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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          U11#_A(x1,x2) = max{48, x2 + 41}
          tt_A = 41
          activate#_A(x1) = max{43, x1 - 51}
          n__isNat_A(x1) = x1 + 43
          isNat#_A(x1) = max{42, x1 - 8}
          n__s_A(x1) = max{92, x1}
          n__plus_A(x1,x2) = max{x1 + 92, x2 + 100}
          plus#_A(x1,x2) = max{48, x1 + 41, x2 + 46}
          s_A(x1) = max{92, x1}
          activate_A(x1) = max{19, x1}
          and#_A(x1,x2) = max{42, x1 + 3, x2 - 3}
          isNat_A(x1) = x1 + 43
          U21#_A(x1,x2,x3) = max{138, x2 + 46, x3 + 41}
          and_A(x1,x2) = max{x1 - 21, x2 + 39}
          |0|_A = 1
          U11_A(x1,x2) = max{91, x1 - 1, x2}
          U21_A(x1,x2,x3) = max{192, x1 + 1, x2 + 100, x3 + 92}
          plus_A(x1,x2) = max{x1 + 92, x2 + 100}
          n__0_A = 0
    
      precedence:
      
        |0| = n__0 > activate# = isNat# = plus# = U21# > U11# > n__isNat = isNat > tt = and > n__plus = and# = plus > U11 > activate = U21 > s > n__s
      
      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) = [1]
        pi(and#) = [1]
        pi(isNat) = []
        pi(U21#) = []
        pi(and) = []
        pi(|0|) = []
        pi(U11) = [2]
        pi(U21) = [1, 2, 3]
        pi(plus) = []
        pi(n__0) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          U11#_A(x1,x2) = 17
          tt_A = 32
          activate#_A(x1) = 18
          n__isNat_A(x1) = 9
          isNat#_A(x1) = 18
          n__s_A(x1) = 33
          n__plus_A(x1,x2) = 17
          plus#_A(x1,x2) = 18
          s_A(x1) = 33
          activate_A(x1) = 33
          and#_A(x1,x2) = max{20, x1 - 13}
          isNat_A(x1) = 33
          U21#_A(x1,x2,x3) = 18
          and_A(x1,x2) = 33
          |0|_A = 34
          U11_A(x1,x2) = 81
          U21_A(x1,x2,x3) = 18
          plus_A(x1,x2) = 18
          n__0_A = 31
    
      precedence:
      
        activate > s > n__s = and# > isNat > tt > activate# = n__isNat = isNat# = plus# = U21# = n__0 > U11# > plus > n__plus > and > |0| > U11 = U21
      
      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:

  p1, p10, p12

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


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

and R consists of:

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

The 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{9, x1 - 14}
          n__isNat_A(x1) = x1 + 19
          isNat#_A(x1) = max{0, x1 - 13}
          n__plus_A(x1,x2) = max{x1 + 34, x2 + 22}
          activate_A(x1) = x1
          plus#_A(x1,x2) = max{x1 + 20, x2 + 1}
          |0|_A = 35
          n__s_A(x1) = max{22, x1}
          s_A(x1) = max{22, x1}
          U21#_A(x1,x2,x3) = max{x1 - 24, x2 + 1, x3 + 20}
          and_A(x1,x2) = max{24, x1 - 17, x2 + 2}
          isNat_A(x1) = x1 + 19
          tt_A = 34
          U11_A(x1,x2) = max{x1 + 15, x2 + 1}
          U21_A(x1,x2,x3) = max{x2 + 22, x3 + 34}
          plus_A(x1,x2) = max{x1 + 34, x2 + 22}
          n__0_A = 35
    
      precedence:
      
        and > activate = plus > U21 > s > isNat# = n__s > activate# = n__isNat = isNat = U11 > n__plus = plus# = U21# > |0| > tt = n__0
      
      partial status:
    
