YES

We show the termination of the TRS R:

  a__and(tt(),T) -> mark(T)
  a__isNatIList(IL) -> a__isNatList(IL)
  a__isNat(|0|()) -> tt()
  a__isNat(s(N)) -> a__isNat(N)
  a__isNat(length(L)) -> a__isNatList(L)
  a__isNatIList(zeros()) -> tt()
  a__isNatIList(cons(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
  a__isNatList(nil()) -> tt()
  a__isNatList(cons(N,L)) -> a__and(a__isNat(N),a__isNatList(L))
  a__isNatList(take(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
  a__zeros() -> cons(|0|(),zeros())
  a__take(|0|(),IL) -> a__uTake1(a__isNatIList(IL))
  a__uTake1(tt()) -> nil()
  a__take(s(M),cons(N,IL)) -> a__uTake2(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
  a__uTake2(tt(),M,N,IL) -> cons(mark(N),take(M,IL))
  a__length(cons(N,L)) -> a__uLength(a__and(a__isNat(N),a__isNatList(L)),L)
  a__uLength(tt(),L) -> s(a__length(mark(L)))
  mark(and(X1,X2)) -> a__and(mark(X1),mark(X2))
  mark(isNatIList(X)) -> a__isNatIList(X)
  mark(isNatList(X)) -> a__isNatList(X)
  mark(isNat(X)) -> a__isNat(X)
  mark(length(X)) -> a__length(mark(X))
  mark(zeros()) -> a__zeros()
  mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
  mark(uTake1(X)) -> a__uTake1(mark(X))
  mark(uTake2(X1,X2,X3,X4)) -> a__uTake2(mark(X1),X2,X3,X4)
  mark(uLength(X1,X2)) -> a__uLength(mark(X1),X2)
  mark(tt()) -> tt()
  mark(|0|()) -> |0|()
  mark(s(X)) -> s(mark(X))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(nil()) -> nil()
  a__and(X1,X2) -> and(X1,X2)
  a__isNatIList(X) -> isNatIList(X)
  a__isNatList(X) -> isNatList(X)
  a__isNat(X) -> isNat(X)
  a__length(X) -> length(X)
  a__zeros() -> zeros()
  a__take(X1,X2) -> take(X1,X2)
  a__uTake1(X) -> uTake1(X)
  a__uTake2(X1,X2,X3,X4) -> uTake2(X1,X2,X3,X4)
  a__uLength(X1,X2) -> uLength(X1,X2)

-- SCC decomposition.

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

p1: a__and#(tt(),T) -> mark#(T)
p2: a__isNatIList#(IL) -> a__isNatList#(IL)
p3: a__isNat#(s(N)) -> a__isNat#(N)
p4: a__isNat#(length(L)) -> a__isNatList#(L)
p5: a__isNatIList#(cons(N,IL)) -> a__and#(a__isNat(N),a__isNatIList(IL))
p6: a__isNatIList#(cons(N,IL)) -> a__isNat#(N)
p7: a__isNatIList#(cons(N,IL)) -> a__isNatIList#(IL)
p8: a__isNatList#(cons(N,L)) -> a__and#(a__isNat(N),a__isNatList(L))
p9: a__isNatList#(cons(N,L)) -> a__isNat#(N)
p10: a__isNatList#(cons(N,L)) -> a__isNatList#(L)
p11: a__isNatList#(take(N,IL)) -> a__and#(a__isNat(N),a__isNatIList(IL))
p12: a__isNatList#(take(N,IL)) -> a__isNat#(N)
p13: a__isNatList#(take(N,IL)) -> a__isNatIList#(IL)
p14: a__take#(|0|(),IL) -> a__uTake1#(a__isNatIList(IL))
p15: a__take#(|0|(),IL) -> a__isNatIList#(IL)
p16: a__take#(s(M),cons(N,IL)) -> a__uTake2#(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
p17: a__take#(s(M),cons(N,IL)) -> a__and#(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL)))
p18: a__take#(s(M),cons(N,IL)) -> a__isNat#(M)
p19: a__take#(s(M),cons(N,IL)) -> a__and#(a__isNat(N),a__isNatIList(IL))
p20: a__take#(s(M),cons(N,IL)) -> a__isNat#(N)
p21: a__take#(s(M),cons(N,IL)) -> a__isNatIList#(IL)
p22: a__uTake2#(tt(),M,N,IL) -> mark#(N)
p23: a__length#(cons(N,L)) -> a__uLength#(a__and(a__isNat(N),a__isNatList(L)),L)
p24: a__length#(cons(N,L)) -> a__and#(a__isNat(N),a__isNatList(L))
p25: a__length#(cons(N,L)) -> a__isNat#(N)
p26: a__length#(cons(N,L)) -> a__isNatList#(L)
p27: a__uLength#(tt(),L) -> a__length#(mark(L))
p28: a__uLength#(tt(),L) -> mark#(L)
p29: mark#(and(X1,X2)) -> a__and#(mark(X1),mark(X2))
p30: mark#(and(X1,X2)) -> mark#(X1)
p31: mark#(and(X1,X2)) -> mark#(X2)
p32: mark#(isNatIList(X)) -> a__isNatIList#(X)
p33: mark#(isNatList(X)) -> a__isNatList#(X)
p34: mark#(isNat(X)) -> a__isNat#(X)
p35: mark#(length(X)) -> a__length#(mark(X))
p36: mark#(length(X)) -> mark#(X)
p37: mark#(zeros()) -> a__zeros#()
p38: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p39: mark#(take(X1,X2)) -> mark#(X1)
p40: mark#(take(X1,X2)) -> mark#(X2)
p41: mark#(uTake1(X)) -> a__uTake1#(mark(X))
p42: mark#(uTake1(X)) -> mark#(X)
p43: mark#(uTake2(X1,X2,X3,X4)) -> a__uTake2#(mark(X1),X2,X3,X4)
p44: mark#(uTake2(X1,X2,X3,X4)) -> mark#(X1)
p45: mark#(uLength(X1,X2)) -> a__uLength#(mark(X1),X2)
p46: mark#(uLength(X1,X2)) -> mark#(X1)
p47: mark#(s(X)) -> mark#(X)
p48: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__and(tt(),T) -> mark(T)
r2: a__isNatIList(IL) -> a__isNatList(IL)
r3: a__isNat(|0|()) -> tt()
r4: a__isNat(s(N)) -> a__isNat(N)
r5: a__isNat(length(L)) -> a__isNatList(L)
r6: a__isNatIList(zeros()) -> tt()
r7: a__isNatIList(cons(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r8: a__isNatList(nil()) -> tt()
r9: a__isNatList(cons(N,L)) -> a__and(a__isNat(N),a__isNatList(L))
r10: a__isNatList(take(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r11: a__zeros() -> cons(|0|(),zeros())
r12: a__take(|0|(),IL) -> a__uTake1(a__isNatIList(IL))
r13: a__uTake1(tt()) -> nil()
r14: a__take(s(M),cons(N,IL)) -> a__uTake2(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
r15: a__uTake2(tt(),M,N,IL) -> cons(mark(N),take(M,IL))
r16: a__length(cons(N,L)) -> a__uLength(a__and(a__isNat(N),a__isNatList(L)),L)
r17: a__uLength(tt(),L) -> s(a__length(mark(L)))
r18: mark(and(X1,X2)) -> a__and(mark(X1),mark(X2))
r19: mark(isNatIList(X)) -> a__isNatIList(X)
r20: mark(isNatList(X)) -> a__isNatList(X)
r21: mark(isNat(X)) -> a__isNat(X)
r22: mark(length(X)) -> a__length(mark(X))
r23: mark(zeros()) -> a__zeros()
r24: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r25: mark(uTake1(X)) -> a__uTake1(mark(X))
r26: mark(uTake2(X1,X2,X3,X4)) -> a__uTake2(mark(X1),X2,X3,X4)
r27: mark(uLength(X1,X2)) -> a__uLength(mark(X1),X2)
r28: mark(tt()) -> tt()
r29: mark(|0|()) -> |0|()
r30: mark(s(X)) -> s(mark(X))
r31: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r32: mark(nil()) -> nil()
r33: a__and(X1,X2) -> and(X1,X2)
r34: a__isNatIList(X) -> isNatIList(X)
r35: a__isNatList(X) -> isNatList(X)
r36: a__isNat(X) -> isNat(X)
r37: a__length(X) -> length(X)
r38: a__zeros() -> zeros()
r39: a__take(X1,X2) -> take(X1,X2)
r40: a__uTake1(X) -> uTake1(X)
r41: a__uTake2(X1,X2,X3,X4) -> uTake2(X1,X2,X3,X4)
r42: a__uLength(X1,X2) -> uLength(X1,X2)

