YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: U11#(tt(),N) -> activate#(N)
p2: U21#(tt(),M,N) -> s#(plus(activate(N),activate(M)))
p3: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p4: U21#(tt(),M,N) -> activate#(N)
p5: U21#(tt(),M,N) -> activate#(M)
p6: and#(tt(),X) -> activate#(X)
p7: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p8: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p9: isNat#(n__plus(V1,V2)) -> activate#(V1)
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,|0|()) -> U11#(isNat(N),N)
p14: plus#(N,|0|()) -> isNat#(N)
p15: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p16: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p17: plus#(N,s(M)) -> isNat#(M)
p18: activate#(n__0()) -> |0|#()
p19: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p20: activate#(n__isNat(X)) -> isNat#(X)
p21: activate#(n__s(X)) -> s#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: U11#(tt(),N) -> activate#(N)
p2: activate#(n__isNat(X)) -> isNat#(X)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p5: plus#(N,s(M)) -> isNat#(M)
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: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p11: and#(tt(),X) -> activate#(X)
p12: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p13: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p14: U21#(tt(),M,N) -> activate#(M)
p15: U21#(tt(),M,N) -> activate#(N)
p16: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p17: plus#(N,|0|()) -> isNat#(N)
p18: plus#(N,|0|()) -> U11#(isNat(N),N)

and R consists of:

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

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() = 6
          activate#_A(x1) = x1 + 2
          n__isNat_A(x1) = x1
          isNat#_A(x1) = x1 + 2
          n__s_A(x1) = x1 + 2
          n__plus_A(x1,x2) = x1 + x2 + 3
          plus#_A(x1,x2) = x1 + x2
          s_A(x1) = x1 + 2
          activate_A(x1) = x1
          and#_A(x1,x2) = x2 + 2
          isNat_A(x1) = x1
          U21#_A(x1,x2,x3) = x2 + x3 + 2
          and_A(x1,x2) = x2 + 1
          |0|_A() = 7
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 5
          plus_A(x1,x2) = x1 + x2 + 3
          n__0_A() = 7
  
    precedence:
    
      U11# = tt = n__s = n__plus = plus# = s = and# = U21# = U11 = U21 = plus > activate# = isNat# > n__isNat = isNat > |0| = n__0 > activate = and
    
    partial status:
  
      pi(U11#) = [2]
      pi(tt) = []
      pi(activate#) = [1]
      pi(n__isNat) = [1]
      pi(isNat#) = [1]
      pi(n__s) = []
      pi(n__plus) = []
      pi(plus#) = [1, 2]
      pi(s) = []
      pi(activate) = [1]
      pi(and#) = []
      pi(isNat) = [1]
      pi(U21#) = []
      pi(and) = [2]
      pi(|0|) = []
      pi(U11) = [2]
      pi(U21) = []
      pi(plus) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

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

p1: U11#(tt(),N) -> activate#(N)
p2: activate#(n__isNat(X)) -> isNat#(X)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p5: plus#(N,s(M)) -> isNat#(M)
p6: isNat#(n__s(V1)) -> isNat#(activate(V1))
p7: isNat#(n__plus(V1,V2)) -> activate#(V2)
p8: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p9: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p10: and#(tt(),X) -> activate#(X)
p11: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p12: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p13: U21#(tt(),M,N) -> activate#(M)
p14: U21#(tt(),M,N) -> activate#(N)
p15: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p16: plus#(N,|0|()) -> isNat#(N)
p17: plus#(N,|0|()) -> U11#(isNat(N),N)

and R consists of:

r1: U11(tt(),N) -> activate(N)
r2: U21(tt(),M,N) -> s(plus(activate(N),activate(M)))
r3: and(tt(),X) -> activate(X)
r4: isNat(n__0()) -> tt()
r5: isNat(n__plus(V1,V2)) -> and(isNat(activate(V1)),n__isNat(activate(V2)))
r6: isNat(n__s(V1)) -> isNat(activate(V1))
r7: plus(N,|0|()) -> U11(isNat(N),N)
r8: plus(N,s(M)) -> U21(and(isNat(M),n__isNat(N)),M,N)
r9: |0|() -> n__0()
r10: plus(X1,X2) -> n__plus(X1,X2)
r11: isNat(X) -> n__isNat(X)
r12: s(X) -> n__s(X)
r13: activate(n__0()) -> |0|()
r14: activate(n__plus(X1,X2)) -> plus(X1,X2)
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(X)
r17: 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: U11#(tt(),N) -> activate#(N)
p2: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p3: plus#(N,|0|()) -> U11#(isNat(N),N)
p4: plus#(N,|0|()) -> isNat#(N)
p5: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p6: and#(tt(),X) -> activate#(X)
p7: activate#(n__isNat(X)) -> isNat#(X)
p8: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
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)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p13: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p14: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p15: plus#(N,s(M)) -> isNat#(M)
p16: U21#(tt(),M,N) -> activate#(N)
p17: U21#(tt(),M,N) -> activate#(M)

and R consists of:

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          U11#_A(x1,x2) = x2 + 7
          tt_A() = 2
          activate#_A(x1) = x1 + 4
          n__plus_A(x1,x2) = x1 + x2
          plus#_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 6
          isNat_A(x1) = x1 + 1
          isNat#_A(x1) = x1 + 5
          and#_A(x1,x2) = x1 + x2 + 3
          activate_A(x1) = x1
          n__isNat_A(x1) = x1 + 1
          n__s_A(x1) = x1 + 6
          s_A(x1) = x1 + 6
          U21#_A(x1,x2,x3) = x2 + x3 + 7
          and_A(x1,x2) = x2
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 6
          plus_A(x1,x2) = x1 + x2
          n__0_A() = 6
  
    precedence:
    
      U11# = n__plus = U21# = U21 = plus > n__s = s > activate# = plus# = isNat = isNat# = and# = n__isNat > |0| = n__0 > activate = and = U11 > tt
    
    partial status:
  
      pi(U11#) = [2]
      pi(tt) = []
      pi(activate#) = [1]
      pi(n__plus) = [1, 2]
      pi(plus#) = [1, 2]
      pi(|0|) = []
      pi(isNat) = [1]
      pi(isNat#) = [1]
      pi(and#) = []
      pi(activate) = [1]
      pi(n__isNat) = [1]
      pi(n__s) = [1]
      pi(s) = [1]
      pi(U21#) = [2, 3]
      pi(and) = [2]
      pi(U11) = [2]
      pi(U21) = [3]
      pi(plus) = [1, 2]
      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: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,|0|()) -> U11#(isNat(N),N)
p3: plus#(N,|0|()) -> isNat#(N)
p4: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p5: and#(tt(),X) -> activate#(X)
p6: activate#(n__isNat(X)) -> isNat#(X)
p7: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p8: isNat#(n__plus(V1,V2)) -> activate#(V2)
p9: isNat#(n__s(V1)) -> isNat#(activate(V1))
p10: isNat#(n__s(V1)) -> activate#(V1)
p11: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p12: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p13: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p14: plus#(N,s(M)) -> isNat#(M)
p15: U21#(tt(),M,N) -> activate#(N)
p16: U21#(tt(),M,N) -> activate#(M)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: activate#(n__isNat(X)) -> isNat#(X)
p5: isNat#(n__s(V1)) -> isNat#(activate(V1))
p6: isNat#(n__plus(V1,V2)) -> activate#(V2)
p7: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p8: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p9: and#(tt(),X) -> activate#(X)
p10: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p11: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p12: U21#(tt(),M,N) -> activate#(M)
p13: U21#(tt(),M,N) -> activate#(N)
p14: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p15: plus#(N,|0|()) -> isNat#(N)

and R consists of:

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

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 + 4
          plus#_A(x1,x2) = x1 + x2 + 2
          s_A(x1) = x1 + 4
          isNat#_A(x1) = x1 + 1
          n__s_A(x1) = x1 + 4
          n__isNat_A(x1) = x1 + 2
          activate_A(x1) = x1
          and#_A(x1,x2) = x2 + 1
          isNat_A(x1) = x1 + 2
          tt_A() = 0
          U21#_A(x1,x2,x3) = x2 + x3 + 5
          and_A(x1,x2) = x2 + 1
          |0|_A() = 0
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 8
          plus_A(x1,x2) = x1 + x2 + 4
          n__0_A() = 0
  
    precedence:
    
      n__isNat = isNat > plus# = U21# > isNat# > tt > activate# = and# > n__plus = and = U11 = plus > activate > s = n__s = |0| = U21 = n__0
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__plus) = []
      pi(plus#) = []
      pi(s) = []
      pi(isNat#) = [1]
      pi(n__s) = []
      pi(n__isNat) = []
      pi(activate) = [1]
      pi(and#) = [2]
      pi(isNat) = []
      pi(tt) = []
      pi(U21#) = []
      pi(and) = []
      pi(|0|) = []
      pi(U11) = []
      pi(U21) = []
      pi(plus) = []
      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: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: isNat#(n__s(V1)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p7: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p8: and#(tt(),X) -> activate#(X)
p9: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p10: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p11: U21#(tt(),M,N) -> activate#(M)
p12: U21#(tt(),M,N) -> activate#(N)
p13: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p14: plus#(N,|0|()) -> isNat#(N)

and R consists of:

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


-- Reduction pair.

