YES

We show the termination of the TRS R:

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: U11#(tt(),V2) -> isNat#(activate(V2))
p2: isNat#(n__s(V1)) -> activate#(V1)
p3: activate#(n__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: isNat#(n__plus(V1,V2)) -> U11#(isNat(activate(V1)),activate(V2))
p10: U11#(tt(),V2) -> activate#(V2)
p11: plus#(N,s(M)) -> U41#(isNat(M),M,N)
p12: U41#(tt(),M,N) -> activate#(M)
p13: U41#(tt(),M,N) -> activate#(N)
p14: U41#(tt(),M,N) -> isNat#(activate(N))
p15: U41#(tt(),M,N) -> U42#(isNat(activate(N)),activate(M),activate(N))
p16: U42#(tt(),M,N) -> activate#(M)
p17: U42#(tt(),M,N) -> activate#(N)
p18: U42#(tt(),M,N) -> plus#(activate(N),activate(M))
p19: plus#(N,|0|()) -> isNat#(N)
p20: plus#(N,|0|()) -> U31#(isNat(N),N)
p21: U31#(tt(),N) -> activate#(N)

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        U11#_A(x1,x2) = x1 + x2
        tt_A() = 1
        isNat#_A(x1) = x1
        activate_A(x1) = x1
        n__s_A(x1) = x1 + 3
        activate#_A(x1) = x1
        n__plus_A(x1,x2) = x1 + x2 + 5
        plus#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 3
        isNat_A(x1) = 4
        U41#_A(x1,x2,x3) = x2 + x3 + 2
        U42#_A(x1,x2,x3) = x2 + x3 + 1
        |0|_A() = 2
        U31#_A(x1,x2) = x2 + 1
        U42_A(x1,x2,x3) = x2 + x3 + 8
        plus_A(x1,x2) = x1 + x2 + 5
        U12_A(x1) = 2
        U31_A(x1,x2) = x2 + 1
        U41_A(x1,x2,x3) = x2 + x3 + 8
        U11_A(x1,x2) = 3
        U21_A(x1) = x1
        n__0_A() = 2
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        U41# > activate > |0| > tt > n__0 > U21 > U12 > isNat > U11 > plus > U41 > U42 > U31 > U31# > U42# > isNat# > s > n__s > plus# > U11# > activate# > n__plus
      
      argument filter:
    
        pi(U11#) = []
        pi(tt) = []
        pi(isNat#) = [1]
        pi(activate) = [1]
        pi(n__s) = []
        pi(activate#) = []
        pi(n__plus) = 2
        pi(plus#) = 1
        pi(s) = []
        pi(isNat) = []
        pi(U41#) = [2, 3]
        pi(U42#) = 3
        pi(|0|) = []
        pi(U31#) = []
        pi(U42) = []
        pi(plus) = [2]
        pi(U12) = []
        pi(U31) = 2
        pi(U41) = []
        pi(U11) = []
        pi(U21) = 1
        pi(n__0) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19, p20, p21

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