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, p26, p27, p28, p29, p30, p31, p32, p33, p34, p35, p36, p38, p39, p40, p42, p43, p44, p45, p46, p47, p48}


-- Reduction pair.

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

p1: a__and#(tt(),T) -> mark#(T)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(uLength(X1,X2)) -> mark#(X1)
p5: mark#(uLength(X1,X2)) -> a__uLength#(mark(X1),X2)
p6: a__uLength#(tt(),L) -> mark#(L)
p7: mark#(uTake2(X1,X2,X3,X4)) -> mark#(X1)
p8: mark#(uTake2(X1,X2,X3,X4)) -> a__uTake2#(mark(X1),X2,X3,X4)
p9: a__uTake2#(tt(),M,N,IL) -> mark#(N)
p10: mark#(uTake1(X)) -> mark#(X)
p11: mark#(take(X1,X2)) -> mark#(X2)
p12: mark#(take(X1,X2)) -> mark#(X1)
p13: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p14: a__take#(s(M),cons(N,IL)) -> a__isNatIList#(IL)
p15: a__isNatIList#(cons(N,IL)) -> a__isNatIList#(IL)
p16: a__isNatIList#(cons(N,IL)) -> a__isNat#(N)
p17: a__isNat#(length(L)) -> a__isNatList#(L)
p18: a__isNatList#(take(N,IL)) -> a__isNatIList#(IL)
p19: a__isNatIList#(cons(N,IL)) -> a__and#(a__isNat(N),a__isNatIList(IL))
p20: a__isNatIList#(IL) -> a__isNatList#(IL)
p21: a__isNatList#(take(N,IL)) -> a__isNat#(N)
p22: a__isNat#(s(N)) -> a__isNat#(N)
p23: a__isNatList#(take(N,IL)) -> a__and#(a__isNat(N),a__isNatIList(IL))
p24: a__isNatList#(cons(N,L)) -> a__isNatList#(L)
p25: a__isNatList#(cons(N,L)) -> a__isNat#(N)
p26: a__isNatList#(cons(N,L)) -> a__and#(a__isNat(N),a__isNatList(L))
p27: a__take#(s(M),cons(N,IL)) -> a__isNat#(N)
p28: a__take#(s(M),cons(N,IL)) -> a__and#(a__isNat(N),a__isNatIList(IL))
p29: a__take#(s(M),cons(N,IL)) -> a__isNat#(M)
p30: a__take#(s(M),cons(N,IL)) -> a__and#(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL)))
p31: a__take#(s(M),cons(N,IL)) -> a__uTake2#(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
p32: a__take#(|0|(),IL) -> a__isNatIList#(IL)
p33: mark#(length(X)) -> mark#(X)
p34: mark#(length(X)) -> a__length#(mark(X))
p35: a__length#(cons(N,L)) -> a__isNatList#(L)
p36: a__length#(cons(N,L)) -> a__isNat#(N)
p37: a__length#(cons(N,L)) -> a__and#(a__isNat(N),a__isNatList(L))
p38: a__length#(cons(N,L)) -> a__uLength#(a__and(a__isNat(N),a__isNatList(L)),L)
p39: a__uLength#(tt(),L) -> a__length#(mark(L))
p40: mark#(isNat(X)) -> a__isNat#(X)
p41: mark#(isNatList(X)) -> a__isNatList#(X)
p42: mark#(isNatIList(X)) -> a__isNatIList#(X)
p43: mark#(and(X1,X2)) -> mark#(X2)
p44: mark#(and(X1,X2)) -> mark#(X1)
p45: mark#(and(X1,X2)) -> a__and#(mark(X1),mark(X2))

