YES

We show the termination of the TRS R:

  U11(tt(),V2) -> U12(isNat(activate(V2)))
  U12(tt()) -> tt()
  U21(tt()) -> tt()
  U31(tt(),N) -> activate(N)
  U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
  U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
  isNat(n__0()) -> tt()
  isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
  isNat(n__s(V1)) -> U21(isNat(activate(V1)))
  plus(N,|0|()) -> U31(isNat(N),N)
  plus(N,s(M)) -> U41(isNat(M),M,N)
  |0|() -> n__0()
  plus(X1,X2) -> n__plus(X1,X2)
  s(X) -> n__s(X)
  activate(n__0()) -> |0|()
  activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
  activate(n__s(X)) -> s(activate(X))
  activate(X) -> X

-- SCC decomposition.

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

p1: U11#(tt(),V2) -> U12#(isNat(activate(V2)))
p2: U11#(tt(),V2) -> isNat#(activate(V2))
p3: U11#(tt(),V2) -> activate#(V2)
p4: U31#(tt(),N) -> activate#(N)
p5: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p6: U41#(tt(),M,N) -> isNat#(activate(N))
p7: U41#(tt(),M,N) -> activate#(N)
p8: U41#(tt(),M,N) -> activate#(M)
p9: U42#(tt(),M,N) -> s#(plus(activate(N),activate(M)))
p10: U42#(tt(),M,N) -> plus#(activate(N),activate(M))
p11: U42#(tt(),M,N) -> activate#(N)
p12: U42#(tt(),M,N) -> activate#(M)
p13: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p14: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p15: isNat#(n__plus(V1,V2)) -> activate#(V1)
p16: isNat#(n__plus(V1,V2)) -> activate#(V2)
p17: isNat#(n__s(V1)) -> U21#(isNat(activate(V1)))
p18: isNat#(n__s(V1)) -> isNat#(activate(V1))
p19: isNat#(n__s(V1)) -> activate#(V1)
p20: plus#(N,|0|()) -> U31#(isNat(N),N)
p21: plus#(N,|0|()) -> isNat#(N)
p22: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p23: plus#(N,s(M)) -> isNat#(M)
p24: activate#(n__0()) -> |0|#()
p25: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p26: activate#(n__plus(X1,X2)) -> activate#(X1)
p27: activate#(n__plus(X1,X2)) -> activate#(X2)
p28: activate#(n__s(X)) -> s#(activate(X))
p29: activate#(n__s(X)) -> activate#(X)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5, p6, p7, p8, p10, p11, p12, p13, p14, p15, p16, p18, p19, p20, p21, p22, p23, p25, p26, p27, p29}


-- Reduction pair.

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

p1: U11#(tt(),V2) -> isNat#(activate(V2))
p2: isNat#(n__s(V1)) -> activate#(V1)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p7: plus#(N,s(M)) -> isNat#(M)
p8: isNat#(n__s(V1)) -> isNat#(activate(V1))
p9: isNat#(n__plus(V1,V2)) -> activate#(V2)
p10: isNat#(n__plus(V1,V2)) -> activate#(V1)
p11: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p12: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p13: U11#(tt(),V2) -> activate#(V2)
p14: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p15: U41#(tt(),M,N) -> activate#(M)
p16: U41#(tt(),M,N) -> activate#(N)
p17: U41#(tt(),M,N) -> isNat#(activate(N))
p18: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p19: U42#(tt(),M,N) -> activate#(M)
p20: U42#(tt(),M,N) -> activate#(N)
p21: U42#(tt(),M,N) -> plus#(activate(N),activate(M))
p22: plus#(N,|0|()) -> isNat#(N)
p23: plus#(N,|0|()) -> U31#(isNat(N),N)
p24: U31#(tt(),N) -> activate#(N)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          U11#_A(x1,x2) = x1 + x2 + 1
          tt_A() = 15
          isNat#_A(x1) = x1 + 7
          activate_A(x1) = x1
          n__s_A(x1) = x1 + 8
          activate#_A(x1) = x1 + 5
          n__plus_A(x1,x2) = x1 + x2 + 6
          plus#_A(x1,x2) = x1 + x2 + 10
          s_A(x1) = x1 + 8
          isNat_A(x1) = x1 + 2
          U41#_A(x1,x2,x3) = x2 + x3 + 17
          U42#_A(x1,x2,x3) = x2 + x3 + 16
          |0|_A() = 16
          U31#_A(x1,x2) = x2 + 17
          U42_A(x1,x2,x3) = x2 + x3 + 14
          plus_A(x1,x2) = x1 + x2 + 6
          U12_A(x1) = x1 + 14
          U31_A(x1,x2) = x2 + 17
          U41_A(x1,x2,x3) = x2 + x3 + 14
          U11_A(x1,x2) = x1 + x2 + 6
          U21_A(x1) = x1 + 7
          n__0_A() = 16
  
    precedence:
    
