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:

  max/plus interpretations on natural numbers:
  
    a__U11#_A(x1,x2) = max{0, x1 - 27, x2 - 15}
    tt_A = 29
    a__length#_A(x1) = max{2, x1 - 30}
    mark_A(x1) = max{20, x1 + 14}
    cons_A(x1,x2) = max{18, x1 + 11, x2 + 16}
    a__isNatList#_A(x1) = 2
    a__isNat#_A(x1) = 2
    s_A(x1) = max{5, x1}
    length_A(x1) = max{25, x1}
    a__and#_A(x1,x2) = max{2, x1 - 31, x2 - 13}
    a__isNat_A(x1) = 29
    isNatList_A(x1) = 15
    mark#_A(x1) = max{2, x1 - 16}
    isNatIList_A(x1) = 15
    a__isNatIList#_A(x1) = 2
    isNat_A(x1) = 15
    and_A(x1,x2) = max{x1, x2 + 4}
    a__isNatList_A(x1) = 29
    a__and_A(x1,x2) = max{x1, x2 + 14}
    U11_A(x1,x2) = max{16, x1 + 10, x2 + 10}
    a__zeros_A = 18
    |0|_A = 1
    zeros_A = 0
    a__U11_A(x1,x2) = max{x1 + 10, x2 + 16}
    a__length_A(x1) = max{39, x1}
    a__isNatIList_A(x1) = 29
    nil_A = 0

The next rules are strictly ordered:

  p24

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: a__U11#(tt(),L) -> mark#(L)
p23: mark#(U11(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, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21, p22, p23}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

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

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    a__U11#_A(x1,x2) = max{1, x1 - 6, x2}
    tt_A = 11
    a__length#_A(x1) = max{0, x1 - 6}
    mark_A(x1) = max{11, x1 + 2}
    cons_A(x1,x2) = max{11, x1, x2 + 7}
    a__and_A(x1,x2) = max{11, x1, x2 + 2}
    a__isNatList_A(x1) = max{11, x1 - 5}
    isNat_A(x1) = max{6, x1 - 5}
    mark#_A(x1) = max{5, x1 - 1}
    U11_A(x1,x2) = max{x1, x2 + 7}
    length_A(x1) = max{7, x1}
    a__and#_A(x1,x2) = max{5, x1 - 18, x2 - 1}
    and_A(x1,x2) = max{6, x1, x2}
    a__isNat#_A(x1) = max{5, x1 - 6}
    a__isNatList#_A(x1) = max{5, x1 - 6}
    a__isNat_A(x1) = max{11, x1 - 5}
    isNatList_A(x1) = max{2, x1 - 5}
    s_A(x1) = max{6, x1}
    isNatIList_A(x1) = max{9, x1 - 5}
    a__isNatIList#_A(x1) = max{8, x1 - 6}
    a__zeros_A = 11
    |0|_A = 7
    zeros_A = 0
    a__U11_A(x1,x2) = max{10, x1, x2 + 7}
    a__length_A(x1) = max{7, x1}
    a__isNatIList_A(x1) = max{11, x1 - 3}
    nil_A = 6

The next rules are strictly ordered:

  p5

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__U11#(tt(),L) -> mark#(L)
p4: mark#(U11(X1,X2)) -> mark#(X1)
p5: a__length#(cons(N,L)) -> a__and#(a__isNatList(L),isNat(N))
p6: a__and#(tt(),X) -> 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#(isNat(X)) -> a__isNat#(X)
p11: a__isNat#(length(V1)) -> a__isNatList#(V1)
p12: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p13: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p14: a__isNat#(s(V1)) -> a__isNat#(V1)
p15: mark#(isNatList(X)) -> a__isNatList#(X)
p16: mark#(isNatIList(X)) -> a__isNatIList#(X)
p17: a__isNatIList#(V) -> a__isNatList#(V)
p18: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p19: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1)
p20: mark#(cons(X1,X2)) -> mark#(X1)
p21: mark#(s(X)) -> mark#(X)
p22: a__length#(cons(N,L)) -> a__isNatList#(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}
  {p4, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21}


-- Reduction pair.