and R consists of:

r1: a__and(tt(),T) -> mark(T)
r2: a__isNatIList(IL) -> a__isNatList(IL)
r3: a__isNat(|0|()) -> tt()
r4: a__isNat(s(N)) -> a__isNat(N)
r5: a__isNat(length(L)) -> a__isNatList(L)
r6: a__isNatIList(zeros()) -> tt()
r7: a__isNatIList(cons(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r8: a__isNatList(nil()) -> tt()
r9: a__isNatList(cons(N,L)) -> a__and(a__isNat(N),a__isNatList(L))
r10: a__isNatList(take(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r11: a__zeros() -> cons(|0|(),zeros())
r12: a__take(|0|(),IL) -> a__uTake1(a__isNatIList(IL))
r13: a__uTake1(tt()) -> nil()
r14: a__take(s(M),cons(N,IL)) -> a__uTake2(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
r15: a__uTake2(tt(),M,N,IL) -> cons(mark(N),take(M,IL))
r16: a__length(cons(N,L)) -> a__uLength(a__and(a__isNat(N),a__isNatList(L)),L)
r17: a__uLength(tt(),L) -> s(a__length(mark(L)))
r18: mark(and(X1,X2)) -> a__and(mark(X1),mark(X2))
r19: mark(isNatIList(X)) -> a__isNatIList(X)
r20: mark(isNatList(X)) -> a__isNatList(X)
r21: mark(isNat(X)) -> a__isNat(X)
r22: mark(length(X)) -> a__length(mark(X))
r23: mark(zeros()) -> a__zeros()
r24: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r25: mark(uTake1(X)) -> a__uTake1(mark(X))
r26: mark(uTake2(X1,X2,X3,X4)) -> a__uTake2(mark(X1),X2,X3,X4)
r27: mark(uLength(X1,X2)) -> a__uLength(mark(X1),X2)
r28: mark(tt()) -> tt()
r29: mark(|0|()) -> |0|()
r30: mark(s(X)) -> s(mark(X))
r31: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r32: mark(nil()) -> nil()
r33: a__and(X1,X2) -> and(X1,X2)
r34: a__isNatIList(X) -> isNatIList(X)
r35: a__isNatList(X) -> isNatList(X)
r36: a__isNat(X) -> isNat(X)
r37: a__length(X) -> length(X)
r38: a__zeros() -> zeros()
r39: a__take(X1,X2) -> take(X1,X2)
r40: a__uTake1(X) -> uTake1(X)
r41: a__uTake2(X1,X2,X3,X4) -> uTake2(X1,X2,X3,X4)
r42: a__uLength(X1,X2) -> uLength(X1,X2)

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, r33, r34, r35, r36, r37, r38, r39, r40, r41, r42

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      a__and#_A(x1,x2) = max{x1 - 17, x2 + 6}
      tt_A = 14
      mark#_A(x1) = x1 + 6
      cons_A(x1,x2) = max{15, x1, x2}
      s_A(x1) = max{45, x1}
      uLength_A(x1,x2) = max{62, x1 + 32, x2 + 44}
      a__uLength#_A(x1,x2) = max{39, x1 - 17, x2 + 16}
      mark_A(x1) = max{17, x1}
      uTake2_A(x1,x2,x3,x4) = max{87, x1, x2 + 42, x3 + 67, x4 + 68}
      a__uTake2#_A(x1,x2,x3,x4) = max{x1 - 12, x2 + 9, x3 + 29, x4 + 29}
      uTake1_A(x1) = max{86, x1 + 58}
      take_A(x1,x2) = max{86, x1 + 42, x2 + 68}
      a__take#_A(x1,x2) = max{75, x1 + 10, x2 + 30}
      a__isNatIList#_A(x1) = x1 + 16
      a__isNat#_A(x1) = max{9, x1 + 8}
      length_A(x1) = max{62, x1 + 44}
      a__isNatList#_A(x1) = x1 + 15
      a__isNat_A(x1) = max{7, x1 + 4}
      a__isNatIList_A(x1) = x1 + 10
      a__isNatList_A(x1) = x1 + 9
      a__and_A(x1,x2) = max{23, x1 + 4, x2}
      |0|_A = 16
      a__length#_A(x1) = max{39, x1 + 16}
      isNat_A(x1) = max{7, x1 + 4}
      isNatList_A(x1) = x1 + 9
      isNatIList_A(x1) = x1 + 10
      and_A(x1,x2) = max{23, x1 + 4, x2}
      a__zeros_A = 16
      zeros_A = 15
      a__take_A(x1,x2) = max{86, x1 + 42, x2 + 68}
      a__uTake1_A(x1) = max{86, x1 + 58}
      nil_A = 73
      a__uTake2_A(x1,x2,x3,x4) = max{87, x1, x2 + 42, x3 + 67, x4 + 68}
      a__length_A(x1) = max{62, x1 + 44}
      a__uLength_A(x1,x2) = max{62, x1 + 32, x2 + 44}
    
    2. max/plus interpretations on natural numbers:
    
      a__and#_A(x1,x2) = 1
      tt_A = 14
      mark#_A(x1) = 1
      cons_A(x1,x2) = 15
      s_A(x1) = 5
      uLength_A(x1,x2) = 17
      a__uLength#_A(x1,x2) = 12
      mark_A(x1) = 24
      uTake2_A(x1,x2,x3,x4) = 4
      a__uTake2#_A(x1,x2,x3,x4) = 2
      uTake1_A(x1) = 5
      take_A(x1,x2) = 5
      a__take#_A(x1,x2) = 3
      a__isNatIList#_A(x1) = 1
      a__isNat#_A(x1) = 0
      length_A(x1) = 5
      a__isNatList#_A(x1) = 1
      a__isNat_A(x1) = 15
      a__isNatIList_A(x1) = 24
      a__isNatList_A(x1) = 24
      a__and_A(x1,x2) = 24
      |0|_A = 10
      a__length#_A(x1) = 12
      isNat_A(x1) = 5
      isNatList_A(x1) = 23
      isNatIList_A(x1) = 24
      and_A(x1,x2) = 6
      a__zeros_A = 16
      zeros_A = 17
      a__take_A(x1,x2) = 17
      a__uTake1_A(x1) = 17
      nil_A = 0
      a__uTake2_A(x1,x2,x3,x4) = 16
      a__length_A(x1) = 23
      a__uLength_A(x1,x2) = 22
    

The next rules are strictly ordered:

  p4, p5, p6, p8, p9, p10, p11, p12, p13, p14, p16, p17, p18, p20, p21, p23, p25, p27, p28, p29, p30, p31, p32, p33, p34, p35, p36, p37, p40, p44

We remove them from the problem.