      U31 > U41# > U42# > U11# = activate# > plus# > U31# > activate > plus > isNat# > n__plus > |0| > U41 > n__0 > tt = n__s = s = isNat = U42 = U12 = U11 > U21
    
    partial status:
  
      pi(U11#) = [2]
      pi(tt) = []
      pi(isNat#) = [1]
      pi(activate) = [1]
      pi(n__s) = []
      pi(activate#) = []
      pi(n__plus) = [2]
      pi(plus#) = []
      pi(s) = []
      pi(isNat) = []
      pi(U41#) = [3]
      pi(U42#) = [3]
      pi(|0|) = []
      pi(U31#) = []
      pi(U42) = []
      pi(plus) = [1, 2]
      pi(U12) = []
      pi(U31) = []
      pi(U41) = [3]
      pi(U11) = []
      pi(U21) = [1]
      pi(n__0) = []

The next rules are strictly ordered:

  p15

We remove them from the problem.

-- SCC decomposition.

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

p1: U11#(tt(),V2) -> isNat#(activate(V2))
p2: isNat#(n__s(V1)) -> activate#(V1)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p7: plus#(N,s(M)) -> isNat#(M)
p8: isNat#(n__s(V1)) -> isNat#(activate(V1))
p9: isNat#(n__plus(V1,V2)) -> activate#(V2)
p10: isNat#(n__plus(V1,V2)) -> activate#(V1)
p11: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p12: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p13: U11#(tt(),V2) -> activate#(V2)
p14: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p15: U41#(tt(),M,N) -> activate#(N)
p16: U41#(tt(),M,N) -> isNat#(activate(N))
p17: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p18: U42#(tt(),M,N) -> activate#(M)
p19: U42#(tt(),M,N) -> activate#(N)
p20: U42#(tt(),M,N) -> plus#(activate(N),activate(M))
p21: plus#(N,|0|()) -> isNat#(N)
p22: plus#(N,|0|()) -> U31#(isNat(N),N)
p23: U31#(tt(),N) -> activate#(N)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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: U11#(tt(),V2) -> isNat#(activate(V2))
p2: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p3: U11#(tt(),V2) -> activate#(V2)
p4: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p5: plus#(N,|0|()) -> U31#(isNat(N),N)
p6: U31#(tt(),N) -> activate#(N)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: activate#(n__s(X)) -> activate#(X)
p10: plus#(N,|0|()) -> isNat#(N)
p11: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p12: isNat#(n__plus(V1,V2)) -> activate#(V1)
p13: isNat#(n__plus(V1,V2)) -> activate#(V2)
p14: isNat#(n__s(V1)) -> isNat#(activate(V1))
p15: isNat#(n__s(V1)) -> activate#(V1)
p16: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p17: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p18: U42#(tt(),M,N) -> plus#(activate(N),activate(M))
p19: plus#(N,s(M)) -> isNat#(M)
p20: U42#(tt(),M,N) -> activate#(N)
p21: U42#(tt(),M,N) -> activate#(M)
p22: U41#(tt(),M,N) -> isNat#(activate(N))
p23: U41#(tt(),M,N) -> activate#(N)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          U11#_A(x1,x2) = x1 + x2 + 6
          tt_A() = 1
          isNat#_A(x1) = x1 + 6
          activate_A(x1) = x1
          n__plus_A(x1,x2) = x1 + x2 + 11
          isNat_A(x1) = 4
          activate#_A(x1) = x1
          plus#_A(x1,x2) = x1 + x2 + 6
          |0|_A() = 0
          U31#_A(x1,x2) = x1 + x2 + 1
          n__s_A(x1) = x1
          s_A(x1) = x1
          U41#_A(x1,x2,x3) = x2 + x3 + 6
          U42#_A(x1,x2,x3) = x2 + x3 + 6
          U42_A(x1,x2,x3) = x2 + x3 + 11
          plus_A(x1,x2) = x1 + x2 + 11
          U12_A(x1) = 2
          U31_A(x1,x2) = x1 + x2 + 1
          U41_A(x1,x2,x3) = x2 + x3 + 11
          U11_A(x1,x2) = 3
          U21_A(x1) = 2
          n__0_A() = 0
  
    precedence:
    
      U11 > n__plus = plus# = U41# = U42# = plus > U41 > n__s = s = U42 > activate = |0| = n__0 > isNat = U21 > U11# = tt = isNat# = activate# = U31# = U12 = U31
    
    partial status:
  
      pi(U11#) = []
      pi(tt) = []
      pi(isNat#) = []
      pi(activate) = [1]
      pi(n__plus) = []
      pi(isNat) = []
      pi(activate#) = []
      pi(plus#) = []
      pi(|0|) = []
      pi(U31#) = []
      pi(n__s) = []
      pi(s) = []
      pi(U41#) = []
      pi(U42#) = []
      pi(U42) = [2]
      pi(plus) = []
      pi(U12) = []
      pi(U31) = []
      pi(U41) = []
      pi(U11) = []
      pi(U21) = []
      pi(n__0) = []