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

p1: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,|0|()) -> isNat#(N)
p3: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p4: and#(tt(),X) -> activate#(X)
p5: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p6: isNat#(n__plus(V1,V2)) -> activate#(V2)
p7: isNat#(n__s(V1)) -> isNat#(activate(V1))
p8: isNat#(n__s(V1)) -> activate#(V1)
p9: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p10: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p11: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p12: plus#(N,s(M)) -> isNat#(M)
p13: U21#(tt(),M,N) -> activate#(N)
p14: U21#(tt(),M,N) -> activate#(M)

and R consists of:

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          activate#_A(x1) = x1 + 3
          n__plus_A(x1,x2) = x1 + x2
          plus#_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 2
          isNat#_A(x1) = x1 + 3
          and#_A(x1,x2) = x2 + 3
          isNat_A(x1) = 0
          activate_A(x1) = x1
          n__isNat_A(x1) = 0
          tt_A() = 0
          n__s_A(x1) = x1
          s_A(x1) = x1
          U21#_A(x1,x2,x3) = x2 + x3 + 3
          and_A(x1,x2) = x2
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3
          plus_A(x1,x2) = x1 + x2
          n__0_A() = 2
  
    precedence:
    
      activate# = n__plus = plus# = |0| = isNat# = and# = isNat = activate = n__isNat = tt = n__s = s = U21# = and = U11 = U21 = plus = n__0
    
    partial status:
  
      pi(activate#) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(|0|) = []
      pi(isNat#) = []
      pi(and#) = []
      pi(isNat) = []
      pi(activate) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(n__s) = []
      pi(s) = []
      pi(U21#) = []
      pi(and) = []
      pi(U11) = []
      pi(U21) = []
      pi(plus) = []
      pi(n__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: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p3: and#(tt(),X) -> activate#(X)
p4: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: isNat#(n__s(V1)) -> isNat#(activate(V1))
p7: isNat#(n__s(V1)) -> activate#(V1)
p8: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p9: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p10: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p11: plus#(N,s(M)) -> isNat#(M)
p12: U21#(tt(),M,N) -> activate#(N)
p13: U21#(tt(),M,N) -> activate#(M)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: isNat#(n__s(V1)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p7: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p8: and#(tt(),X) -> activate#(X)
p9: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p10: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p11: U21#(tt(),M,N) -> activate#(M)
p12: U21#(tt(),M,N) -> activate#(N)
p13: U21#(tt(),M,N) -> plus#(activate(N),activate(M))

and R consists of:

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

Take the 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
          plus#_A(x1,x2) = x1 + x2 + 1
          s_A(x1) = x1
          isNat#_A(x1) = x1 + 1
          n__s_A(x1) = x1
          activate_A(x1) = x1
          and#_A(x1,x2) = x2 + 1
          isNat_A(x1) = 0
          n__isNat_A(x1) = 0
          tt_A() = 0
          U21#_A(x1,x2,x3) = x2 + x3 + 1
          and_A(x1,x2) = x2
          U11_A(x1,x2) = x2 + 2
          U21_A(x1,x2,x3) = x2 + x3 + 2
          plus_A(x1,x2) = x1 + x2 + 2
          |0|_A() = 1
          n__0_A() = 1
  
    precedence:
    
      activate# = n__plus = plus# = s = isNat# = n__s = activate = and# = isNat = n__isNat = tt = U21# = and = U11 = U21 = plus = |0| = n__0
    
    partial status:
  
      pi(activate#) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(s) = []
      pi(isNat#) = []
      pi(n__s) = []
      pi(activate) = []
      pi(and#) = []
      pi(isNat) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(U21#) = []
      pi(and) = []
      pi(U11) = []
      pi(U21) = []
      pi(plus) = []
      pi(|0|) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: isNat#(n__s(V1)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p7: and#(tt(),X) -> activate#(X)
p8: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p9: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p10: U21#(tt(),M,N) -> activate#(M)
p11: U21#(tt(),M,N) -> activate#(N)
p12: U21#(tt(),M,N) -> plus#(activate(N),activate(M))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p3: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p4: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p5: and#(tt(),X) -> activate#(X)
p6: plus#(N,s(M)) -> isNat#(M)
p7: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p8: isNat#(n__plus(V1,V2)) -> activate#(V2)
p9: isNat#(n__s(V1)) -> isNat#(activate(V1))
p10: isNat#(n__s(V1)) -> activate#(V1)
p11: U21#(tt(),M,N) -> activate#(N)
p12: U21#(tt(),M,N) -> activate#(M)

and R consists of:

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

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 + 10
          plus#_A(x1,x2) = x1 + x2 + 9
          s_A(x1) = x1 + 3
          U21#_A(x1,x2,x3) = x2 + x3 + 10
          and_A(x1,x2) = x2 + 7
          isNat_A(x1) = x1 + 4
          n__isNat_A(x1) = x1 + 4
          tt_A() = 14
          activate_A(x1) = x1
          and#_A(x1,x2) = x1 + x2 + 1
          isNat#_A(x1) = x1
          n__s_A(x1) = x1 + 3
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 13
          plus_A(x1,x2) = x1 + x2 + 10
          |0|_A() = 15
          n__0_A() = 15
  
    precedence:
    
      n__plus = plus > s = and = isNat = n__isNat = tt = activate = and# = n__s = U11 = U21 > |0| > n__0 > activate# > plus# = U21# > isNat#
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__plus) = [1]
      pi(plus#) = []
      pi(s) = []
      pi(U21#) = []
      pi(and) = [2]
      pi(isNat) = [1]
      pi(n__isNat) = [1]
      pi(tt) = []
      pi(activate) = [1]
      pi(and#) = []
      pi(isNat#) = [1]
      pi(n__s) = []
      pi(U11) = [2]
      pi(U21) = []
      pi(plus) = [1]
      pi(|0|) = []
      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: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p3: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p4: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p5: plus#(N,s(M)) -> isNat#(M)
p6: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p7: isNat#(n__plus(V1,V2)) -> activate#(V2)
p8: isNat#(n__s(V1)) -> isNat#(activate(V1))
p9: isNat#(n__s(V1)) -> activate#(V1)
p10: U21#(tt(),M,N) -> activate#(N)
p11: U21#(tt(),M,N) -> activate#(M)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p5, p7, p8, p9, p10, p11}


-- Reduction pair.

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

p1: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: isNat#(n__s(V1)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p7: U21#(tt(),M,N) -> activate#(M)
p8: U21#(tt(),M,N) -> activate#(N)
p9: U21#(tt(),M,N) -> plus#(activate(N),activate(M))

and R consists of:

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

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 + 3
          plus#_A(x1,x2) = x1 + x2 + 2
          s_A(x1) = x1 + 3
          isNat#_A(x1) = x1 + 1
          n__s_A(x1) = x1 + 3
          activate_A(x1) = x1
          U21#_A(x1,x2,x3) = x2 + x3 + 4
          and_A(x1,x2) = x2
          isNat_A(x1) = 2
          n__isNat_A(x1) = 2
          tt_A() = 0
          U11_A(x1,x2) = x2 + 3
          U21_A(x1,x2,x3) = x2 + x3 + 6
          plus_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 1
          n__0_A() = 1
  
    precedence:
    
      s = n__s = activate = and = isNat = U11 = U21 = plus > n__isNat > n__plus > activate# = plus# = isNat# = U21# = tt = |0| = n__0
    
    partial status:
  
      pi(activate#) = []
      pi(n__plus) = []
      pi(plus#) = [1]
      pi(s) = []
      pi(isNat#) = []
      pi(n__s) = []
      pi(activate) = []
      pi(U21#) = [3]
      pi(and) = []
      pi(isNat) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(U11) = []
      pi(U21) = []
      pi(plus) = []
      pi(|0|) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p9

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)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: isNat#(n__s(V1)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p7: U21#(tt(),M,N) -> activate#(M)
p8: U21#(tt(),M,N) -> activate#(N)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

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

p1: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p3: U21#(tt(),M,N) -> activate#(N)
p4: U21#(tt(),M,N) -> activate#(M)
p5: plus#(N,s(M)) -> isNat#(M)
p6: isNat#(n__plus(V1,V2)) -> activate#(V2)
p7: isNat#(n__s(V1)) -> isNat#(activate(V1))
p8: isNat#(n__s(V1)) -> activate#(V1)

and R consists of:

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

Take the reduction pair:

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

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p3: U21#(tt(),M,N) -> activate#(M)
p4: plus#(N,s(M)) -> isNat#(M)
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: isNat#(n__s(V1)) -> isNat#(activate(V1))
p7: isNat#(n__s(V1)) -> activate#(V1)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