-- SCC decomposition.

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

p1: a__and#(tt(),T) -> mark#(T)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(uTake2(X1,X2,X3,X4)) -> mark#(X1)
p5: a__isNatIList#(cons(N,IL)) -> a__isNatIList#(IL)
p6: a__isNatIList#(cons(N,IL)) -> a__and#(a__isNat(N),a__isNatIList(IL))
p7: a__isNat#(s(N)) -> a__isNat#(N)
p8: a__isNatList#(cons(N,L)) -> a__isNatList#(L)
p9: a__isNatList#(cons(N,L)) -> a__and#(a__isNat(N),a__isNatList(L))
p10: a__length#(cons(N,L)) -> a__uLength#(a__and(a__isNat(N),a__isNatList(L)),L)
p11: a__uLength#(tt(),L) -> a__length#(mark(L))
p12: mark#(isNatList(X)) -> a__isNatList#(X)
p13: mark#(isNatIList(X)) -> a__isNatIList#(X)
p14: mark#(and(X1,X2)) -> mark#(X2)
p15: mark#(and(X1,X2)) -> a__and#(mark(X1),mark(X2))

and R consists of:

r1: a__and(tt(),T) -> mark(T)
r2: a__isNatIList(IL) -> a__isNatList(IL)
r3: a__isNat(|0|()) -> tt()
r4: a__isNat(s(N)) -> a__isNat(N)
r5: a__isNat(length(L)) -> a__isNatList(L)
r6: a__isNatIList(zeros()) -> tt()
r7: a__isNatIList(cons(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r8: a__isNatList(nil()) -> tt()
r9: a__isNatList(cons(N,L)) -> a__and(a__isNat(N),a__isNatList(L))
r10: a__isNatList(take(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r11: a__zeros() -> cons(|0|(),zeros())
r12: a__take(|0|(),IL) -> a__uTake1(a__isNatIList(IL))
r13: a__uTake1(tt()) -> nil()
r14: a__take(s(M),cons(N,IL)) -> a__uTake2(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
r15: a__uTake2(tt(),M,N,IL) -> cons(mark(N),take(M,IL))
r16: a__length(cons(N,L)) -> a__uLength(a__and(a__isNat(N),a__isNatList(L)),L)
r17: a__uLength(tt(),L) -> s(a__length(mark(L)))
r18: mark(and(X1,X2)) -> a__and(mark(X1),mark(X2))
r19: mark(isNatIList(X)) -> a__isNatIList(X)
r20: mark(isNatList(X)) -> a__isNatList(X)
r21: mark(isNat(X)) -> a__isNat(X)
r22: mark(length(X)) -> a__length(mark(X))
r23: mark(zeros()) -> a__zeros()
r24: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r25: mark(uTake1(X)) -> a__uTake1(mark(X))
r26: mark(uTake2(X1,X2,X3,X4)) -> a__uTake2(mark(X1),X2,X3,X4)
r27: mark(uLength(X1,X2)) -> a__uLength(mark(X1),X2)
r28: mark(tt()) -> tt()
r29: mark(|0|()) -> |0|()
r30: mark(s(X)) -> s(mark(X))
r31: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r32: mark(nil()) -> nil()
r33: a__and(X1,X2) -> and(X1,X2)
r34: a__isNatIList(X) -> isNatIList(X)
r35: a__isNatList(X) -> isNatList(X)
r36: a__isNat(X) -> isNat(X)
r37: a__length(X) -> length(X)
r38: a__zeros() -> zeros()
r39: a__take(X1,X2) -> take(X1,X2)
r40: a__uTake1(X) -> uTake1(X)
r41: a__uTake2(X1,X2,X3,X4) -> uTake2(X1,X2,X3,X4)
r42: a__uLength(X1,X2) -> uLength(X1,X2)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: a__and#(tt(),T) -> mark#(T)
p2: mark#(and(X1,X2)) -> a__and#(mark(X1),mark(X2))
p3: mark#(and(X1,X2)) -> mark#(X2)
p4: mark#(isNatIList(X)) -> a__isNatIList#(X)
p5: a__isNatIList#(cons(N,IL)) -> a__and#(a__isNat(N),a__isNatIList(IL))
p6: a__isNatIList#(cons(N,IL)) -> a__isNatIList#(IL)
p7: mark#(isNatList(X)) -> a__isNatList#(X)
p8: a__isNatList#(cons(N,L)) -> a__and#(a__isNat(N),a__isNatList(L))
p9: a__isNatList#(cons(N,L)) -> a__isNatList#(L)
p10: mark#(uTake2(X1,X2,X3,X4)) -> mark#(X1)
p11: mark#(s(X)) -> mark#(X)
p12: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__and(tt(),T) -> mark(T)
r2: a__isNatIList(IL) -> a__isNatList(IL)
r3: a__isNat(|0|()) -> tt()
r4: a__isNat(s(N)) -> a__isNat(N)
r5: a__isNat(length(L)) -> a__isNatList(L)
r6: a__isNatIList(zeros()) -> tt()
r7: a__isNatIList(cons(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r8: a__isNatList(nil()) -> tt()
r9: a__isNatList(cons(N,L)) -> a__and(a__isNat(N),a__isNatList(L))
r10: a__isNatList(take(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r11: a__zeros() -> cons(|0|(),zeros())
r12: a__take(|0|(),IL) -> a__uTake1(a__isNatIList(IL))
r13: a__uTake1(tt()) -> nil()
r14: a__take(s(M),cons(N,IL)) -> a__uTake2(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
r15: a__uTake2(tt(),M,N,IL) -> cons(mark(N),take(M,IL))
r16: a__length(cons(N,L)) -> a__uLength(a__and(a__isNat(N),a__isNatList(L)),L)
r17: a__uLength(tt(),L) -> s(a__length(mark(L)))
r18: mark(and(X1,X2)) -> a__and(mark(X1),mark(X2))
r19: mark(isNatIList(X)) -> a__isNatIList(X)
r20: mark(isNatList(X)) -> a__isNatList(X)
r21: mark(isNat(X)) -> a__isNat(X)
r22: mark(length(X)) -> a__length(mark(X))
r23: mark(zeros()) -> a__zeros()
r24: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r25: mark(uTake1(X)) -> a__uTake1(mark(X))
r26: mark(uTake2(X1,X2,X3,X4)) -> a__uTake2(mark(X1),X2,X3,X4)
r27: mark(uLength(X1,X2)) -> a__uLength(mark(X1),X2)
r28: mark(tt()) -> tt()
r29: mark(|0|()) -> |0|()
r30: mark(s(X)) -> s(mark(X))
r31: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r32: mark(nil()) -> nil()
r33: a__and(X1,X2) -> and(X1,X2)
r34: a__isNatIList(X) -> isNatIList(X)
r35: a__isNatList(X) -> isNatList(X)
r36: a__isNat(X) -> isNat(X)
r37: a__length(X) -> length(X)
r38: a__zeros() -> zeros()
r39: a__take(X1,X2) -> take(X1,X2)
r40: a__uTake1(X) -> uTake1(X)
r41: a__uTake2(X1,X2,X3,X4) -> uTake2(X1,X2,X3,X4)
r42: a__uLength(X1,X2) -> uLength(X1,X2)

