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, M, N) → U12(tt, activate(M), activate(N))
U12(tt, M, N) → s(plus(activate(N), activate(M)))
plus(N, 0) → N
plus(N, s(M)) → U11(tt, M, N)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

U11¹(tt, M, N) → U12¹(tt, activate(M), activate(N))
U11¹(tt, M, N) → ACTIVATE(M)
U11¹(tt, M, N) → ACTIVATE(N)
U12¹(tt, M, N) → PLUS(activate(N), activate(M))
U12¹(tt, M, N) → ACTIVATE(N)
U12¹(tt, M, N) → ACTIVATE(M)
PLUS(N, s(M)) → U11¹(tt, M, N)

The TRS R consists of the following rules:

U11(tt, M, N) → U12(tt, activate(M), activate(N))
U12(tt, M, N) → s(plus(activate(N), activate(M)))
plus(N, 0) → N
plus(N, s(M)) → U11(tt, M, N)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

U12¹(tt, M, N) → PLUS(activate(N), activate(M))
PLUS(N, s(M)) → U11¹(tt, M, N)
U11¹(tt, M, N) → U12¹(tt, activate(M), activate(N))

The TRS R consists of the following rules:

U11(tt, M, N) → U12(tt, activate(M), activate(N))
U12(tt, M, N) → s(plus(activate(N), activate(M)))
plus(N, 0) → N
plus(N, s(M)) → U11(tt, M, N)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PLUS(N, s(M)) → U11¹(tt, M, N)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(PLUS(x₁, x₂)) = 1 + x₂   
POL(U11¹(x₁, x₂, x₃)) = 1 + x₂   
POL(U12¹(x₁, x₂, x₃)) = 1 + x₂   
POL(activate(x₁)) = x₁   
POL(s(x₁)) = 1 + x₁   
POL(tt) = 0   

The following usable rules [FROCOS05] were oriented:

activate(X) → X

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

U12¹(tt, M, N) → PLUS(activate(N), activate(M))
U11¹(tt, M, N) → U12¹(tt, activate(M), activate(N))

The TRS R consists of the following rules:

U11(tt, M, N) → U12(tt, activate(M), activate(N))
U12(tt, M, N) → s(plus(activate(N), activate(M)))
plus(N, 0) → N
plus(N, s(M)) → U11(tt, M, N)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(8) TRUE