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^2
        order: lexicographic order
        interpretations:
          U11#_A(x1,x2) = ((0,0),(1,0)) x2 + (6,6)
          tt_A() = (1,4)
          activate#_A(x1) = ((0,0),(1,0)) x1 + (6,5)
          n__isNat_A(x1) = ((1,0),(0,0)) x1 + (3,5)
          isNat#_A(x1) = ((0,0),(1,0)) x1 + (6,1)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (5,0)
          n__plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (7,8)
          plus#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (6,7)
          s_A(x1) = ((1,0),(0,0)) x1 + (5,7)
          activate_A(x1) = ((1,0),(1,1)) x1 + (0,3)
          and#_A(x1,x2) = ((0,0),(1,0)) x2 + (6,5)
          isNat_A(x1) = ((1,0),(0,0)) x1 + (3,6)
          U21#_A(x1,x2,x3) = ((0,0),(1,0)) x2 + ((0,0),(1,0)) x3 + (6,8)
          and_A(x1,x2) = ((1,0),(0,0)) x2 + (1,4)
          |0|_A() = (0,7)
          U11_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (1,1)
          U21_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (12,8)
          plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (7,9)
          n__0_A() = (0,6)
  
    precedence:
    
      U11# = activate# = isNat# = n__s = n__plus = plus# = s = activate = and# = isNat = U21# = |0| = U11 = U21 = plus > and > tt = n__isNat = n__0
    
    partial status:
  
      pi(U11#) = []
      pi(tt) = []
      pi(activate#) = []
      pi(n__isNat) = []
      pi(isNat#) = []
      pi(n__s) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(s) = []
      pi(activate) = []
      pi(and#) = []
      pi(isNat) = []
      pi(U21#) = []
      pi(and) = []
      pi(|0|) = []
      pi(U11) = []
      pi(U21) = []
      pi(plus) = []
      pi(n__0) = []

The next rules are strictly ordered:

  p14

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)) -> 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#(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#(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,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N)
p14: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p15: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N))
p16: plus#(N,s(M)) -> isNat#(M)
p17: 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 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^2
        order: lexicographic order
        interpretations:
          U11#_A(x1,x2) = ((1,0),(0,0)) x2 + (12,16)
          tt_A() = (10,33)
          activate#_A(x1) = ((1,0),(0,0)) x1 + (9,9)
          n__plus_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (0,20)
          plus#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (8,9)
          |0|_A() = (11,17)
          isNat_A(x1) = x1 + (6,18)
          isNat#_A(x1) = ((1,0),(0,0)) x1 + (13,9)
          and#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2
          activate_A(x1) = x1 + (0,11)
          n__isNat_A(x1) = x1 + (6,8)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (11,10)
          s_A(x1) = ((1,0),(0,0)) x1 + (11,19)
          U21#_A(x1,x2,x3) = ((1,0),(1,0)) x2 + ((1,0),(0,0)) x3 + (12,21)
          and_A(x1,x2) = x2 + (0,18)
          U11_A(x1,x2) = ((1,0),(0,0)) x2 + (1,18)
          U21_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (11,33)
          plus_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (0,21)
          n__0_A() = (11,16)
  
    precedence:
    
      activate# = plus# = isNat# = and# = U21# > U11# > n__s = s > n__plus = activate = U11 = U21 = plus > isNat = n__isNat = and > tt = |0| = n__0
    
    partial status:
  
      pi(U11#) = []
      pi(tt) = []
      pi(activate#) = []
      pi(n__plus) = []
      pi(plus#) = []
      pi(|0|) = []
      pi(isNat) = [1]
      pi(isNat#) = []
      pi(and#) = []
      pi(activate) = []
      pi(n__isNat) = [1]
      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: U11#(tt(),N) -> activate#(N)
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#(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)

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:

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


-- Reduction pair.