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, r33, r34, r35, r36, r37, r38, r39, r40, r41, r42

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      a__and#_A(x1,x2) = x2 + 33
      tt_A = 2
      mark#_A(x1) = x1 + 33
      and_A(x1,x2) = x2 + 15
      mark_A(x1) = x1 + 15
      isNatIList_A(x1) = x1 + 2
      a__isNatIList#_A(x1) = x1 + 34
      cons_A(x1,x2) = max{x1 + 19, x2 + 15}
      a__isNat_A(x1) = x1 + 35
      a__isNatIList_A(x1) = max{9, x1 + 6}
      isNatList_A(x1) = x1 + 4
      a__isNatList#_A(x1) = x1 + 36
      a__isNatList_A(x1) = x1 + 5
      uTake2_A(x1,x2,x3,x4) = max{x1 + 33, x3 + 73, x4 + 69}
      s_A(x1) = x1
      a__and_A(x1,x2) = x2 + 15
      a__zeros_A = 22
      |0|_A = 2
      zeros_A = 7
      a__take_A(x1,x2) = max{66, x2 + 54}
      a__uTake1_A(x1) = max{1, x1 - 8}
      nil_A = 0
      a__uTake2_A(x1,x2,x3,x4) = max{x1 + 33, x3 + 73, x4 + 69}
      take_A(x1,x2) = x2 + 54
      a__length_A(x1) = x1 + 18
      a__uLength_A(x1,x2) = max{x1 + 2, x2 + 33}
      length_A(x1) = x1 + 18
      uTake1_A(x1) = max{0, x1 - 8}
      uLength_A(x1,x2) = max{x1 + 2, x2 + 19}
      isNat_A(x1) = x1 + 35
    
    2. max/plus interpretations on natural numbers:
    
      a__and#_A(x1,x2) = 0
      tt_A = 9
      mark#_A(x1) = 0
      and_A(x1,x2) = 9
      mark_A(x1) = 10
      isNatIList_A(x1) = 7
      a__isNatIList#_A(x1) = max{1, x1 - 2}
      cons_A(x1,x2) = 3
      a__isNat_A(x1) = max{15, x1 + 8}
      a__isNatIList_A(x1) = 17
      isNatList_A(x1) = x1 + 9
      a__isNatList#_A(x1) = x1 + 4
      a__isNatList_A(x1) = max{16, x1 + 8}
      uTake2_A(x1,x2,x3,x4) = 4
      s_A(x1) = max{5, x1}
      a__and_A(x1,x2) = 10
      a__zeros_A = 4
      |0|_A = 2
      zeros_A = 5
      a__take_A(x1,x2) = 8
      a__uTake1_A(x1) = 9
      nil_A = 1
      a__uTake2_A(x1,x2,x3,x4) = 4
      take_A(x1,x2) = 3
      a__length_A(x1) = max{10, x1}
      a__uLength_A(x1,x2) = 10
      length_A(x1) = max{3, x1}
      uTake1_A(x1) = 2
      uLength_A(x1,x2) = 5
      isNat_A(x1) = x1 + 7
    

The next rules are strictly ordered:

  p3, p4, p5, p6, p7, p8, p9, p10, p12

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__and(tt(),T) -> mark(T)
r2: a__isNatIList(IL) -> a__isNatList(IL)
r3: a__isNat(|0|()) -> tt()
r4: a__isNat(s(N)) -> a__isNat(N)
r5: a__isNat(length(L)) -> a__isNatList(L)
r6: a__isNatIList(zeros()) -> tt()
r7: a__isNatIList(cons(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r8: a__isNatList(nil()) -> tt()
r9: a__isNatList(cons(N,L)) -> a__and(a__isNat(N),a__isNatList(L))
r10: a__isNatList(take(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r11: a__zeros() -> cons(|0|(),zeros())
r12: a__take(|0|(),IL) -> a__uTake1(a__isNatIList(IL))
r13: a__uTake1(tt()) -> nil()
r14: a__take(s(M),cons(N,IL)) -> a__uTake2(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
r15: a__uTake2(tt(),M,N,IL) -> cons(mark(N),take(M,IL))
r16: a__length(cons(N,L)) -> a__uLength(a__and(a__isNat(N),a__isNatList(L)),L)
r17: a__uLength(tt(),L) -> s(a__length(mark(L)))
r18: mark(and(X1,X2)) -> a__and(mark(X1),mark(X2))
r19: mark(isNatIList(X)) -> a__isNatIList(X)
r20: mark(isNatList(X)) -> a__isNatList(X)
r21: mark(isNat(X)) -> a__isNat(X)
r22: mark(length(X)) -> a__length(mark(X))
r23: mark(zeros()) -> a__zeros()
r24: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r25: mark(uTake1(X)) -> a__uTake1(mark(X))
r26: mark(uTake2(X1,X2,X3,X4)) -> a__uTake2(mark(X1),X2,X3,X4)
r27: mark(uLength(X1,X2)) -> a__uLength(mark(X1),X2)
r28: mark(tt()) -> tt()
r29: mark(|0|()) -> |0|()
r30: mark(s(X)) -> s(mark(X))
r31: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r32: mark(nil()) -> nil()
r33: a__and(X1,X2) -> and(X1,X2)
r34: a__isNatIList(X) -> isNatIList(X)
r35: a__isNatList(X) -> isNatList(X)
r36: a__isNat(X) -> isNat(X)
r37: a__length(X) -> length(X)
r38: a__zeros() -> zeros()
r39: a__take(X1,X2) -> take(X1,X2)
r40: a__uTake1(X) -> uTake1(X)
r41: a__uTake2(X1,X2,X3,X4) -> uTake2(X1,X2,X3,X4)
r42: a__uLength(X1,X2) -> uLength(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

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

and R consists of:

r1: a__and(tt(),T) -> mark(T)
r2: a__isNatIList(IL) -> a__isNatList(IL)
r3: a__isNat(|0|()) -> tt()
r4: a__isNat(s(N)) -> a__isNat(N)
r5: a__isNat(length(L)) -> a__isNatList(L)
r6: a__isNatIList(zeros()) -> tt()
r7: a__isNatIList(cons(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r8: a__isNatList(nil()) -> tt()
r9: a__isNatList(cons(N,L)) -> a__and(a__isNat(N),a__isNatList(L))
r10: a__isNatList(take(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r11: a__zeros() -> cons(|0|(),zeros())
r12: a__take(|0|(),IL) -> a__uTake1(a__isNatIList(IL))
r13: a__uTake1(tt()) -> nil()
r14: a__take(s(M),cons(N,IL)) -> a__uTake2(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
r15: a__uTake2(tt(),M,N,IL) -> cons(mark(N),take(M,IL))
r16: a__length(cons(N,L)) -> a__uLength(a__and(a__isNat(N),a__isNatList(L)),L)
r17: a__uLength(tt(),L) -> s(a__length(mark(L)))
r18: mark(and(X1,X2)) -> a__and(mark(X1),mark(X2))
r19: mark(isNatIList(X)) -> a__isNatIList(X)
r20: mark(isNatList(X)) -> a__isNatList(X)
r21: mark(isNat(X)) -> a__isNat(X)
r22: mark(length(X)) -> a__length(mark(X))
r23: mark(zeros()) -> a__zeros()
r24: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r25: mark(uTake1(X)) -> a__uTake1(mark(X))
r26: mark(uTake2(X1,X2,X3,X4)) -> a__uTake2(mark(X1),X2,X3,X4)
r27: mark(uLength(X1,X2)) -> a__uLength(mark(X1),X2)
r28: mark(tt()) -> tt()
r29: mark(|0|()) -> |0|()
r30: mark(s(X)) -> s(mark(X))
r31: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r32: mark(nil()) -> nil()
r33: a__and(X1,X2) -> and(X1,X2)
r34: a__isNatIList(X) -> isNatIList(X)
r35: a__isNatList(X) -> isNatList(X)
r36: a__isNat(X) -> isNat(X)
r37: a__length(X) -> length(X)
r38: a__zeros() -> zeros()
r39: a__take(X1,X2) -> take(X1,X2)
r40: a__uTake1(X) -> uTake1(X)
r41: a__uTake2(X1,X2,X3,X4) -> uTake2(X1,X2,X3,X4)
r42: a__uLength(X1,X2) -> uLength(X1,X2)

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, r33, r34, r35, r36, r37, r38, r39, r40, r41, r42

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      a__and#_A(x1,x2) = max{43, x2 + 2}
      tt_A = 119
      mark#_A(x1) = max{1, x1}
      s_A(x1) = max{50, x1 + 10}
      and_A(x1,x2) = max{43, x2 + 5}
      mark_A(x1) = max{38, x1}
      a__and_A(x1,x2) = max{43, x2 + 5}
      a__isNatIList_A(x1) = max{98, x1 + 93}
      a__isNatList_A(x1) = max{70, x1 + 68}
      a__isNat_A(x1) = max{137, x1 + 69}
      |0|_A = 0
      length_A(x1) = max{0, x1 - 1}
      zeros_A = 27
      cons_A(x1,x2) = x2 + 10
      nil_A = 57
      take_A(x1,x2) = max{113, x1 + 74, x2 + 36}
      a__zeros_A = 37
      a__take_A(x1,x2) = max{113, x1 + 74, x2 + 36}
      a__uTake1_A(x1) = max{58, x1 - 63}
      a__uTake2_A(x1,x2,x3,x4) = max{123, x1 - 65, x2 + 84, x4 + 46}
      a__length_A(x1) = max{0, x1 - 1}
      a__uLength_A(x1,x2) = max{x1 - 69, x2 + 9}
      isNatIList_A(x1) = max{98, x1 + 93}
      isNatList_A(x1) = max{70, x1 + 68}
      isNat_A(x1) = max{137, x1 + 69}
      uTake1_A(x1) = max{58, x1 - 63}
      uTake2_A(x1,x2,x3,x4) = max{123, x1 - 65, x2 + 84, x4 + 46}
      uLength_A(x1,x2) = max{x1 - 69, x2 + 9}
    
