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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      U11#_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + (0,6,9,0)
      tt_A() = (9,1,1,7)
      activate#_A(x1) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(1,0,1,0)) x1 + (0,5,10,10)
      n__isNat_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,1,0,0)) x1 + (16,21,19,2)
      isNat#_A(x1) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(1,0,1,0)) x1 + (0,21,31,7)
      n__s_A(x1) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,1,1,0)) x1 + (16,10,0,12)
      n__plus_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(1,1,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,1,1,0),(1,1,1,0)) x2 + (12,10,4,0)
      plus#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x2 + (0,17,20,8)
      s_A(x1) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,0,0)) x1 + (16,11,0,11)
      activate_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(0,0,1,0)) x1 + (0,10,16,10)
      and#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,0,0,0),(0,1,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x2
      isNat_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,0,0,0)) x1 + (16,22,18,1)
      U21#_A(x1,x2,x3) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,0,0)) x2 + ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,0,0,0)) x3 + (0,18,22,9)
      and_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + (1,1,17,0)
      |0|_A() = (5,1,2,1)
      U11_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,0,0)) x2 + (1,0,0,0)
      U21_A(x1,x2,x3) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + x3 + (28,3,19,16)
      plus_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,0)) x2 + (12,10,27,15)
      n__0_A() = (5,0,1,0)

The next rules are strictly ordered:

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

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