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

p1: U21#(tt(),M,N) -> plus#(activate(N),activate(M))
p2: 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 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^2
        order: lexicographic order
        interpretations:
          U21#_A(x1,x2,x3) = ((1,0),(0,0)) x2 + (5,0)
          tt_A() = (3,2)
          plus#_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (4,12)
          activate_A(x1) = x1 + (0,8)
          s_A(x1) = ((1,0),(0,0)) x1 + (9,10)
          and_A(x1,x2) = ((1,0),(0,0)) x2 + (6,13)
          isNat_A(x1) = ((1,0),(0,0)) x1 + (8,15)
          n__isNat_A(x1) = ((1,0),(0,0)) x1 + (8,11)
          U11_A(x1,x2) = ((1,0),(1,1)) x2 + (3,9)
          U21_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(1,0)) x3 + (16,11)
          plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (7,14)
          |0|_A() = (2,2)
          n__0_A() = (2,1)
          n__plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (7,7)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (9,9)
  
    precedence:
    
      U21# > activate = s = isNat = n__isNat > plus# = U11 > tt = U21 = plus > and = |0| = n__0 = n__plus = n__s
    
    partial status:
  
      pi(U21#) = []
      pi(tt) = []
      pi(plus#) = []
      pi(activate) = [1]
      pi(s) = []
      pi(and) = []
      pi(isNat) = []
      pi(n__isNat) = []
      pi(U11) = []
      pi(U21) = []
      pi(plus) = [1]
      pi(|0|) = []
      pi(n__0) = []
      pi(n__plus) = []
      pi(n__s) = []

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)) -> 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:

  (no SCCs)

-- Reduction pair.

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

p1: activate#(n__isNat(X)) -> isNat#(X)
p2: isNat#(n__s(V1)) -> activate#(V1)
p3: isNat#(n__s(V1)) -> isNat#(activate(V1))
p4: isNat#(n__plus(V1,V2)) -> activate#(V2)
p5: isNat#(n__plus(V1,V2)) -> activate#(V1)
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)

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^2
        order: lexicographic order
        interpretations:
          activate#_A(x1) = ((1,0),(1,0)) x1 + (17,9)
          n__isNat_A(x1) = ((1,0),(0,0)) x1 + (3,0)
          isNat#_A(x1) = ((1,0),(1,0)) x1 + (4,1)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (18,19)
          activate_A(x1) = ((1,0),(1,1)) x1 + (0,20)
          n__plus_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (13,0)
          and#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (6,10)
          isNat_A(x1) = ((1,0),(0,0)) x1 + (3,22)
          tt_A() = (12,21)
          U11_A(x1,x2) = ((1,0),(0,0)) x2 + (11,21)
          U21_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (31,24)
          s_A(x1) = ((1,0),(0,0)) x1 + (18,23)
          plus_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (13,25)
          and_A(x1,x2) = ((1,0),(0,0)) x2 + (9,21)
          |0|_A() = (10,1)
          n__0_A() = (10,0)
  
    precedence:
    
      n__s = activate = isNat = tt = U11 = U21 = s = plus = and > n__plus > activate# = n__isNat = isNat# = and# > |0| = n__0
    
    partial status:
  
      pi(activate#) = []
      pi(n__isNat) = []
      pi(isNat#) = []
      pi(n__s) = []
      pi(activate) = []
      pi(n__plus) = []
      pi(and#) = []
      pi(isNat) = []
      pi(tt) = []
      pi(U11) = []
      pi(U21) = []
      pi(s) = []
      pi(plus) = []
      pi(and) = []
      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__isNat(X)) -> isNat#(X)
p2: isNat#(n__s(V1)) -> activate#(V1)
p3: isNat#(n__plus(V1,V2)) -> activate#(V2)
p4: isNat#(n__plus(V1,V2)) -> activate#(V1)
p5: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p6: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p7: and#(tt(),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(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__isNat(X)) -> isNat#(X)
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#(V1)
p6: isNat#(n__plus(V1,V2)) -> activate#(V2)
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 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^2
        order: lexicographic order
        interpretations:
          activate#_A(x1) = ((0,0),(1,0)) x1 + (10,11)
          n__isNat_A(x1) = ((1,0),(0,0)) x1 + (1,6)
          isNat#_A(x1) = ((0,0),(1,0)) x1 + (10,12)
          n__plus_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (0,6)
          and#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (10,5)
          isNat_A(x1) = ((1,0),(1,0)) x1 + (1,13)
          activate_A(x1) = ((1,0),(1,1)) x1 + (0,11)
          tt_A() = (8,1)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (9,12)
          U11_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (1,2)
          U21_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (9,15)
          s_A(x1) = ((1,0),(0,0)) x1 + (9,14)
          plus_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (0,16)
          and_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (0,4)
          |0|_A() = (9,2)
          n__0_A() = (9,2)
  