    2. max/plus interpretations on natural numbers:
    
      a__and#_A(x1,x2) = 0
      tt_A = 12
      mark#_A(x1) = 0
      s_A(x1) = 15
      and_A(x1,x2) = 9
      mark_A(x1) = 15
      a__and_A(x1,x2) = 12
      a__isNatIList_A(x1) = 13
      a__isNatList_A(x1) = 11
      a__isNat_A(x1) = 13
      |0|_A = 4
      length_A(x1) = 0
      zeros_A = 15
      cons_A(x1,x2) = 10
      nil_A = 10
      take_A(x1,x2) = 12
      a__zeros_A = 14
      a__take_A(x1,x2) = 13
      a__uTake1_A(x1) = 14
      a__uTake2_A(x1,x2,x3,x4) = 10
      a__length_A(x1) = 15
      a__uLength_A(x1,x2) = 15
      isNatIList_A(x1) = 4
      isNatList_A(x1) = 0
      isNat_A(x1) = 6
      uTake1_A(x1) = 9
      uTake2_A(x1,x2,x3,x4) = 9
      uLength_A(x1,x2) = 0
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__and(tt(),T) -> mark(T)
r2: a__isNatIList(IL) -> a__isNatList(IL)
r3: a__isNat(|0|()) -> tt()
r4: a__isNat(s(N)) -> a__isNat(N)
r5: a__isNat(length(L)) -> a__isNatList(L)
r6: a__isNatIList(zeros()) -> tt()
r7: a__isNatIList(cons(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r8: a__isNatList(nil()) -> tt()
r9: a__isNatList(cons(N,L)) -> a__and(a__isNat(N),a__isNatList(L))
r10: a__isNatList(take(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r11: a__zeros() -> cons(|0|(),zeros())
r12: a__take(|0|(),IL) -> a__uTake1(a__isNatIList(IL))
r13: a__uTake1(tt()) -> nil()
r14: a__take(s(M),cons(N,IL)) -> a__uTake2(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
r15: a__uTake2(tt(),M,N,IL) -> cons(mark(N),take(M,IL))
r16: a__length(cons(N,L)) -> a__uLength(a__and(a__isNat(N),a__isNatList(L)),L)
r17: a__uLength(tt(),L) -> s(a__length(mark(L)))
r18: mark(and(X1,X2)) -> a__and(mark(X1),mark(X2))
r19: mark(isNatIList(X)) -> a__isNatIList(X)
r20: mark(isNatList(X)) -> a__isNatList(X)
r21: mark(isNat(X)) -> a__isNat(X)
r22: mark(length(X)) -> a__length(mark(X))
r23: mark(zeros()) -> a__zeros()
r24: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r25: mark(uTake1(X)) -> a__uTake1(mark(X))
r26: mark(uTake2(X1,X2,X3,X4)) -> a__uTake2(mark(X1),X2,X3,X4)
r27: mark(uLength(X1,X2)) -> a__uLength(mark(X1),X2)
r28: mark(tt()) -> tt()
r29: mark(|0|()) -> |0|()
r30: mark(s(X)) -> s(mark(X))
r31: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r32: mark(nil()) -> nil()
r33: a__and(X1,X2) -> and(X1,X2)
r34: a__isNatIList(X) -> isNatIList(X)
r35: a__isNatList(X) -> isNatList(X)
r36: a__isNat(X) -> isNat(X)
r37: a__length(X) -> length(X)
r38: a__zeros() -> zeros()
r39: a__take(X1,X2) -> take(X1,X2)
r40: a__uTake1(X) -> uTake1(X)
r41: a__uTake2(X1,X2,X3,X4) -> uTake2(X1,X2,X3,X4)
r42: a__uLength(X1,X2) -> uLength(X1,X2)

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

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

and R consists of:

r1: a__and(tt(),T) -> mark(T)
r2: a__isNatIList(IL) -> a__isNatList(IL)
r3: a__isNat(|0|()) -> tt()
r4: a__isNat(s(N)) -> a__isNat(N)
r5: a__isNat(length(L)) -> a__isNatList(L)
r6: a__isNatIList(zeros()) -> tt()
r7: a__isNatIList(cons(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r8: a__isNatList(nil()) -> tt()
r9: a__isNatList(cons(N,L)) -> a__and(a__isNat(N),a__isNatList(L))
r10: a__isNatList(take(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r11: a__zeros() -> cons(|0|(),zeros())
r12: a__take(|0|(),IL) -> a__uTake1(a__isNatIList(IL))
r13: a__uTake1(tt()) -> nil()
r14: a__take(s(M),cons(N,IL)) -> a__uTake2(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
r15: a__uTake2(tt(),M,N,IL) -> cons(mark(N),take(M,IL))
r16: a__length(cons(N,L)) -> a__uLength(a__and(a__isNat(N),a__isNatList(L)),L)
r17: a__uLength(tt(),L) -> s(a__length(mark(L)))
r18: mark(and(X1,X2)) -> a__and(mark(X1),mark(X2))
r19: mark(isNatIList(X)) -> a__isNatIList(X)
r20: mark(isNatList(X)) -> a__isNatList(X)
r21: mark(isNat(X)) -> a__isNat(X)
r22: mark(length(X)) -> a__length(mark(X))
r23: mark(zeros()) -> a__zeros()
r24: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r25: mark(uTake1(X)) -> a__uTake1(mark(X))
r26: mark(uTake2(X1,X2,X3,X4)) -> a__uTake2(mark(X1),X2,X3,X4)
r27: mark(uLength(X1,X2)) -> a__uLength(mark(X1),X2)
r28: mark(tt()) -> tt()
r29: mark(|0|()) -> |0|()
r30: mark(s(X)) -> s(mark(X))
r31: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r32: mark(nil()) -> nil()
r33: a__and(X1,X2) -> and(X1,X2)
r34: a__isNatIList(X) -> isNatIList(X)
r35: a__isNatList(X) -> isNatList(X)
r36: a__isNat(X) -> isNat(X)
r37: a__length(X) -> length(X)
r38: a__zeros() -> zeros()
r39: a__take(X1,X2) -> take(X1,X2)
r40: a__uTake1(X) -> uTake1(X)
r41: a__uTake2(X1,X2,X3,X4) -> uTake2(X1,X2,X3,X4)
r42: a__uLength(X1,X2) -> uLength(X1,X2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      a__isNat#_A(x1) = x1
      s_A(x1) = max{2, x1 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      a__isNat#_A(x1) = x1
      s_A(x1) = 0
    

