,0),(0,0)) x1 + (6,0)
          tt_A() = (0,3)
          mark#_A(x1) = ((1,0),(0,0)) x1 + (0,1)
          length_A(x1) = x1 + (14,0)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,6)
          a__zeros_A() = (19,10)
          |0|_A() = (0,1)
          zeros_A() = (11,0)
          a__U11_A(x1,x2) = ((1,0),(0,0)) x2 + (22,7)
          a__length_A(x1) = x1 + (14,0)
          mark_A(x1) = ((1,0),(1,1)) x1 + (8,0)
          a__isNatIList_A(x1) = ((1,0),(0,0)) x1 + (14,15)
          a__isNatList_A(x1) = ((1,0),(0,0)) x1 + (13,4)
          a__and_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (9,5)
          isNatIList_A(x1) = ((1,0),(0,0)) x1 + (7,9)
          nil_A() = (1,2)
          isNat_A(x1) = ((1,0),(0,0)) x1 + (4,2)
          U11_A(x1,x2) = ((1,0),(0,0)) x2 + (21,0)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (9,0)
  
    precedence:
    
      a__and > a__U11 = mark = a__isNatIList > length = a__length > a__and# > mark# = a__zeros > isNatIList = nil > cons > |0| > a__isNat = zeros > s > a__isNatList# = isNatList = a__isNatList = U11 > isNat > tt > and
    
    partial status:
  
      pi(a__isNatList#) = []
      pi(cons) = [1]
      pi(a__and#) = [2]
      pi(a__isNat) = []
      pi(isNatList) = []
      pi(tt) = []
      pi(mark#) = []
      pi(length) = [1]
      pi(s) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__U11) = []
      pi(a__length) = [1]
      pi(mark) = []
      pi(a__isNatIList) = []
      pi(a__isNatList) = []
      pi(a__and) = []
      pi(isNatIList) = []
      pi(nil) = []
      pi(isNat) = []
      pi(U11) = []
      pi(and) = []

The next rules are strictly ordered:

  p4

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__and#(a__isNat(V1),isNatList(V2))
p2: a__and#(tt(),X) -> mark#(X)
p3: mark#(length(X)) -> mark#(X)
p4: mark#(cons(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 estimated dependency graph contains the following SCCs:

  {p3, p4, p5}


-- 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#(cons(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: lexicographic order
        interpretations:
          mark#_A(x1) = x1 + (1,2)
          length_A(x1) = ((1,0),(1,1)) x1 + (2,3)
          s_A(x1) = x1 + (2,1)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,3)
  
    precedence:
    
      length = s > mark# = cons
    
    partial status:
  
      pi(mark#) = [1]
      pi(length) = [1]
      pi(s) = [1]
      pi(cons) = [2]

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: mark#(length(X)) -> mark#(X)
p2: mark#(cons(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, p2}


-- Reduction pair.

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

p1: mark#(length(X)) -> mark#(X)
p2: mark#(cons(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: lexicographic order
        interpretations:
          mark#_A(x1) = ((1,0),(1,0)) x1 + (1,2)
          length_A(x1) = ((1,0),(0,0)) x1 + (2,1)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (2,1)
  
    precedence:
    
      mark# = length = cons
    
    partial status:
  
      pi(mark#) = []
      pi(length) = []
      pi(cons) = [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#(cons(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#(cons(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: lexicographic order
        interpretations:
          mark#_A(x1) = ((1,0),(1,1)) x1
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
  
    precedence:
    
      mark# = cons
    
    partial status:
  
      pi(mark#) = [1]
      pi(cons) = [1]

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__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: lexicographic 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.