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(activate(X1),activate(X2))
  activate(n__isNat(X)) -> isNat(X)
  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(),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#(activate(X1),activate(X2))
p20: activate#(n__plus(X1,X2)) -> activate#(X1)
p21: activate#(n__plus(X1,X2)) -> activate#(X2)
p22: activate#(n__isNat(X)) -> isNat#(X)
p23: activate#(n__s(X)) -> s#(activate(X))
p24: activate#(n__s(X)) -> activate#(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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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, p21, p22, p24}


-- Reduction pair.

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

p1: U11#(tt(),N) -> activate#(N)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__isNat(X)) -> isNat#(X)
p4: isNat#(n__s(V1)) -> activate#(V1)
p5: activate#(n__plus(X1,X2)) -> activate#(X2)
p6: activate#(n__plus(X1,X2)) -> activate#(X1)
p7: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p8: plus#(N,s(M)) -> isNat#(M)
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: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p14: and#(tt(),X) -> activate#(X)
p15: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p16: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p17: U21#(tt(),M,N) -> activate#(M)
p18: U21#(tt(),M,N) -> activate#(N)
p19: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p20: plus#(N,|0|()) -> isNat#(N)
p21: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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 + 1
          tt_A() = 1
          activate#_A(x1) = x1
          n__s_A(x1) = x1
          n__isNat_A(x1) = x1 + 2
          isNat#_A(x1) = x1 + 1
          n__plus_A(x1,x2) = x1 + x2 + 5
          plus#_A(x1,x2) = x1 + x2 + 4
          activate_A(x1) = x1
          s_A(x1) = x1
          and#_A(x1,x2) = x2 + 1
          isNat_A(x1) = x1 + 2
          U21#_A(x1,x2,x3) = x2 + x3 + 4
          and_A(x1,x2) = x2 + 1
          |0|_A() = 0
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 5
          plus_A(x1,x2) = x1 + x2 + 5
          n__0_A() = 0
  
    precedence:
    
      isNat# > plus# = U21# > U11# > and > |0| = n__0 > tt = n__isNat = isNat > n__plus = plus > U21 > n__s = s > activate = U11 > activate# = and#
    
    partial status:
  
      pi(U11#) = [2]
      pi(tt) = []
      pi(activate#) = []
      pi(n__s) = []
      pi(n__isNat) = []
      pi(isNat#) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(activate) = [1]
      pi(s) = []
      pi(and#) = [2]
      pi(isNat) = []
      pi(U21#) = []
      pi(and) = [2]
      pi(|0|) = []
      pi(U11) = []
      pi(U21) = [3]
      pi(plus) = []
      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: U11#(tt(),N) -> activate#(N)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__isNat(X)) -> isNat#(X)
p4: isNat#(n__s(V1)) -> activate#(V1)
p5: activate#(n__plus(X1,X2)) -> activate#(X2)
p6: activate#(n__plus(X1,X2)) -> activate#(X1)
p7: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p8: plus#(N,s(M)) -> isNat#(M)
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: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p14: and#(tt(),X) -> activate#(X)
p15: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p16: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p17: U21#(tt(),M,N) -> activate#(M)
p18: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p19: plus#(N,|0|()) -> isNat#(N)
p20: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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, p18, p19, p20}


-- 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#(activate(X1),activate(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__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: activate#(n__isNat(X)) -> isNat#(X)
p10: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p11: isNat#(n__plus(V1,V2)) -> activate#(V1)
p12: activate#(n__s(X)) -> activate#(X)
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)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p17: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p18: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p19: plus#(N,s(M)) -> isNat#(M)
p20: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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 + 2
          tt_A() = 3
          activate#_A(x1) = x1 + 1
          n__plus_A(x1,x2) = x1 + x2 + 7
          plus#_A(x1,x2) = x1 + x2 + 2
          activate_A(x1) = x1
          |0|_A() = 0
          isNat_A(x1) = x1 + 4
          isNat#_A(x1) = x1 + 1
          and#_A(x1,x2) = x2 + 2
          n__isNat_A(x1) = x1 + 4
          n__s_A(x1) = x1 + 9
          s_A(x1) = x1 + 9
          U21#_A(x1,x2,x3) = x2 + x3 + 3
          and_A(x1,x2) = x1 + x2
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 16
          plus_A(x1,x2) = x1 + x2 + 7
          n__0_A() = 0
  