    precedence:
    
      activate# = isNat# = and# = isNat = activate = tt = n__s = U21 = s = plus = and > n__plus = U11 > |0| = n__0 > n__isNat
    
    partial status:
  
      pi(activate#) = []
      pi(n__isNat) = []
      pi(isNat#) = []
      pi(n__plus) = []
      pi(and#) = []
      pi(isNat) = []
      pi(activate) = []
      pi(tt) = []
      pi(n__s) = []
      pi(U11) = []
      pi(U21) = []
      pi(s) = []
      pi(plus) = []
      pi(and) = []
      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__isNat(X)) -> isNat#(X)
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)) -> 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}


-- Reduction pair.

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

p1: activate#(n__isNat(X)) -> isNat#(X)
p2: isNat#(n__s(V1)) -> activate#(V1)
p3: isNat#(n__plus(V1,V2)) -> activate#(V2)
p4: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p5: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2)))
p6: and#(tt(),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(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^2
        order: lexicographic order
        interpretations:
          activate#_A(x1) = ((0,0),(1,0)) x1 + (10,7)
          n__isNat_A(x1) = x1 + (1,6)
          isNat#_A(x1) = ((0,0),(1,0)) x1 + (10,1)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (6,6)
          n__plus_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (9,0)
          activate_A(x1) = ((1,0),(1,1)) x1 + (0,5)
          and#_A(x1,x2) = ((0,0),(1,0)) x2 + (10,8)
          isNat_A(x1) = x1 + (1,12)
          tt_A() = (0,0)
          U11_A(x1,x2) = ((1,0),(0,0)) x2 + (1,3)
          U21_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (15,14)
          s_A(x1) = ((1,0),(0,0)) x1 + (6,13)
          plus_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (9,13)
          and_A(x1,x2) = ((1,0),(1,1)) x2 + (0,5)
          |0|_A() = (0,2)
          n__0_A() = (0,1)
  
    precedence:
    
      activate = U11 = U21 = plus = and > isNat = tt > activate# = isNat# = and# > n__plus = |0| > n__isNat > n__s = s = n__0
    
    partial status:
  
      pi(activate#) = []
      pi(n__isNat) = []
      pi(isNat#) = []
      pi(n__s) = []
      pi(n__plus) = []
      pi(activate) = []
      pi(and#) = []
      pi(isNat) = [1]
      pi(tt) = []
      pi(U11) = []
      pi(U21) = []
      pi(s) = []
      pi(plus) = []
      pi(and) = []
      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__isNat(X)) -> isNat#(X)
p2: isNat#(n__s(V1)) -> activate#(V1)
p3: isNat#(n__plus(V1,V2)) -> activate#(V2)
p4: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p5: and#(tt(),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(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}


-- Reduction pair.

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

p1: activate#(n__isNat(X)) -> isNat#(X)
p2: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p3: isNat#(n__plus(V1,V2)) -> activate#(V2)
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 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^2
        order: lexicographic order
        interpretations:
          activate#_A(x1) = ((1,0),(0,0)) x1 + (8,0)
          n__isNat_A(x1) = x1 + (9,10)
          isNat#_A(x1) = x1
          n__plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (20,0)
          activate_A(x1) = x1 + (0,3)
          n__s_A(x1) = x1 + (8,0)
          U11_A(x1,x2) = ((1,0),(1,1)) x2 + (1,14)
          tt_A() = (4,1)
          U21_A(x1,x2,x3) = ((1,0),(1,1)) x2 + ((1,0),(1,1)) x3 + (28,10)
          s_A(x1) = x1 + (8,1)
          plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (20,2)
          and_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (1,0)
          isNat_A(x1) = x1 + (9,12)
          n__0_A() = (0,11)
          |0|_A() = (0,13)
  