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: U11#(tt(),V2) -> isNat#(activate(V2))
p2: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p3: U11#(tt(),V2) -> activate#(V2)
p4: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p5: U31#(tt(),N) -> activate#(N)
p6: activate#(n__plus(X1,X2)) -> activate#(X1)
p7: activate#(n__plus(X1,X2)) -> activate#(X2)
p8: activate#(n__s(X)) -> activate#(X)
p9: plus#(N,|0|()) -> isNat#(N)
p10: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p11: isNat#(n__plus(V1,V2)) -> activate#(V1)
p12: isNat#(n__plus(V1,V2)) -> activate#(V2)
p13: isNat#(n__s(V1)) -> isNat#(activate(V1))
p14: isNat#(n__s(V1)) -> activate#(V1)
p15: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p16: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p17: U42#(tt(),M,N) -> plus#(activate(N),activate(M))
p18: plus#(N,s(M)) -> isNat#(M)
p19: U42#(tt(),M,N) -> activate#(N)
p20: U42#(tt(),M,N) -> activate#(M)
p21: U41#(tt(),M,N) -> isNat#(activate(N))
p22: U41#(tt(),M,N) -> activate#(N)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: U11#(tt(),V2) -> isNat#(activate(V2))
p2: isNat#(n__s(V1)) -> activate#(V1)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p7: plus#(N,s(M)) -> isNat#(M)
p8: isNat#(n__s(V1)) -> isNat#(activate(V1))
p9: isNat#(n__plus(V1,V2)) -> activate#(V2)
p10: isNat#(n__plus(V1,V2)) -> activate#(V1)
p11: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p12: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p13: U11#(tt(),V2) -> activate#(V2)
p14: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p15: U41#(tt(),M,N) -> activate#(N)
p16: U41#(tt(),M,N) -> isNat#(activate(N))
p17: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p18: U42#(tt(),M,N) -> activate#(M)
p19: U42#(tt(),M,N) -> activate#(N)
p20: U42#(tt(),M,N) -> plus#(activate(N),activate(M))
p21: plus#(N,|0|()) -> isNat#(N)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          U11#_A(x1,x2) = x2 + 11
          tt_A() = 6
          isNat#_A(x1) = x1 + 11
          activate_A(x1) = x1
          n__s_A(x1) = x1 + 5
          activate#_A(x1) = x1 + 11
          n__plus_A(x1,x2) = x1 + x2
          plus#_A(x1,x2) = x1 + x2 + 10
          s_A(x1) = x1 + 5
          isNat_A(x1) = 9
          U41#_A(x1,x2,x3) = x1 + x2 + x3 + 5
          U42#_A(x1,x2,x3) = x2 + x3 + 11
          |0|_A() = 10
          U42_A(x1,x2,x3) = x2 + x3 + 5
          plus_A(x1,x2) = x1 + x2
          U12_A(x1) = 7
          U31_A(x1,x2) = x2 + 10
          U41_A(x1,x2,x3) = x2 + x3 + 5
          U11_A(x1,x2) = 8
          U21_A(x1) = 9
          n__0_A() = 10
  
    precedence:
    
      activate = plus = U31 > isNat > |0| > U11# = isNat# = activate# = plus# = U41# = U42# = U12 = U11 = U21 > tt > n__s = n__plus = s = U42 = U41 = n__0
    
    partial status:
  
      pi(U11#) = []
      pi(tt) = []
      pi(isNat#) = []
      pi(activate) = []
      pi(n__s) = []
      pi(activate#) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(s) = []
      pi(isNat) = []
      pi(U41#) = []
      pi(U42#) = []
      pi(|0|) = []
      pi(U42) = []
      pi(plus) = []
      pi(U12) = []
      pi(U31) = []
      pi(U41) = []
      pi(U11) = []
      pi(U21) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p20

We remove them from the problem.

-- SCC decomposition.