    precedence:
    
      activate > U11# = activate# = plus# = |0| = and# = U21# > tt = isNat# > isNat = n__isNat = and > plus > U21 > s = n__0 > n__plus = n__s > U11
    
    partial status:
  
      pi(U11#) = []
      pi(tt) = []
      pi(activate#) = []
      pi(n__plus) = [1]
      pi(plus#) = []
      pi(activate) = []
      pi(|0|) = []
      pi(isNat) = []
      pi(isNat#) = [1]
      pi(and#) = []
      pi(n__isNat) = []
      pi(n__s) = [1]
      pi(s) = []
      pi(U21#) = []
      pi(and) = []
      pi(U11) = []
      pi(U21) = [2]
      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: U11#(tt(),N) -> activate#(N)
p2: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p3: plus#(N,|0|()) -> U11#(isNat(N),N)
p4: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p5: and#(tt(),X) -> activate#(X)
p6: activate#(n__plus(X1,X2)) -> activate#(X1)
p7: activate#(n__plus(X1,X2)) -> activate#(X2)
p8: activate#(n__isNat(X)) -> isNat#(X)
p9: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p10: isNat#(n__plus(V1,V2)) -> activate#(V1)
p11: activate#(n__s(X)) -> activate#(X)
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)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p16: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p17: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p18: plus#(N,s(M)) -> isNat#(M)
p19: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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, p18, p19}


-- Reduction pair.

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

p1: U11#(tt(),N) -> activate#(N)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__isNat(X)) -> isNat#(X)
p4: isNat#(n__s(V1)) -> activate#(V1)
p5: activate#(n__plus(X1,X2)) -> activate#(X2)
p6: activate#(n__plus(X1,X2)) -> activate#(X1)
p7: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p8: plus#(N,s(M)) -> isNat#(M)
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: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p14: and#(tt(),X) -> activate#(X)
p15: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p16: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p17: U21#(tt(),M,N) -> activate#(M)
p18: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p19: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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 + 5
          tt_A() = 2
          activate#_A(x1) = x1 + 4
          n__s_A(x1) = x1 + 2
          n__isNat_A(x1) = x1
          isNat#_A(x1) = x1 + 4
          n__plus_A(x1,x2) = x1 + x2
          plus#_A(x1,x2) = x1 + x2 + 3
          activate_A(x1) = x1
          s_A(x1) = x1 + 2
          and#_A(x1,x2) = x1 + x2 + 3
          isNat_A(x1) = x1
          U21#_A(x1,x2,x3) = x2 + x3 + 5
          and_A(x1,x2) = x1 + x2
          |0|_A() = 3
          U11_A(x1,x2) = x2 + 2
          U21_A(x1,x2,x3) = x2 + x3 + 2
          plus_A(x1,x2) = x1 + x2
          n__0_A() = 3
  
    precedence:
    
      U11 > |0| = n__0 > activate# = isNat# > n__plus = and = U21 = plus > n__isNat = activate = isNat > tt > s > n__s > U11# = plus# = and# = U21#
    
    partial status:
  
      pi(U11#) = []
      pi(tt) = []
      pi(activate#) = []
      pi(n__s) = []
      pi(n__isNat) = [1]
      pi(isNat#) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(activate) = [1]
      pi(s) = []
      pi(and#) = []
      pi(isNat) = [1]
      pi(U21#) = []
      pi(and) = []
      pi(|0|) = []
      pi(U11) = []
      pi(U21) = []
      pi(plus) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p19