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

p1: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: isNat#(n__s(V1)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p7: U21#(tt(),M,N) -> activate#(M)

and R consists of:

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

Take the 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 + 5
          plus#_A(x1,x2) = x2 + 1
          s_A(x1) = x1 + 4
          isNat#_A(x1) = x1 + 1
          n__s_A(x1) = x1 + 4
          activate_A(x1) = x1
          U21#_A(x1,x2,x3) = x2 + 5
          and_A(x1,x2) = x2
          isNat_A(x1) = 3
          n__isNat_A(x1) = 3
          tt_A() = 2
          U11_A(x1,x2) = x1 + x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 9
          plus_A(x1,x2) = x1 + x2 + 5
          |0|_A() = 0
          n__0_A() = 0
  
    precedence:
    
      activate = and = isNat > isNat# > s = U11 = U21 = plus > n__plus > plus# > n__s = U21# > activate# > |0| > n__0 > n__isNat = tt
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__plus) = [1, 2]
      pi(plus#) = [2]
      pi(s) = []
      pi(isNat#) = [1]
      pi(n__s) = [1]
      pi(activate) = []
      pi(U21#) = []
      pi(and) = []
      pi(isNat) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(U11) = []
      pi(U21) = []
      pi(plus) = [1, 2]
      pi(|0|) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p7

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)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: isNat#(n__s(V1)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: activate#(n__plus(X1,X2)) -> plus#(X1,X2)
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__plus(V1,V2)) -> activate#(V2)
p4: isNat#(n__s(V1)) -> isNat#(activate(V1))
p5: isNat#(n__s(V1)) -> activate#(V1)

and R consists of:

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

Take the 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 + 4
          plus#_A(x1,x2) = x2 + 3
          s_A(x1) = x1
          isNat#_A(x1) = x1 + 2
          n__s_A(x1) = x1
          activate_A(x1) = x1
          U11_A(x1,x2) = x1 + x2 + 1
          tt_A() = 1
          U21_A(x1,x2,x3) = x2 + x3 + 4
          plus_A(x1,x2) = x1 + x2 + 4
          and_A(x1,x2) = x2
          isNat_A(x1) = 3
          n__0_A() = 5
          n__isNat_A(x1) = 3
          |0|_A() = 5
  
    precedence:
    
      plus# = s = activate = U11 = U21 = plus = and = isNat > n__s > tt = n__0 = |0| > n__plus = n__isNat > isNat# > activate#
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__plus) = [2]
      pi(plus#) = []
      pi(s) = []
      pi(isNat#) = []
      pi(n__s) = []
      pi(activate) = []
      pi(U11) = []
      pi(tt) = []
      pi(U21) = []
      pi(plus) = []
      pi(and) = []
      pi(isNat) = []
      pi(n__0) = []
      pi(n__isNat) = []
      pi(|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__plus(V1,V2)) -> activate#(V2)
p3: isNat#(n__s(V1)) -> isNat#(activate(V1))
p4: isNat#(n__s(V1)) -> activate#(V1)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p3}


-- Reduction pair.

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

p1: isNat#(n__s(V1)) -> isNat#(activate(V1))

and R consists of:

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          isNat#_A(x1) = x1 + 1
          n__s_A(x1) = x1 + 2
          activate_A(x1) = x1
          U11_A(x1,x2) = x2 + 1
          tt_A() = 0
          U21_A(x1,x2,x3) = x2 + x3 + 4
          s_A(x1) = x1 + 2
          plus_A(x1,x2) = x1 + x2 + 2
          and_A(x1,x2) = x1 + x2 + 1
          isNat_A(x1) = x1 + 1
          n__0_A() = 0
          n__plus_A(x1,x2) = x1 + x2 + 2
          n__isNat_A(x1) = x1 + 1
          |0|_A() = 0
  
    precedence:
    
      isNat# > tt = n__0 = |0| > U11 = U21 = plus = n__plus > and = isNat = n__isNat > n__s = s > activate
    
    partial status:
  
      pi(isNat#) = [1]
      pi(n__s) = []
      pi(activate) = [1]
      pi(U11) = []
      pi(tt) = []
      pi(U21) = []
      pi(s) = []
      pi(plus) = []
      pi(and) = [1, 2]
      pi(isNat) = [1]
      pi(n__0) = []
      pi(n__plus) = []
      pi(n__isNat) = [1]
      pi(|0|) = []

The next rules are strictly ordered:

  p1

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