        pi(activate#) = []
        pi(n__isNat) = []
        pi(isNat#) = []
        pi(n__plus) = [2]
        pi(activate) = [1]
        pi(plus#) = []
        pi(|0|) = []
        pi(n__s) = []
        pi(s) = []
        pi(U21#) = []
        pi(and) = [2]
        pi(isNat) = []
        pi(tt) = []
        pi(U11) = [1, 2]
        pi(U21) = []
        pi(plus) = []
        pi(n__0) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = 55
          n__isNat_A(x1) = 34
          isNat#_A(x1) = 55
          n__plus_A(x1,x2) = 34
          activate_A(x1) = max{41, x1 - 1}
          plus#_A(x1,x2) = 17
          |0|_A = 76
          n__s_A(x1) = 41
          s_A(x1) = 41
          U21#_A(x1,x2,x3) = 17
          and_A(x1,x2) = max{42, x2 - 1}
          isNat_A(x1) = 34
          tt_A = 22
          U11_A(x1,x2) = max{56, x1 + 20, x2 + 42}
          U21_A(x1,x2,x3) = 41
          plus_A(x1,x2) = 41
          n__0_A = 21
    
      precedence:
      
        U11 > activate# = isNat# > plus# = U21# > n__0 > activate = and = isNat > plus > n__plus > |0| > n__isNat = n__s = s = tt = U21
      
      partial status:
    
        pi(activate#) = []
        pi(n__isNat) = []
        pi(isNat#) = []
        pi(n__plus) = []
        pi(activate) = []
        pi(plus#) = []
        pi(|0|) = []
        pi(n__s) = []
        pi(s) = []
        pi(U21#) = []
        pi(and) = []
        pi(isNat) = []
        pi(tt) = []
        pi(U11) = [1, 2]
        pi(U21) = []
        pi(plus) = []
        pi(n__0) = []
    

The next rules are strictly ordered:

  p5, p11, p12, p13

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- 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))

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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = x1 + 8
          n__isNat_A(x1) = x1 + 10
          isNat#_A(x1) = x1 + 9
          n__s_A(x1) = max{11, x1}
          activate_A(x1) = x1
          n__plus_A(x1,x2) = max{30, x1 + 7, x2 + 19}
          U11_A(x1,x2) = max{6, x1 - 4, x2}
          tt_A = 29
          U21_A(x1,x2,x3) = max{30, x1 - 11, x2 + 19, x3 + 7}
          s_A(x1) = max{11, x1}
          plus_A(x1,x2) = max{30, x1 + 7, x2 + 19}
          and_A(x1,x2) = x2 + 8
          isNat_A(x1) = x1 + 10
          n__0_A = 30
          |0|_A = 30
    
      precedence:
      
        and > isNat# > activate# = n__plus = tt = plus > U11 = U21 > activate > |0| > s > n__0 > n__isNat = n__s = isNat
      
      partial status:
    
        pi(activate#) = [1]
        pi(n__isNat) = []
        pi(isNat#) = []
        pi(n__s) = []
        pi(activate) = [1]
        pi(n__plus) = []
        pi(U11) = [2]
        pi(tt) = []
        pi(U21) = []
        pi(s) = []
        pi(plus) = []
        pi(and) = []
        pi(isNat) = []
        pi(n__0) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = 26
          n__isNat_A(x1) = 12
          isNat#_A(x1) = 26
          n__s_A(x1) = 13
          activate_A(x1) = max{25, x1 + 2}
          n__plus_A(x1,x2) = 27
          U11_A(x1,x2) = max{55, x2 + 26}
          tt_A = 13
          U21_A(x1,x2,x3) = 26
          s_A(x1) = 25
          plus_A(x1,x2) = 28
          and_A(x1,x2) = 55
          isNat_A(x1) = 13
          n__0_A = 27
          |0|_A = 26
    
      precedence:
      
        n__isNat = activate = U11 = tt = |0| > U21 > s = plus = and > isNat# = n__plus > activate# = n__s > isNat > n__0
      
      partial status:
    
        pi(activate#) = []
        pi(n__isNat) = []
        pi(isNat#) = []
        pi(n__s) = []
        pi(activate) = [1]
        pi(n__plus) = []
        pi(U11) = [2]
        pi(tt) = []
        pi(U21) = []
        pi(s) = []
        pi(plus) = []
        pi(and) = []
        pi(isNat) = []
        pi(n__0) = []
        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__s(V1)) -> activate#(V1)
p2: isNat#(n__s(V1)) -> isNat#(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))

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, p5}


-- Reduction pair.