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__s(X)) -> activate#(X)
p3: activate#(n__isNat(X)) -> isNat#(X)
p4: isNat#(n__s(V1)) -> activate#(V1)
p5: activate#(n__plus(X1,X2)) -> activate#(X2)
p6: activate#(n__plus(X1,X2)) -> activate#(X1)
p7: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p8: plus#(N,s(M)) -> isNat#(M)
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: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p14: and#(tt(),X) -> activate#(X)
p15: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p16: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p17: U21#(tt(),M,N) -> activate#(M)
p18: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(X))
r17: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {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: activate#(n__s(X)) -> activate#(X)
p2: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p3: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p4: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p5: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p6: and#(tt(),X) -> activate#(X)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: activate#(n__isNat(X)) -> isNat#(X)
p10: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
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)) -> isNat#(M)
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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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 + 5
          n__s_A(x1) = x1
          n__plus_A(x1,x2) = x1 + x2
          plus#_A(x1,x2) = x1 + x2 + 5
          activate_A(x1) = x1
          s_A(x1) = x1
          U21#_A(x1,x2,x3) = x2 + x3 + 5
          and_A(x1,x2) = x2
          isNat_A(x1) = x1 + 1
          n__isNat_A(x1) = x1 + 1
          tt_A() = 4
          and#_A(x1,x2) = x1 + x2 + 2
          isNat#_A(x1) = x1 + 5
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3
          plus_A(x1,x2) = x1 + x2
          |0|_A() = 5
          n__0_A() = 5
  
    precedence:
    
      activate# = n__s = n__plus = plus# = activate = s = U21# = and = isNat = n__isNat = tt = and# = isNat# = U11 = U21 = plus = |0| = n__0
    
    partial status:
  
      pi(activate#) = []
      pi(n__s) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(activate) = []
      pi(s) = []
      pi(U21#) = []
      pi(and) = []
      pi(isNat) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(and#) = []
      pi(isNat#) = []
      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__s(X)) -> activate#(X)
p2: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p3: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p4: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p5: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p6: and#(tt(),X) -> activate#(X)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
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)) -> isNat#(M)
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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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}


-- 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#(X2)
p3: activate#(n__plus(X1,X2)) -> activate#(X1)
p4: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p5: plus#(N,s(M)) -> isNat#(M)
p6: isNat#(n__s(V1)) -> activate#(V1)
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: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p12: and#(tt(),X) -> activate#(X)
p13: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p14: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p15: U21#(tt(),M,N) -> activate#(M)
p16: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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__s_A(x1) = x1
          n__plus_A(x1,x2) = x1 + x2 + 5
          plus#_A(x1,x2) = x2 + 4
          activate_A(x1) = x1
          s_A(x1) = x1
          isNat#_A(x1) = x1 + 4
          and#_A(x1,x2) = x2 + 1
          isNat_A(x1) = 2
          n__isNat_A(x1) = 2
          tt_A() = 1
          U21#_A(x1,x2,x3) = x2 + 4
          and_A(x1,x2) = x2
          U11_A(x1,x2) = x2 + 4
          U21_A(x1,x2,x3) = x2 + x3 + 5
          plus_A(x1,x2) = x1 + x2 + 5
          |0|_A() = 3
          n__0_A() = 3
  
    precedence:
    
      n__s = n__plus = activate = s = isNat = n__isNat = tt = and = U21 = plus > activate# = plus# = isNat# = and# = U21# = U11 = |0| = n__0
    
    partial status:
  
      pi(activate#) = []
      pi(n__s) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(activate) = []
      pi(s) = []
      pi(isNat#) = []
      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:

  p11

We remove them from the problem.

-- SCC decomposition.

