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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      a__U11#_A(x1,x2) = ((0,0),(1,0)) x2 + (0,2)
      tt_A() = (2,3)
      a__length#_A(x1) = ((0,0),(1,0)) x1
      mark_A(x1) = x1 + (2,3)
      cons_A(x1,x2) = x1 + x2 + (2,1)
      a__isNatList#_A(x1) = (0,1)
      a__isNat#_A(x1) = (0,1)
      s_A(x1) = x1 + (0,3)
      length_A(x1) = x1 + (2,0)
      a__and#_A(x1,x2) = ((0,0),(1,0)) x2 + (0,1)
      a__isNat_A(x1) = (2,3)
      isNatList_A(x1) = (0,1)
      mark#_A(x1) = ((0,0),(1,0)) x1 + (0,1)
      isNatIList_A(x1) = x1 + (3,0)
      a__isNatIList#_A(x1) = ((0,0),(1,0)) x1 + (0,3)
      isNat_A(x1) = (0,0)
      and_A(x1,x2) = x1 + x2
      a__isNatList_A(x1) = (2,3)
      a__and_A(x1,x2) = x1 + x2
      U11_A(x1,x2) = x1 + x2 + (2,0)
      a__zeros_A() = (3,2)
      |0|_A() = (0,4)
      zeros_A() = (1,3)
      a__U11_A(x1,x2) = x1 + x2 + (2,0)
      a__length_A(x1) = x1 + (2,3)
      a__isNatIList_A(x1) = x1 + (3,3)
      nil_A() = (1,0)

The next rules are strictly ordered:

  p2, p9, p10, p11, p12, p13, p18, p19, p20, p22, p23, 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__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p3: a__isNat#(s(V1)) -> a__isNat#(V1)
p4: a__isNat#(length(V1)) -> a__isNatList#(V1)
p5: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2))
p6: a__and#(tt(),X) -> mark#(X)
p7: mark#(s(X)) -> mark#(X)
p8: mark#(isNatList(X)) -> a__isNatList#(X)
p9: mark#(isNat(X)) -> a__isNat#(X)
p10: mark#(and(X1,X2)) -> mark#(X1)
p11: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
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, p12}
  {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__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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      a__U11#_A(x1,x2) = x1 + x2
      tt_A() = (17,20)
      a__length#_A(x1) = x1 + (2,6)
      mark_A(x1) = x1 + (15,13)
      cons_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (14,0)
      a__and_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (15,0)
      a__isNatList_A(x1) = x1 + (3,17)
      isNat_A(x1) = x1 + (1,1)
      a__zeros_A() = (16,14)
      |0|_A() = (1,11)
      zeros_A() = (1,1)
      a__U11_A(x1,x2) = x1 + x2 + (1,1)
      s_A(x1) = x1 + (0,1)
      a__length_A(x1) = x1 + (3,6)
      a__isNat_A(x1) = x1 + (16,10)
      length_A(x1) = x1 + (3,6)
      a__isNatIList_A(x1) = ((1,0),(1,0)) x1 + (17,18)
      isNatIList_A(x1) = x1 + (15,1)
      nil_A() = (15,1)
      U11_A(x1,x2) = x1 + x2 + (1,1)
      isNatList_A(x1) = x1 + (1,2)
      and_A(x1,x2) = ((1,0),(1,0)) x2 + (14,1)

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

and R consists of:

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

The 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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      a__isNatList#_A(x1) = ((1,0),(1,1)) x1 + (2,17)
      cons_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (8,0)
      a__isNat#_A(x1) = x1 + (10,25)
      length_A(x1) = ((1,0),(1,0)) x1 + (1,1)
      a__and#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2
      a__isNat_A(x1) = x1 + (9,12)
      isNatList_A(x1) = ((1,0),(1,1)) x1 + (1,1)
      tt_A() = (8,19)
      mark#_A(x1) = ((1,0),(1,1)) x1 + (8,16)
      and_A(x1,x2) = x1 + x2
      mark_A(x1) = x1 + (8,8)
      isNat_A(x1) = ((1,0),(1,1)) x1 + (3,5)
      s_A(x1) = x1
      a__zeros_A() = (9,4)
      |0|_A() = (0,1)
      zeros_A() = (1,1)
      a__U11_A(x1,x2) = ((1,0),(1,0)) x2 + (9,9)
      a__length_A(x1) = ((1,0),(1,0)) x1 + (1,1)
      a__and_A(x1,x2) = x1 + x2
      a__isNatIList_A(x1) = x1 + (8,11)
      a__isNatList_A(x1) = ((1,0),(1,1)) x1 + (2,10)
      isNatIList_A(x1) = x1 + (6,0)
      nil_A() = (7,1)
      U11_A(x1,x2) = x2 + (9,9)

The next rules are strictly ordered:

  p2, p3, p4, p7, p9

We remove them from the problem.

-- SCC decomposition.

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

p1: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1)
p2: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p3: mark#(and(X1,X2)) -> mark#(X1)
p4: a__isNat#(s(V1)) -> a__isNat#(V1)
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:

  {p4}
  {p3, p5}


-- 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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      a__isNat#_A(x1) = x1
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1
  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

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

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      mark#_A(x1) = ((1,0),(1,1)) x1
      and_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1, p2
  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

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