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

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

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:

  max/plus interpretations on natural numbers:
  
    a__U11#_A(x1,x2) = max{x1 - 4, x2 + 14}
    tt_A = 38
    a__length#_A(x1) = max{1, x1 - 5}
    mark_A(x1) = max{38, x1 + 18}
    cons_A(x1,x2) = max{x1 + 16, x2 + 38}
    a__and_A(x1,x2) = max{x1 + 1, x2 + 21}
    a__isNatList_A(x1) = max{13, x1 + 1}
    isNat_A(x1) = max{0, x1 - 7}
    a__zeros_A = 38
    |0|_A = 22
    zeros_A = 0
    a__U11_A(x1,x2) = x2 + 28
    s_A(x1) = max{19, x1}
    a__length_A(x1) = max{21, x1 - 10}
    a__isNat_A(x1) = max{38, x1 + 11}
    length_A(x1) = max{1, x1 - 10}
    a__isNatIList_A(x1) = x1 + 38
    isNatIList_A(x1) = max{38, x1 + 20}
    nil_A = 39
    U11_A(x1,x2) = x2 + 28
    isNatList_A(x1) = max{4, x1 + 1}
    and_A(x1,x2) = max{x1 + 1, x2 + 21}

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

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

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:

  max/plus interpretations on natural numbers:
  
    mark#_A(x1) = max{0, x1 - 2}
    U11_A(x1,x2) = max{x1 + 3, x2 + 11}
    s_A(x1) = max{2, x1}
    cons_A(x1,x2) = max{x1 + 5, x2 + 9}
    isNatIList_A(x1) = 4
    a__isNatIList#_A(x1) = 2
    a__isNat#_A(x1) = 2
    length_A(x1) = max{12, x1 + 11}
    a__isNatList#_A(x1) = 2
    a__and#_A(x1,x2) = max{0, x1 - 11, x2 - 2}
    a__isNat_A(x1) = 13
    isNatList_A(x1) = 4
    tt_A = 10
    isNat_A(x1) = 4
    and_A(x1,x2) = max{4, x1, x2 + 3}
    mark_A(x1) = x1 + 9
    a__zeros_A = 18
    |0|_A = 13
    zeros_A = 9
    a__U11_A(x1,x2) = max{x1 + 3, x2 + 20}
    a__length_A(x1) = max{13, x1 + 11}
    a__and_A(x1,x2) = max{13, x1, x2 + 9}
    a__isNatIList_A(x1) = 13
    a__isNatList_A(x1) = 13
    nil_A = 1

The next rules are strictly ordered:

  p1, p3, p15

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(isNatIList(X)) -> a__isNatIList#(X)
p3: a__isNatIList#(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__isNat#(V1)
p7: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p8: a__and#(tt(),X) -> mark#(X)
p9: mark#(isNatList(X)) -> a__isNatList#(X)
p10: mark#(isNat(X)) -> a__isNat#(X)
p11: mark#(and(X1,X2)) -> mark#(X1)
p12: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p13: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p14: a__isNatIList#(V) -> a__isNatList#(V)

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}


-- Reduction pair.

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

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

  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:

  max/plus interpretations on natural numbers:
  
    mark#_A(x1) = max{0, x1 - 5}
    s_A(x1) = max{5, x1}
    and_A(x1,x2) = max{6, x1, x2}
    a__and#_A(x1,x2) = max{0, x2 - 5}
    mark_A(x1) = x1 + 6
    tt_A = 1
    isNat_A(x1) = 6
    a__isNat#_A(x1) = 0
    length_A(x1) = x1 + 7
    a__isNatList#_A(x1) = 0
    cons_A(x1,x2) = max{x1 + 8, x2 + 6}
    a__isNat_A(x1) = 7
    isNatList_A(x1) = 1
    isNatIList_A(x1) = max{4, x1 + 3}
    a__isNatIList#_A(x1) = max{0, x1 - 8}
    a__zeros_A = 12
    |0|_A = 4
    zeros_A = 6
    a__U11_A(x1,x2) = x2 + 13
    a__length_A(x1) = x1 + 7
    a__and_A(x1,x2) = max{7, x1, x2 + 6}
    a__isNatIList_A(x1) = max{9, x1 + 3}
    a__isNatList_A(x1) = 7
    nil_A = 1
    U11_A(x1,x2) = x2 + 7