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#(X2)
p3: activate#(n__plus(X1,X2)) -> activate#(X1)
p4: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p5: plus#(N,s(M)) -> isNat#(M)
p6: isNat#(n__s(V1)) -> activate#(V1)
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: 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) -> 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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}


-- 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)) -> plus#(activate(X1),activate(X2))
p3: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p4: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p5: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p6: and#(tt(),X) -> activate#(X)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: plus#(N,s(M)) -> isNat#(M)
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: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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 + 6
          n__s_A(x1) = x1
          n__plus_A(x1,x2) = x1 + x2
          plus#_A(x1,x2) = x1 + x2 + 6
          activate_A(x1) = x1
          s_A(x1) = x1
          U21#_A(x1,x2,x3) = x2 + x3 + 6
          and_A(x1,x2) = x2
          isNat_A(x1) = x1 + 2
          n__isNat_A(x1) = x1 + 2
          tt_A() = 7
          and#_A(x1,x2) = x1 + x2 + 1
          isNat#_A(x1) = x1 + 6
          U11_A(x1,x2) = x2 + 5
          U21_A(x1,x2,x3) = x2 + x3
          plus_A(x1,x2) = x1 + x2
          |0|_A() = 6
          n__0_A() = 6
  
    precedence:
    
      activate# = n__s = n__plus = plus# = activate = s = U21# = and = isNat = n__isNat = tt = and# = isNat# = U11 = U21 = plus = |0| = n__0
    
    partial status:
  
      pi(activate#) = []
      pi(n__s) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(activate) = []
      pi(s) = []
      pi(U21#) = []
      pi(and) = []
      pi(isNat) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(and#) = []
      pi(isNat#) = []
      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__s(X)) -> activate#(X)
p2: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p3: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p4: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p5: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p6: activate#(n__plus(X1,X2)) -> activate#(X1)
p7: activate#(n__plus(X1,X2)) -> activate#(X2)
p8: plus#(N,s(M)) -> isNat#(M)
p9: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p10: isNat#(n__plus(V1,V2)) -> activate#(V1)
p11: isNat#(n__plus(V1,V2)) -> activate#(V2)
p12: isNat#(n__s(V1)) -> isNat#(activate(V1))
p13: isNat#(n__s(V1)) -> activate#(V1)
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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(X))
r17: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, 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__s(X)) -> activate#(X)
p2: activate#(n__plus(X1,X2)) -> activate#(X2)
p3: activate#(n__plus(X1,X2)) -> activate#(X1)
p4: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p5: plus#(N,s(M)) -> isNat#(M)
p6: isNat#(n__s(V1)) -> activate#(V1)
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)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p12: U21#(tt(),M,N) -> activate#(M)
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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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__s_A(x1) = x1 + 4
          n__plus_A(x1,x2) = x1 + x2 + 3
          plus#_A(x1,x2) = x2 + 2
          activate_A(x1) = x1
          s_A(x1) = x1 + 4
          isNat#_A(x1) = x1 + 1
          U21#_A(x1,x2,x3) = x2 + 3
          and_A(x1,x2) = x2
          isNat_A(x1) = 2
          n__isNat_A(x1) = 2
          tt_A() = 1
          U11_A(x1,x2) = x2 + 4
          U21_A(x1,x2,x3) = x2 + x3 + 7
          plus_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 3
          n__0_A() = 3
  
    precedence:
    