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

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

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          isNat#_A(x1) = max{0, x1 - 1}
          n__s_A(x1) = x1
          activate_A(x1) = x1
          n__plus_A(x1,x2) = max{x1 + 7, x2 + 11}
          U11_A(x1,x2) = max{8, x1 - 11, x2 + 1}
          tt_A = 12
          U21_A(x1,x2,x3) = max{x1 - 11, x2 + 11, x3 + 7}
          s_A(x1) = x1
          plus_A(x1,x2) = max{x1 + 7, x2 + 11}
          and_A(x1,x2) = max{12, x1 - 3, x2 + 1}
          isNat_A(x1) = x1 + 17
          n__0_A = 13
          n__isNat_A(x1) = x1 + 17
          |0|_A = 13
    
      precedence:
      
        isNat# = n__s = activate = n__plus = U11 = tt = U21 = s = plus = and = isNat = n__0 = n__isNat = |0|
      
      partial status:
    
        pi(isNat#) = []
        pi(n__s) = []
        pi(activate) = []
        pi(n__plus) = []
        pi(U11) = []
        pi(tt) = []
        pi(U21) = []
        pi(s) = []
        pi(plus) = []
        pi(and) = []
        pi(isNat) = []
        pi(n__0) = []
        pi(n__isNat) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          isNat#_A(x1) = 22
          n__s_A(x1) = 20
          activate_A(x1) = max{14, x1 + 2}
          n__plus_A(x1,x2) = 20
          U11_A(x1,x2) = 23
          tt_A = 30
          U21_A(x1,x2,x3) = 21
          s_A(x1) = 20
          plus_A(x1,x2) = 22
          and_A(x1,x2) = 42
          isNat_A(x1) = 20
          n__0_A = 10
          n__isNat_A(x1) = 20
          |0|_A = 13
    
      precedence:
      
        isNat# > activate = n__plus = tt = U21 = s = n__0 = |0| > n__s = and > isNat > U11 = plus = n__isNat
      
      partial status:
    
        pi(isNat#) = []
        pi(n__s) = []
        pi(activate) = [1]
        pi(n__plus) = []
        pi(U11) = []
        pi(tt) = []
        pi(U21) = []
        pi(s) = []
        pi(plus) = []
        pi(and) = []
        pi(isNat) = []
        pi(n__0) = []
        pi(n__isNat) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            isNat#_A(x1) = ((1,1),(1,1)) x1 + (1,26)
            n__s_A(x1) = ((0,1),(1,0)) x1 + (1,31)
            activate_A(x1) = x1 + (4,2)
            U11_A(x1,x2) = ((1,1),(1,1)) x2 + (5,2)
            tt_A() = (3,1)
            U21_A(x1,x2,x3) = ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (31,43)
            s_A(x1) = ((0,1),(1,0)) x1 + (1,31)
            plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (0,17)
            and_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (2,11)
            isNat_A(x1) = ((1,1),(1,1)) x1 + (3,0)
            n__0_A() = (2,0)
            n__plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (0,17)
            n__isNat_A(x1) = ((1,1),(1,1)) x1 + (2,0)
            |0|_A() = (5,0)
    
      precedence:
      