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

p1: U11#(tt(),V2) -> isNat#(activate(V2))
p2: isNat#(n__s(V1)) -> activate#(V1)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p7: plus#(N,s(M)) -> isNat#(M)
p8: isNat#(n__s(V1)) -> isNat#(activate(V1))
p9: isNat#(n__plus(V1,V2)) -> activate#(V2)
p10: isNat#(n__plus(V1,V2)) -> activate#(V1)
p11: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p12: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p13: U11#(tt(),V2) -> activate#(V2)
p14: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p15: U41#(tt(),M,N) -> activate#(N)
p16: U41#(tt(),M,N) -> isNat#(activate(N))
p17: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p18: U42#(tt(),M,N) -> activate#(M)
p19: U42#(tt(),M,N) -> activate#(N)
p20: plus#(N,|0|()) -> isNat#(N)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: U11#(tt(),V2) -> isNat#(activate(V2))
p2: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p3: U11#(tt(),V2) -> activate#(V2)
p4: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p5: plus#(N,|0|()) -> isNat#(N)
p6: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p7: isNat#(n__plus(V1,V2)) -> activate#(V1)
p8: activate#(n__plus(X1,X2)) -> activate#(X1)
p9: activate#(n__plus(X1,X2)) -> activate#(X2)
p10: activate#(n__s(X)) -> activate#(X)
p11: isNat#(n__plus(V1,V2)) -> activate#(V2)
p12: isNat#(n__s(V1)) -> isNat#(activate(V1))
p13: isNat#(n__s(V1)) -> activate#(V1)
p14: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p15: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p16: U42#(tt(),M,N) -> activate#(N)
p17: U42#(tt(),M,N) -> activate#(M)
p18: U41#(tt(),M,N) -> isNat#(activate(N))
p19: U41#(tt(),M,N) -> activate#(N)
p20: plus#(N,s(M)) -> isNat#(M)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          U11#_A(x1,x2) = x2 + 6
          tt_A() = 4
          isNat#_A(x1) = x1 + 5
          activate_A(x1) = x1
          n__plus_A(x1,x2) = x1 + x2 + 3
          isNat_A(x1) = 9
          activate#_A(x1) = x1
          plus#_A(x1,x2) = x1 + x2 + 2
          |0|_A() = 10
          n__s_A(x1) = x1 + 10
          s_A(x1) = x1 + 10
          U41#_A(x1,x2,x3) = x2 + x3 + 11
          U42#_A(x1,x2,x3) = x2 + x3 + 5
          U42_A(x1,x2,x3) = x2 + x3 + 13
          plus_A(x1,x2) = x1 + x2 + 3
          U12_A(x1) = 5
          U31_A(x1,x2) = x1 + x2 + 2
          U41_A(x1,x2,x3) = x2 + x3 + 13
          U11_A(x1,x2) = 6
          U21_A(x1) = x1
          n__0_A() = 10
  
    precedence:
    
      n__plus = U41# = U42# = plus > U31 = U41 > isNat = n__s = s = U42 > tt = activate = |0| = U12 = U11 = U21 = n__0 > U11# > isNat# = activate# = plus#
    
    partial status:
  
      pi(U11#) = []
      pi(tt) = []
      pi(isNat#) = []
      pi(activate) = [1]
      pi(n__plus) = []
      pi(isNat) = []
      pi(activate#) = []
      pi(plus#) = []
      pi(|0|) = []
      pi(n__s) = []
      pi(s) = []
      pi(U41#) = [3]
      pi(U42#) = []
      pi(U42) = []
      pi(plus) = []
      pi(U12) = []
      pi(U31) = [2]
      pi(U41) = []
      pi(U11) = []
      pi(U21) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p2: U11#(tt(),V2) -> activate#(V2)
p3: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p4: plus#(N,|0|()) -> isNat#(N)
p5: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p6: isNat#(n__plus(V1,V2)) -> activate#(V1)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: activate#(n__s(X)) -> activate#(X)
p10: isNat#(n__plus(V1,V2)) -> activate#(V2)
p11: isNat#(n__s(V1)) -> isNat#(activate(V1))
p12: isNat#(n__s(V1)) -> activate#(V1)
p13: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p14: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p15: U42#(tt(),M,N) -> activate#(N)
p16: U42#(tt(),M,N) -> activate#(M)
p17: U41#(tt(),M,N) -> isNat#(activate(N))
p18: U41#(tt(),M,N) -> activate#(N)
p19: plus#(N,s(M)) -> isNat#(M)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p2: U11#(tt(),V2) -> activate#(V2)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p7: plus#(N,s(M)) -> isNat#(M)
p8: isNat#(n__s(V1)) -> activate#(V1)
p9: isNat#(n__s(V1)) -> isNat#(activate(V1))
p10: isNat#(n__plus(V1,V2)) -> activate#(V2)
p11: isNat#(n__plus(V1,V2)) -> activate#(V1)
p12: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p13: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p14: U41#(tt(),M,N) -> activate#(N)
p15: U41#(tt(),M,N) -> isNat#(activate(N))
p16: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p17: U42#(tt(),M,N) -> activate#(M)
p18: U42#(tt(),M,N) -> activate#(N)
p19: plus#(N,|0|()) -> isNat#(N)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          isNat#_A(x1) = x1 + 6
          n__plus_A(x1,x2) = x1 + x2 + 6
          U11#_A(x1,x2) = x2 + 2
          isNat_A(x1) = 7
          activate_A(x1) = x1
          tt_A() = 3
          activate#_A(x1) = x1 + 1
          n__s_A(x1) = x1 + 13
          plus#_A(x1,x2) = x1 + x2 + 5
          s_A(x1) = x1 + 13
          U41#_A(x1,x2,x3) = x1 + x2 + x3 + 5
          U42#_A(x1,x2,x3) = x2 + x3 + 2
          |0|_A() = 8
          U42_A(x1,x2,x3) = x2 + x3 + 19
          plus_A(x1,x2) = x1 + x2 + 6
          U12_A(x1) = 4
          U31_A(x1,x2) = x2 + 1
          U41_A(x1,x2,x3) = x2 + x3 + 19
          U11_A(x1,x2) = 5
          U21_A(x1) = 4
          n__0_A() = 8
  