      activate = and = isNat = n__isNat = tt = |0| = n__0 > isNat# > U21# > n__plus = U21 = plus > activate# = n__s = plus# = s > U11
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__s) = [1]
      pi(n__plus) = [2]
      pi(plus#) = [2]
      pi(activate) = []
      pi(s) = [1]
      pi(isNat#) = [1]
      pi(U21#) = [2]
      pi(and) = []
      pi(isNat) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(U11) = []
      pi(U21) = []
      pi(plus) = [2]
      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__s(X)) -> activate#(X)
p2: activate#(n__plus(X1,X2)) -> activate#(X2)
p3: activate#(n__plus(X1,X2)) -> activate#(X1)
p4: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p5: plus#(N,s(M)) -> isNat#(M)
p6: isNat#(n__s(V1)) -> activate#(V1)
p7: isNat#(n__s(V1)) -> isNat#(activate(V1))
p8: isNat#(n__plus(V1,V2)) -> activate#(V2)
p9: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
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) -> 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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__s(X)) -> activate#(X)
p2: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2))
p3: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p4: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p5: plus#(N,s(M)) -> isNat#(M)
p6: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p7: isNat#(n__plus(V1,V2)) -> activate#(V2)
p8: activate#(n__plus(X1,X2)) -> activate#(X1)
p9: activate#(n__plus(X1,X2)) -> activate#(X2)
p10: isNat#(n__s(V1)) -> isNat#(activate(V1))
p11: isNat#(n__s(V1)) -> activate#(V1)
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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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__s_A(x1) = x1 + 5
          n__plus_A(x1,x2) = x1 + x2
          plus#_A(x1,x2) = x2 + 1
          activate_A(x1) = x1
          s_A(x1) = x1 + 5
          U21#_A(x1,x2,x3) = x2 + 3
          and_A(x1,x2) = x2
          isNat_A(x1) = 4
          n__isNat_A(x1) = 4
          tt_A() = 3
          isNat#_A(x1) = x1 + 3
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 5
          plus_A(x1,x2) = x1 + x2
          |0|_A() = 5
          n__0_A() = 5
  
    precedence:
    
      n__plus = activate = and = isNat = tt = isNat# = plus = |0| = n__0 > activate# > n__s = plus# = s = U21# = n__isNat = U21 > U11
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__s) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(activate) = []
      pi(s) = []
      pi(U21#) = []
      pi(and) = []
      pi(isNat) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(isNat#) = []
      pi(U11) = [2]
      pi(U21) = []
      pi(plus) = []
      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: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(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)) -> isNat#(M)
p5: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p6: isNat#(n__plus(V1,V2)) -> activate#(V2)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: isNat#(n__s(V1)) -> isNat#(activate(V1))
p10: isNat#(n__s(V1)) -> activate#(V1)
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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(X))
r17: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, 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#(activate(X1),activate(X2))
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
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)) -> isNat#(activate(V1))
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) -> 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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
          plus#_A(x1,x2) = x2
          activate_A(x1) = x1
          s_A(x1) = x1 + 3
          isNat#_A(x1) = x1 + 2
          n__s_A(x1) = x1 + 3
          U21#_A(x1,x2,x3) = x2 + 2
          and_A(x1,x2) = x1 + x2
          isNat_A(x1) = 0
          n__isNat_A(x1) = 0
          tt_A() = 0
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 3
          plus_A(x1,x2) = x1 + x2
          |0|_A() = 2
          n__0_A() = 2
  
    precedence:
    
      activate# = n__plus = plus# = activate = s = isNat# = n__s = U21# = and = isNat = n__isNat = tt = U11 = U21 = plus = |0| = n__0
    
    partial status:
  
      pi(activate#) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(activate) = []
      pi(s) = []
      pi(isNat#) = []
      pi(n__s) = []
      pi(U21#) = []
      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:

  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#(activate(X1),activate(X2))
p2: plus#(N,s(M)) -> isNat#(M)
p3: isNat#(n__s(V1)) -> activate#(V1)
p4: activate#(n__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: isNat#(n__plus(V1,V2)) -> activate#(V2)
p7: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p8: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p9: U21#(tt(),M,N) -> activate#(M)
p10: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(X))
r17: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- 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)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p3: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p4: plus#(N,s(M)) -> isNat#(M)
p5: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p6: isNat#(n__plus(V1,V2)) -> activate#(V2)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: isNat#(n__s(V1)) -> activate#(V1)
p10: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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 + 4
          n__plus_A(x1,x2) = x1 + x2 + 5
          plus#_A(x1,x2) = x1 + x2 + 2
          activate_A(x1) = x1
          s_A(x1) = x1 + 10
          U21#_A(x1,x2,x3) = x2 + x3 + 11
          and_A(x1,x2) = x2 + 5
          isNat_A(x1) = x1 + 4
          n__isNat_A(x1) = x1 + 4
          tt_A() = 3
          isNat#_A(x1) = x1 + 1
          n__s_A(x1) = x1 + 10
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 15
          plus_A(x1,x2) = x1 + x2 + 5
          |0|_A() = 0
          n__0_A() = 0
  