    precedence:
    
      and > n__isNat = activate = U11 = tt = U21 = s = plus = isNat > n__s > activate# > isNat# = |0| > n__0 > n__plus
    
    partial status:
  
      pi(activate#) = []
      pi(n__isNat) = []
      pi(isNat#) = [1]
      pi(n__plus) = []
      pi(activate) = [1]
      pi(n__s) = [1]
      pi(U11) = [2]
      pi(tt) = []
      pi(U21) = []
      pi(s) = []
      pi(plus) = [1, 2]
      pi(and) = []
      pi(isNat) = [1]
      pi(n__0) = []
      pi(|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__isNat(X)) -> isNat#(X)
p2: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1))
p3: isNat#(n__plus(V1,V2)) -> activate#(V2)

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}


-- Reduction pair.

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

p1: activate#(n__isNat(X)) -> isNat#(X)
p2: isNat#(n__plus(V1,V2)) -> activate#(V2)
p3: 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(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^2
        order: lexicographic order
        interpretations:
          activate#_A(x1) = ((1,0),(0,0)) x1 + (0,9)
          n__isNat_A(x1) = x1 + (3,6)
          isNat#_A(x1) = ((1,0),(0,0)) x1 + (2,9)
          n__plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (0,8)
          activate_A(x1) = x1 + (0,3)
          U11_A(x1,x2) = x2 + (0,4)
          tt_A() = (1,1)
          U21_A(x1,x2,x3) = ((1,0),(1,0)) x2 + x3 + (2,14)
          s_A(x1) = ((1,0),(0,0)) x1 + (2,13)
          plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (0,9)
          and_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (0,3)
          isNat_A(x1) = x1 + (3,8)
          n__0_A() = (0,0)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (2,12)
          |0|_A() = (0,0)
  
    precedence:
    
      n__isNat = activate = U11 = tt = U21 = plus = and = isNat > isNat# = s > activate# = |0| > n__0 > n__plus = n__s
    
    partial status:
  
      pi(activate#) = []
      pi(n__isNat) = []
      pi(isNat#) = []
      pi(n__plus) = []
      pi(activate) = []
      pi(U11) = []
      pi(tt) = []
      pi(U21) = []
      pi(s) = []
      pi(plus) = []
      pi(and) = []
      pi(isNat) = []
      pi(n__0) = []
      pi(n__s) = []
      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: isNat#(n__plus(V1,V2)) -> activate#(V2)
p2: 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(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:

  {p2}


-- 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(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^2
        order: lexicographic order
        interpretations:
          isNat#_A(x1) = ((1,0),(1,1)) x1 + (1,0)
          n__plus_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (5,3)
          activate_A(x1) = ((1,0),(1,1)) x1 + (0,2)
          U11_A(x1,x2) = ((1,0),(1,1)) x2 + (1,3)
          tt_A() = (3,3)
          U21_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (5,14)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,13)
          plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (5,4)
          and_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2
          isNat_A(x1) = ((1,0),(0,0)) x1 + (4,10)
          n__0_A() = (0,10)
          n__isNat_A(x1) = ((1,0),(0,0)) x1 + (4,4)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (0,12)
          |0|_A() = (0,11)
  
    precedence:
    
      s = n__s > U11 > n__plus = activate = U21 = plus > isNat > and > isNat# = tt = n__isNat > n__0 = |0|
    
    partial status:
  
      pi(isNat#) = [1]
      pi(n__plus) = [1, 2]
      pi(activate) = [1]
      pi(U11) = [2]
      pi(tt) = []
      pi(U21) = []
      pi(s) = []
      pi(plus) = [1, 2]
      pi(and) = [2]
      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.