    precedence:
    
      activate > U41# = U42# > n__plus = plus# = plus = U21 > U11# > activate# = |0| = U31 = n__0 > isNat# > U12 = U11 > U42 = U41 > isNat = tt = n__s = s
    
    partial status:
  
      pi(isNat#) = [1]
      pi(n__plus) = []
      pi(U11#) = []
      pi(isNat) = []
      pi(activate) = [1]
      pi(tt) = []
      pi(activate#) = [1]
      pi(n__s) = [1]
      pi(plus#) = []
      pi(s) = [1]
      pi(U41#) = [3]
      pi(U42#) = []
      pi(|0|) = []
      pi(U42) = []
      pi(plus) = []
      pi(U12) = []
      pi(U31) = [2]
      pi(U41) = []
      pi(U11) = []
      pi(U21) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p18

We remove them from the problem.

-- SCC decomposition.

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

p1: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p2: U11#(tt(),V2) -> activate#(V2)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p7: plus#(N,s(M)) -> isNat#(M)
p8: isNat#(n__s(V1)) -> activate#(V1)
p9: isNat#(n__s(V1)) -> isNat#(activate(V1))
p10: isNat#(n__plus(V1,V2)) -> activate#(V2)
p11: isNat#(n__plus(V1,V2)) -> activate#(V1)
p12: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p13: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p14: U41#(tt(),M,N) -> activate#(N)
p15: U41#(tt(),M,N) -> isNat#(activate(N))
p16: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p17: U42#(tt(),M,N) -> activate#(M)
p18: plus#(N,|0|()) -> isNat#(N)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p2: U11#(tt(),V2) -> activate#(V2)
p3: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p4: plus#(N,|0|()) -> isNat#(N)
p5: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p6: isNat#(n__plus(V1,V2)) -> activate#(V1)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: activate#(n__s(X)) -> activate#(X)
p10: isNat#(n__plus(V1,V2)) -> activate#(V2)
p11: isNat#(n__s(V1)) -> isNat#(activate(V1))
p12: isNat#(n__s(V1)) -> activate#(V1)
p13: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p14: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p15: U42#(tt(),M,N) -> activate#(M)
p16: U41#(tt(),M,N) -> isNat#(activate(N))
p17: U41#(tt(),M,N) -> activate#(N)
p18: plus#(N,s(M)) -> isNat#(M)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          isNat#_A(x1) = x1 + 7
          n__plus_A(x1,x2) = x1 + x2
          U11#_A(x1,x2) = x1 + x2 + 3
          isNat_A(x1) = x1 + 1
          activate_A(x1) = x1
          tt_A() = 5
          activate#_A(x1) = x1 + 7
          plus#_A(x1,x2) = x1 + x2 + 7
          |0|_A() = 6
          n__s_A(x1) = x1
          s_A(x1) = x1
          U41#_A(x1,x2,x3) = x2 + x3 + 7
          U42#_A(x1,x2,x3) = x1 + x2 + 3
          U42_A(x1,x2,x3) = x2 + x3
          plus_A(x1,x2) = x1 + x2
          U12_A(x1) = x1
          U31_A(x1,x2) = x2 + 5
          U41_A(x1,x2,x3) = x2 + x3
          U11_A(x1,x2) = x1 + x2
          U21_A(x1) = x1
          n__0_A() = 6
  
    precedence:
    
      n__plus = activate = |0| = n__s = s = U42 = plus = U31 = U41 = n__0 > isNat# = U11# = isNat = tt = activate# = plus# = U41# = U42# = U12 = U11 = U21
    
    partial status:
  