    precedence:
    
      activate = plus > |0| > n__0 > U21 > s > n__s > isNat > n__plus = and = n__isNat > U11 > tt > activate# = U21# = isNat# > plus#
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__plus) = []
      pi(plus#) = [2]
      pi(activate) = []
      pi(s) = []
      pi(U21#) = [2, 3]
      pi(and) = []
      pi(isNat) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(isNat#) = [1]
      pi(n__s) = []
      pi(U11) = [2]
      pi(U21) = []
      pi(plus) = []
      pi(|0|) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p10

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#(activate(X1),activate(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)) -> isNat#(M)
p5: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p6: isNat#(n__plus(V1,V2)) -> activate#(V2)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)
p9: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(X))
r17: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- 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__plus(X1,X2)) -> activate#(X2)
p5: activate#(n__plus(X1,X2)) -> activate#(X1)
p6: isNat#(n__plus(V1,V2)) -> activate#(V2)
p7: isNat#(n__plus(V1,V2)) -> isNat#(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))

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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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 + 12
          plus#_A(x1,x2) = x2 + 2
          activate_A(x1) = x1
          s_A(x1) = x1
          isNat#_A(x1) = x1 + 1
          n__s_A(x1) = x1
          U21#_A(x1,x2,x3) = x2 + 2
          and_A(x1,x2) = x1 + x2 + 1
          isNat_A(x1) = x1 + 5
          n__isNat_A(x1) = x1 + 5
          tt_A() = 3
          U11_A(x1,x2) = x2 + 5
          U21_A(x1,x2,x3) = x2 + x3 + 12
          plus_A(x1,x2) = x1 + x2 + 12
          |0|_A() = 4
          n__0_A() = 4
  
    precedence:
    
      activate# > plus# = U21# = isNat = n__isNat > isNat# > activate = and > tt > n__plus = plus > U21 > |0| = n__0 > s = n__s = U11
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__plus) = [2]
      pi(plus#) = []
      pi(activate) = [1]
      pi(s) = []
      pi(isNat#) = [1]
      pi(n__s) = []
      pi(U21#) = []
      pi(and) = [2]
      pi(isNat) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(U11) = []
      pi(U21) = [3]
      pi(plus) = [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#(activate(X1),activate(X2))
p2: plus#(N,s(M)) -> isNat#(M)
p3: activate#(n__plus(X1,X2)) -> activate#(X2)
p4: activate#(n__plus(X1,X2)) -> activate#(X1)
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p7: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p8: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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#(activate(X1),activate(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)) -> isNat#(M)
p5: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p6: isNat#(n__plus(V1,V2)) -> activate#(V2)
p7: activate#(n__plus(X1,X2)) -> activate#(X1)
p8: activate#(n__plus(X1,X2)) -> activate#(X2)

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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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 + 8
          n__plus_A(x1,x2) = x1 + x2 + 8
          plus#_A(x1,x2) = x2
          activate_A(x1) = x1
          s_A(x1) = x1 + 2
          U21#_A(x1,x2,x3) = x2
          and_A(x1,x2) = x1 + x2 + 1
          isNat_A(x1) = x1 + 3
          n__isNat_A(x1) = x1 + 3
          tt_A() = 1
          isNat#_A(x1) = x1 + 1
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 10
          plus_A(x1,x2) = x1 + x2 + 8
          |0|_A() = 0
          n__0_A() = 0
          n__s_A(x1) = x1 + 2
  
    precedence:
    