        U11 = U21 = s = plus = n__plus > activate > n__0 = |0| > isNat > n__isNat > tt = and > isNat# = n__s
      
      partial status:
    
        pi(isNat#) = []
        pi(n__s) = []
        pi(activate) = []
        pi(U11) = []
        pi(tt) = []
        pi(U21) = [3]
        pi(s) = []
        pi(plus) = [1, 2]
        pi(and) = []
        pi(isNat) = []
        pi(n__0) = []
        pi(n__plus) = [1, 2]
        pi(n__isNat) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            isNat#_A(x1) = (0,0)
            n__s_A(x1) = (0,0)
            activate_A(x1) = (2,0)
            U11_A(x1,x2) = (0,0)
            tt_A() = (7,2)
            U21_A(x1,x2,x3) = x3 + (3,2)
            s_A(x1) = (2,0)
            plus_A(x1,x2) = x1 + ((0,0),(0,1)) x2 + (4,2)
            and_A(x1,x2) = (5,1)
            isNat_A(x1) = (8,2)
            n__0_A() = (8,3)
            n__plus_A(x1,x2) = x1
            n__isNat_A(x1) = (3,1)
            |0|_A() = (1,2)
    
      precedence:
      
        n__0 = |0| > tt > isNat > and > activate = U21 = plus > n__plus = n__isNat > s > n__s > isNat# = U11
      
      partial status:
    
        pi(isNat#) = []
        pi(n__s) = []
        pi(activate) = []
        pi(U11) = []
        pi(tt) = []
        pi(U21) = [3]
        pi(s) = []
        pi(plus) = []
        pi(and) = []
        pi(isNat) = []
        pi(n__0) = []
        pi(n__plus) = [1]
        pi(n__isNat) = []
        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: 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(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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            plus#_A(x1,x2) = ((1,1),(0,1)) x2 + (1,0)
            s_A(x1) = x1 + (10,12)
            U21#_A(x1,x2,x3) = ((1,1),(0,1)) x2 + (11,3)
            and_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,0)
            isNat_A(x1) = x1 + (2,4)
            n__isNat_A(x1) = x1 + (2,4)
            tt_A() = (7,2)
            activate_A(x1) = x1 + (2,3)
            U11_A(x1,x2) = ((1,1),(0,1)) x2 + (8,3)
            U21_A(x1,x2,x3) = ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (39,44)
            plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (17,22)
            |0|_A() = (7,1)
            n__0_A() = (6,1)
            n__plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (17,22)
            n__s_A(x1) = x1 + (9,9)
    
      precedence:
      
        plus# > plus = n__plus > s = U21 > |0| > n__0 > U11 = n__s > and > isNat = n__isNat = tt > U21# = activate
      
      partial status:
    
        pi(plus#) = [2]
        pi(s) = []
        pi(U21#) = []
        pi(and) = []
        pi(isNat) = [1]
        pi(n__isNat) = [1]
        pi(tt) = []
        pi(activate) = []
        pi(U11) = []
        pi(U21) = [3]
        pi(plus) = [1, 2]
        pi(|0|) = []
        pi(n__0) = []
        pi(n__plus) = [1]
        pi(n__s) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            plus#_A(x1,x2) = ((1,1),(0,1)) x2 + (1,0)
            s_A(x1) = (12,4)
            U21#_A(x1,x2,x3) = (11,2)
            and_A(x1,x2) = (17,2)
            isNat_A(x1) = ((1,1),(0,0)) x1 + (6,2)
            n__isNat_A(x1) = (1,1)
            tt_A() = (11,2)
            activate_A(x1) = (7,2)
            U11_A(x1,x2) = (8,2)
            U21_A(x1,x2,x3) = (12,4)
            plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (18,2)
            |0|_A() = (5,2)
            n__0_A() = (4,2)
            n__plus_A(x1,x2) = x1 + (18,2)
            n__s_A(x1) = (6,4)
    
      precedence:
      
        isNat = plus > U11 > U21 > s > U21# > plus# = and = n__plus > n__isNat > n__s > tt = n__0 > activate = |0|
      
      partial status:
    
        pi(plus#) = [2]
        pi(s) = []
        pi(U21#) = []
        pi(and) = []
        pi(isNat) = []
        pi(n__isNat) = []
        pi(tt) = []
        pi(activate) = []
        pi(U11) = []
        pi(U21) = []
        pi(plus) = [2]
        pi(|0|) = []
        pi(n__0) = []
        pi(n__plus) = [1]
        pi(n__s) = []
    

The next rules are strictly ordered:

  p1, p2

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