      pi(isNat#) = []
      pi(n__plus) = []
      pi(U11#) = []
      pi(isNat) = [1]
      pi(activate) = []
      pi(tt) = []
      pi(activate#) = []
      pi(plus#) = []
      pi(|0|) = []
      pi(n__s) = []
      pi(s) = []
      pi(U41#) = []
      pi(U42#) = []
      pi(U42) = []
      pi(plus) = []
      pi(U12) = []
      pi(U31) = []
      pi(U41) = []
      pi(U11) = []
      pi(U21) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p2: U11#(tt(),V2) -> activate#(V2)
p3: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p4: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> activate#(V1)
p6: activate#(n__plus(X1,X2)) -> activate#(X1)
p7: activate#(n__plus(X1,X2)) -> activate#(X2)
p8: activate#(n__s(X)) -> activate#(X)
p9: isNat#(n__plus(V1,V2)) -> activate#(V2)
p10: isNat#(n__s(V1)) -> isNat#(activate(V1))
p11: isNat#(n__s(V1)) -> activate#(V1)
p12: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p13: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p14: U42#(tt(),M,N) -> activate#(M)
p15: U41#(tt(),M,N) -> isNat#(activate(N))
p16: U41#(tt(),M,N) -> activate#(N)
p17: plus#(N,s(M)) -> isNat#(M)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p2: U11#(tt(),V2) -> activate#(V2)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p7: plus#(N,s(M)) -> isNat#(M)
p8: isNat#(n__s(V1)) -> activate#(V1)
p9: isNat#(n__s(V1)) -> isNat#(activate(V1))
p10: isNat#(n__plus(V1,V2)) -> activate#(V2)
p11: isNat#(n__plus(V1,V2)) -> activate#(V1)
p12: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p13: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p14: U41#(tt(),M,N) -> activate#(N)
p15: U41#(tt(),M,N) -> isNat#(activate(N))
p16: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p17: U42#(tt(),M,N) -> activate#(M)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          isNat#_A(x1) = x1
          n__plus_A(x1,x2) = x1 + x2 + 7
          U11#_A(x1,x2) = x2 + 5
          isNat_A(x1) = 6
          activate_A(x1) = x1
          tt_A() = 3
          activate#_A(x1) = x1 + 4
          n__s_A(x1) = x1 + 11
          plus#_A(x1,x2) = x1 + x2 + 1
          s_A(x1) = x1 + 11
          U41#_A(x1,x2,x3) = x1 + x2 + x3 + 4
          U42#_A(x1,x2,x3) = x2 + 5
          U42_A(x1,x2,x3) = x2 + x3 + 18
          plus_A(x1,x2) = x1 + x2 + 7
          U12_A(x1) = 4
          U31_A(x1,x2) = x2 + 1
          U41_A(x1,x2,x3) = x2 + x3 + 18
          U11_A(x1,x2) = 5
          U21_A(x1) = 5
          |0|_A() = 2
          n__0_A() = 2
  
    precedence:
    
      n__plus = plus > activate > tt > isNat = activate# = plus# = U41# = U42# > U31 > U41 > U12 = U11 = |0| = n__0 > U21 > isNat# = U11# = n__s = s = U42
    
    partial status:
  
      pi(isNat#) = []
      pi(n__plus) = []
      pi(U11#) = []
      pi(isNat) = []
      pi(activate) = [1]
      pi(tt) = []
      pi(activate#) = []
      pi(n__s) = []
      pi(plus#) = [2]
      pi(s) = []
      pi(U41#) = [1, 3]
      pi(U42#) = []
      pi(U42) = []
      pi(plus) = []
      pi(U12) = []
      pi(U31) = [2]
      pi(U41) = [2]
      pi(U11) = []
      pi(U21) = []
      pi(|0|) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p16

We remove them from the problem.

-- SCC decomposition.

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

p1: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p2: U11#(tt(),V2) -> activate#(V2)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p7: plus#(N,s(M)) -> isNat#(M)
p8: isNat#(n__s(V1)) -> activate#(V1)
p9: isNat#(n__s(V1)) -> isNat#(activate(V1))
p10: isNat#(n__plus(V1,V2)) -> activate#(V2)
p11: isNat#(n__plus(V1,V2)) -> activate#(V1)
p12: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p13: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p14: U41#(tt(),M,N) -> activate#(N)
p15: U41#(tt(),M,N) -> isNat#(activate(N))
p16: U42#(tt(),M,N) -> activate#(M)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p2: U11#(tt(),V2) -> activate#(V2)
p3: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p4: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p5: U41#(tt(),M,N) -> isNat#(activate(N))
p6: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p7: isNat#(n__plus(V1,V2)) -> activate#(V1)
p8: activate#(n__plus(X1,X2)) -> activate#(X1)
p9: activate#(n__plus(X1,X2)) -> activate#(X2)
p10: activate#(n__s(X)) -> activate#(X)
p11: isNat#(n__plus(V1,V2)) -> activate#(V2)
p12: isNat#(n__s(V1)) -> isNat#(activate(V1))
p13: isNat#(n__s(V1)) -> activate#(V1)
p14: U41#(tt(),M,N) -> activate#(N)
p15: plus#(N,s(M)) -> isNat#(M)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          isNat#_A(x1) = x1 + 5
          n__plus_A(x1,x2) = x1 + x2 + 4
          U11#_A(x1,x2) = x1 + x2
          isNat_A(x1) = 4
          activate_A(x1) = x1
          tt_A() = 2
          activate#_A(x1) = x1 + 1
          plus#_A(x1,x2) = x1 + x2 + 3
          s_A(x1) = x1 + 6
          U41#_A(x1,x2,x3) = x1 + x3 + 4
          n__s_A(x1) = x1 + 6
          U42_A(x1,x2,x3) = x2 + x3 + 10
          plus_A(x1,x2) = x1 + x2 + 4
          U12_A(x1) = x1
          U31_A(x1,x2) = x2 + 4
          U41_A(x1,x2,x3) = x2 + x3 + 10
          U11_A(x1,x2) = 4
          U21_A(x1) = 3
          |0|_A() = 5
          n__0_A() = 5
  