      U21# = U11 > n__plus = plus > and > activate > |0| > n__0 > s = isNat = n__isNat = tt = U21 = n__s > activate# = plus# = isNat#
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__plus) = []
      pi(plus#) = []
      pi(activate) = [1]
      pi(s) = []
      pi(U21#) = [2]
      pi(and) = [2]
      pi(isNat) = [1]
      pi(n__isNat) = []
      pi(tt) = []
      pi(isNat#) = [1]
      pi(U11) = [2]
      pi(U21) = []
      pi(plus) = []
      pi(|0|) = []
      pi(n__0) = []
      pi(n__s) = []

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#(activate(X1),activate(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: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> activate#(V2)
p6: activate#(n__plus(X1,X2)) -> activate#(X1)
p7: activate#(n__plus(X1,X2)) -> activate#(X2)

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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(X))
r17: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- 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(),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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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__plus_A(x1,x2) = x1 + 10
          activate_A(x1) = x1 + 4
          U11_A(x1,x2) = x2 + 8
          tt_A() = 5
          U21_A(x1,x2,x3) = 9
          s_A(x1) = 8
          plus_A(x1,x2) = x1 + 10
          and_A(x1,x2) = x2 + 4
          isNat_A(x1) = 6
          n__0_A() = 4
          n__isNat_A(x1) = 2
          n__s_A(x1) = 7
          |0|_A() = 7
  
    precedence:
    
      activate = U11 = tt = and = isNat > isNat# = U21 = s = plus = n__0 = n__isNat > n__plus = n__s = |0|
    
    partial status:
  
      pi(isNat#) = [1]
      pi(n__plus) = [1]
      pi(activate) = []
      pi(U11) = []
      pi(tt) = []
      pi(U21) = []
      pi(s) = []
      pi(plus) = [1]
      pi(and) = []
      pi(isNat) = []
      pi(n__0) = []
      pi(n__isNat) = []
      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__plus(X1,X2)) -> activate#(X2)
p2: activate#(n__plus(X1,X2)) -> activate#(X1)

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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(X))
r17: 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 + 1
          n__plus_A(x1,x2) = x1 + x2 + 2
  
    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.

-- SCC decomposition.

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

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

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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(X))
r17: 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#(X1)

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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(X))
r17: 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.

-- Reduction pair.

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

p1: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p2: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(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:
          plus#_A(x1,x2) = x2 + 2
          s_A(x1) = x1 + 5
          U21#_A(x1,x2,x3) = x1 + x2 + 2
          and_A(x1,x2) = x2
          isNat_A(x1) = 2
          n__isNat_A(x1) = 2
          tt_A() = 1
          activate_A(x1) = x1
          U11_A(x1,x2) = x2 + 1
          U21_A(x1,x2,x3) = x2 + x3 + 8
          plus_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 3
          n__0_A() = 3
          n__plus_A(x1,x2) = x1 + x2 + 3
          n__s_A(x1) = x1 + 5
  
    precedence:
    
      and = isNat = activate > U11 > plus > U21 > s = U21# > plus# = n__isNat > |0| = n__0 = n__plus = n__s > tt
    
    partial status:
  
      pi(plus#) = [2]
      pi(s) = [1]
      pi(U21#) = []
      pi(and) = []
      pi(isNat) = []
      pi(n__isNat) = []
      pi(tt) = []
      pi(activate) = []
      pi(U11) = [2]
      pi(U21) = []
      pi(plus) = []
      pi(|0|) = []
      pi(n__0) = []
      pi(n__plus) = [2]
      pi(n__s) = [1]

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: 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(activate(X1),activate(X2))
r15: activate(n__isNat(X)) -> isNat(X)
r16: activate(n__s(X)) -> s(activate(X))
r17: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  (no SCCs)