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__length#(cons(N,L)) -> a__uLength#(a__and(a__isNat(N),a__isNatList(L)),L)
p2: a__uLength#(tt(),L) -> a__length#(mark(L))

and R consists of:

r1: a__and(tt(),T) -> mark(T)
r2: a__isNatIList(IL) -> a__isNatList(IL)
r3: a__isNat(|0|()) -> tt()
r4: a__isNat(s(N)) -> a__isNat(N)
r5: a__isNat(length(L)) -> a__isNatList(L)
r6: a__isNatIList(zeros()) -> tt()
r7: a__isNatIList(cons(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r8: a__isNatList(nil()) -> tt()
r9: a__isNatList(cons(N,L)) -> a__and(a__isNat(N),a__isNatList(L))
r10: a__isNatList(take(N,IL)) -> a__and(a__isNat(N),a__isNatIList(IL))
r11: a__zeros() -> cons(|0|(),zeros())
r12: a__take(|0|(),IL) -> a__uTake1(a__isNatIList(IL))
r13: a__uTake1(tt()) -> nil()
r14: a__take(s(M),cons(N,IL)) -> a__uTake2(a__and(a__isNat(M),a__and(a__isNat(N),a__isNatIList(IL))),M,N,IL)
r15: a__uTake2(tt(),M,N,IL) -> cons(mark(N),take(M,IL))
r16: a__length(cons(N,L)) -> a__uLength(a__and(a__isNat(N),a__isNatList(L)),L)
r17: a__uLength(tt(),L) -> s(a__length(mark(L)))
r18: mark(and(X1,X2)) -> a__and(mark(X1),mark(X2))
r19: mark(isNatIList(X)) -> a__isNatIList(X)
r20: mark(isNatList(X)) -> a__isNatList(X)
r21: mark(isNat(X)) -> a__isNat(X)
r22: mark(length(X)) -> a__length(mark(X))
r23: mark(zeros()) -> a__zeros()
r24: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r25: mark(uTake1(X)) -> a__uTake1(mark(X))
r26: mark(uTake2(X1,X2,X3,X4)) -> a__uTake2(mark(X1),X2,X3,X4)
r27: mark(uLength(X1,X2)) -> a__uLength(mark(X1),X2)
r28: mark(tt()) -> tt()
r29: mark(|0|()) -> |0|()
r30: mark(s(X)) -> s(mark(X))
r31: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r32: mark(nil()) -> nil()
r33: a__and(X1,X2) -> and(X1,X2)
r34: a__isNatIList(X) -> isNatIList(X)
r35: a__isNatList(X) -> isNatList(X)
r36: a__isNat(X) -> isNat(X)
r37: a__length(X) -> length(X)
r38: a__zeros() -> zeros()
r39: a__take(X1,X2) -> take(X1,X2)
r40: a__uTake1(X) -> uTake1(X)
r41: a__uTake2(X1,X2,X3,X4) -> uTake2(X1,X2,X3,X4)
r42: a__uLength(X1,X2) -> uLength(X1,X2)

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, r33, r34, r35, r36, r37, r38, r39, r40, r41, r42

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      a__length#_A(x1) = x1 + 27
      cons_A(x1,x2) = max{x1 - 13, x2 + 16}
      a__uLength#_A(x1,x2) = max{x1 - 3, x2 + 42}
      a__and_A(x1,x2) = max{31, x1 - 26, x2}
      a__isNat_A(x1) = x1 + 42
      a__isNatList_A(x1) = x1 + 45
      tt_A = 84
      mark_A(x1) = max{32, x1}
      a__isNatIList_A(x1) = x1 + 71
      zeros_A = 14
      a__zeros_A = 31
      |0|_A = 43
      a__take_A(x1,x2) = max{62, x1 - 1, x2 + 29}
      a__uTake1_A(x1) = max{41, x1 - 43}
      nil_A = 40
      s_A(x1) = max{80, x1 + 16}
      a__uTake2_A(x1,x2,x3,x4) = max{78, x2 + 15, x3 + 16, x4 + 45}
      take_A(x1,x2) = max{62, x1 - 1, x2 + 29}
      a__length_A(x1) = max{64, x1 + 32}
      a__uLength_A(x1,x2) = max{x1 - 3, x2 + 48}
      isNatIList_A(x1) = x1 + 71
      length_A(x1) = max{64, x1 + 32}
      uTake1_A(x1) = max{41, x1 - 43}
      uTake2_A(x1,x2,x3,x4) = max{78, x2 + 15, x3 + 16, x4 + 45}
      uLength_A(x1,x2) = max{x1 - 3, x2 + 48}
      and_A(x1,x2) = max{1, x1 - 26, x2}
      isNatList_A(x1) = x1 + 45
      isNat_A(x1) = x1 + 42
    
    2. max/plus interpretations on natural numbers:
    
      a__length#_A(x1) = max{0, x1 - 2}
      cons_A(x1,x2) = 2
      a__uLength#_A(x1,x2) = 1
      a__and_A(x1,x2) = 4
      a__isNat_A(x1) = 1
      a__isNatList_A(x1) = 2
      tt_A = 0
      mark_A(x1) = 4
      a__isNatIList_A(x1) = 3
      zeros_A = 0
      a__zeros_A = 1
      |0|_A = 3
      a__take_A(x1,x2) = 4
      a__uTake1_A(x1) = 1
      nil_A = 0
      s_A(x1) = 0
      a__uTake2_A(x1,x2,x3,x4) = 4
      take_A(x1,x2) = 0
      a__length_A(x1) = 3
      a__uLength_A(x1,x2) = 1
      isNatIList_A(x1) = 0
      length_A(x1) = 1
      uTake1_A(x1) = 1
      uTake2_A(x1,x2,x3,x4) = 0
      uLength_A(x1,x2) = 0
      and_A(x1,x2) = 0
      isNatList_A(x1) = 2
      isNat_A(x1) = 0
    

The next rules are strictly ordered:

  p1, p2

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