    precedence:
    
      plus# > n__plus = isNat = U41# = plus = U41 > isNat# = U11 > U11# > activate# > s = n__s = U42 > U12 > |0| = n__0 > activate > tt = U31 = U21
    
    partial status:
  
      pi(isNat#) = [1]
      pi(n__plus) = []
      pi(U11#) = []
      pi(isNat) = []
      pi(activate) = [1]
      pi(tt) = []
      pi(activate#) = [1]
      pi(plus#) = [1, 2]
      pi(s) = []
      pi(U41#) = []
      pi(n__s) = []
      pi(U42) = [3]
      pi(plus) = []
      pi(U12) = []
      pi(U31) = []
      pi(U41) = []
      pi(U11) = []
      pi(U21) = []
      pi(|0|) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: U11#(tt(),V2) -> activate#(V2)
p2: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p3: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p4: U41#(tt(),M,N) -> isNat#(activate(N))
p5: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p6: isNat#(n__plus(V1,V2)) -> activate#(V1)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: activate#(n__s(X)) -> activate#(X)
p10: isNat#(n__plus(V1,V2)) -> activate#(V2)
p11: isNat#(n__s(V1)) -> isNat#(activate(V1))
p12: isNat#(n__s(V1)) -> activate#(V1)
p13: U41#(tt(),M,N) -> activate#(N)
p14: plus#(N,s(M)) -> isNat#(M)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {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: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: activate#(n__s(X)) -> activate#(X)
p5: activate#(n__plus(X1,X2)) -> activate#(X2)
p6: activate#(n__plus(X1,X2)) -> activate#(X1)
p7: isNat#(n__s(V1)) -> isNat#(activate(V1))
p8: isNat#(n__plus(V1,V2)) -> activate#(V2)
p9: isNat#(n__plus(V1,V2)) -> activate#(V1)
p10: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p11: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p12: U41#(tt(),M,N) -> activate#(N)
p13: U41#(tt(),M,N) -> isNat#(activate(N))

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          activate#_A(x1) = x1
          n__plus_A(x1,x2) = x1 + x2 + 5
          plus#_A(x1,x2) = x1 + x2 + 1
          activate_A(x1) = x1
          s_A(x1) = x1 + 5
          isNat#_A(x1) = x1 + 3
          n__s_A(x1) = x1 + 5
          U41#_A(x1,x2,x3) = x3 + 4
          isNat_A(x1) = 4
          tt_A() = 2
          U42_A(x1,x2,x3) = x2 + x3 + 10
          plus_A(x1,x2) = x1 + x2 + 5
          U12_A(x1) = 3
          U31_A(x1,x2) = x2 + 6
          U41_A(x1,x2,x3) = x2 + x3 + 10
          U11_A(x1,x2) = 3
          U21_A(x1) = 3
          |0|_A() = 1
          n__0_A() = 1
  
    precedence:
    
      plus# = U41# = isNat > activate# > U12 = U11 > tt = U21 > n__plus = plus > activate > isNat# > U41 = |0| = n__0 > U42 > s > n__s = U31
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__plus) = []
      pi(plus#) = [2]
      pi(activate) = [1]
      pi(s) = []
      pi(isNat#) = [1]
      pi(n__s) = [1]
      pi(U41#) = []
      pi(isNat) = []
      pi(tt) = []
      pi(U42) = [3]
      pi(plus) = []
      pi(U12) = []
      pi(U31) = []
      pi(U41) = []
      pi(U11) = []
      pi(U21) = []
      pi(|0|) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: plus#(N,s(M)) -> isNat#(M)
p2: isNat#(n__s(V1)) -> activate#(V1)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: isNat#(n__s(V1)) -> isNat#(activate(V1))
p7: isNat#(n__plus(V1,V2)) -> activate#(V2)
p8: isNat#(n__plus(V1,V2)) -> activate#(V1)
p9: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p10: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p11: U41#(tt(),M,N) -> activate#(N)
p12: U41#(tt(),M,N) -> isNat#(activate(N))

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p6, p9}
  {p3, p4, p5}


-- Reduction pair.

