Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
activate(x1)  =  x1
U21(x1, x2, x3)  =  U21(x1, x2, x3)
s(x1)  =  s(x1)
plus(x1, x2)  =  plus(x1, x2)
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  x1
n__0  =  n__0
n__plus(x1, x2)  =  n__plus(x1, x2)
n__isNat(x1)  =  x1
n__s(x1)  =  n__s(x1)
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:

[U21₃, plus₂, n_plus2] > [s₁, n_s1] > U11₂
[U21₃, plus₂, n_plus2] > and₂ > U11₂
[n₀, 0] > tt > [s₁, n_s1] > U11₂

Status:

n_plus2: [1,2]
plus₂: [1,2]
tt: multiset
n₀: multiset
s₁: [1]
U11₂: [2,1]
and₂: multiset
n_s1: [1]
U21₃: [3,2,1]
0: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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)

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(activate(x₁)) = 2 + x₁   
POL(isNat(x₁)) = 1 + x₁   
POL(n__0) = 0   
POL(n__isNat(x₁)) = x₁   
POL(n__plus(x₁, x₂)) = x₁ + x₂   
POL(n__s(x₁)) = x₁   
POL(plus(x₁, x₂)) = 1 + x₁ + x₂   
POL(s(x₁)) = 1 + x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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

(4) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(5) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(6) TRUE

(7) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(8) TRUE