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(isNatIList(X)) -> a__isNatIList#(X)
p3: a__isNatIList#(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__isNat#(V1)
p7: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p8: a__and#(tt(),X) -> mark#(X)
p9: mark#(isNatList(X)) -> a__isNatList#(X)
p10: mark#(and(X1,X2)) -> mark#(X1)
p11: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p12: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2))
p13: a__isNatIList#(V) -> a__isNatList#(V)

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:

  max/plus interpretations on natural numbers:
  
    mark#_A(x1) = x1 + 5
    s_A(x1) = max{2, x1}
    isNatIList_A(x1) = max{2, x1}
    a__isNatIList#_A(x1) = max{4, x1 + 3}
    cons_A(x1,x2) = max{x1 + 11, x2 + 10}
    a__isNat#_A(x1) = max{12, x1 - 1}
    length_A(x1) = max{7, x1 + 6}
    a__isNatList#_A(x1) = max{4, x1 + 3}
    a__and#_A(x1,x2) = max{12, x1 + 1, x2 + 5}
    a__isNat_A(x1) = max{4, x1 + 2}
    isNatList_A(x1) = max{2, x1}
    tt_A = 4
    and_A(x1,x2) = max{x1 + 7, x2 + 7}
    mark_A(x1) = x1 + 10
    a__zeros_A = 12
    |0|_A = 1
    zeros_A = 2
    a__U11_A(x1,x2) = max{x1 + 2, x2 + 16}
    a__length_A(x1) = max{7, x1 + 6}
    a__and_A(x1,x2) = max{x1 + 7, x2 + 10}
    a__isNatIList_A(x1) = max{5, x1 + 1}
    a__isNatList_A(x1) = max{5, x1 + 1}
    nil_A = 1
    isNat_A(x1) = max{4, x1 - 6}
    U11_A(x1,x2) = max{x1 + 2, x2 + 6}

The next rules are strictly ordered:

  p2, p3, p5, p6, p7, p9, p10, p12

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(s(X)) -> mark#(X)
p2: a__isNat#(s(V1)) -> a__isNat#(V1)
p3: a__and#(tt(),X) -> mark#(X)
p4: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p5: a__isNatIList#(V) -> a__isNatList#(V)

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, p3, p4}
  {p2}


-- Reduction pair.

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

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

  max/plus interpretations on natural numbers:
  
    mark#_A(x1) = max{0, x1 - 2}
    s_A(x1) = max{2, x1}
    and_A(x1,x2) = max{5, x1, x2 + 4}
    a__and#_A(x1,x2) = max{3, x1 - 7, x2 + 2}
    mark_A(x1) = x1 + 5
    tt_A = 8
    a__zeros_A = 14
    cons_A(x1,x2) = max{x1 + 2, x2 + 5}
    |0|_A = 12
    zeros_A = 9
    a__U11_A(x1,x2) = max{x1 + 2, x2 + 15}
    a__length_A(x1) = max{11, x1 + 10}
    a__and_A(x1,x2) = max{10, x1, x2 + 5}
    a__isNat_A(x1) = 10
    length_A(x1) = max{11, x1 + 10}
    a__isNatList_A(x1) = 10
    a__isNatIList_A(x1) = 10
    isNatIList_A(x1) = 5
    nil_A = 15
    isNatList_A(x1) = 5
    isNat_A(x1) = 5
    U11_A(x1,x2) = max{x1 + 2, x2 + 10}

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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

  (no rules)

Take the monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    mark#_A(x1) = x1
    s_A(x1) = x1 + 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 monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    a__isNat#_A(x1) = x1
    s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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