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

p1: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p2: isNat#(n__s(V1)) -> isNat#(activate(V1))

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          isNat#_A(x1) = x1 + 1
          n__plus_A(x1,x2) = x1 + x2 + 4
          activate_A(x1) = x1
          n__s_A(x1) = x1 + 3
          U12_A(x1) = 2
          tt_A() = 1
          U11_A(x1,x2) = 3
          isNat_A(x1) = x1 + 2
          U21_A(x1) = x1
          U42_A(x1,x2,x3) = x2 + x3 + 7
          s_A(x1) = x1 + 3
          plus_A(x1,x2) = x1 + x2 + 4
          U31_A(x1,x2) = x2 + 4
          U41_A(x1,x2,x3) = x2 + x3 + 7
          n__0_A() = 3
          |0|_A() = 3
  
    precedence:
    
      isNat# = U11 = isNat > activate = n__s = U12 = tt = U21 = U42 = s = plus = U31 = U41 = n__0 = |0| > n__plus
    
    partial status:
  
      pi(isNat#) = [1]
      pi(n__plus) = [2]
      pi(activate) = [1]
      pi(n__s) = []
      pi(U12) = []
      pi(tt) = []
      pi(U11) = []
      pi(isNat) = [1]
      pi(U21) = [1]
      pi(U42) = []
      pi(s) = []
      pi(plus) = [1, 2]
      pi(U31) = [2]
      pi(U41) = []
      pi(n__0) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          isNat#_A(x1) = x1 + 1
          n__plus_A(x1,x2) = x1 + 10
          activate_A(x1) = x1 + 2
          U12_A(x1) = 5
          tt_A() = 4
          U11_A(x1,x2) = x1 + 2
          isNat_A(x1) = x1 + 1
          U21_A(x1) = 5
          U42_A(x1,x2,x3) = 8
          s_A(x1) = 7
          plus_A(x1,x2) = x1 + 10
          U31_A(x1,x2) = x2 + 3
          U41_A(x1,x2,x3) = 9
          n__0_A() = 5
          n__s_A(x1) = 6
          |0|_A() = 6
  
    precedence:
    
      tt = U21 = U42 = s = n__s > isNat# > n__plus = activate = U11 = plus = U31 = U41 > isNat > U12 = n__0 = |0|
    
    partial status:
  
      pi(isNat#) = [1]
      pi(n__plus) = []
      pi(activate) = [1]
      pi(U12) = []
      pi(tt) = []
      pi(U11) = [1]
      pi(isNat) = [1]
      pi(U21) = []
      pi(U42) = []
      pi(s) = []
      pi(plus) = [1]
      pi(U31) = [2]
      pi(U41) = []
      pi(n__0) = []
      pi(n__s) = []
      pi(|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: activate#(n__s(X)) -> activate#(X)
p2: activate#(n__plus(X1,X2)) -> activate#(X1)
p3: activate#(n__plus(X1,X2)) -> activate#(X2)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          activate#_A(x1) = x1
          n__s_A(x1) = x1 + 1
          n__plus_A(x1,x2) = x1 + x2 + 1
  
    precedence:
    
      activate# = n__s = n__plus
    
    partial status:
  
      pi(activate#) = []
      pi(n__s) = [1]
      pi(n__plus) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__plus(X1,X2)) -> activate#(X1)
p2: activate#(n__plus(X1,X2)) -> activate#(X2)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: activate#(n__plus(X1,X2)) -> activate#(X1)
p2: activate#(n__plus(X1,X2)) -> activate#(X2)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          activate#_A(x1) = x1 + 1
          n__plus_A(x1,x2) = x1 + x2 + 2
  
    precedence:
    
      activate# = n__plus
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__plus) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__plus(X1,X2)) -> activate#(X2)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: activate#(n__plus(X1,X2)) -> activate#(X2)

and R consists of:

r1: U11(tt(),V2) -> U12(isNat(activate(V2)))
r2: U12(tt()) -> tt()
r3: U21(tt()) -> tt()
r4: U31(tt(),N) -> activate(N)
r5: U41(tt(),M,N) -> U42(isNat(activate(N)),activate(M),activate(N))
r6: U42(tt(),M,N) -> s(plus(activate(N),activate(M)))
r7: isNat(n__0()) -> tt()
r8: isNat(n__plus(V1,V2)) -> U11(isNat(activate(V1)),activate(V2))
r9: isNat(n__s(V1)) -> U21(isNat(activate(V1)))
r10: plus(N,|0|()) -> U31(isNat(N),N)
r11: plus(N,s(M)) -> U41(isNat(M),M,N)
r12: |0|() -> n__0()
r13: plus(X1,X2) -> n__plus(X1,X2)
r14: s(X) -> n__s(X)
r15: activate(n__0()) -> |0|()
r16: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          activate#_A(x1) = x1
          n__plus_A(x1,x2) = x1 + x2 + 1
  
    precedence:
    
      activate# = n__plus
    
    partial status:
  
      pi(activate#) = []
      pi(n__plus) = [2]

The next rules are strictly ordered:

  p1

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