Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

PLUS(N, s(M)) → U111(tt, M, N)
U111(tt, M, N) → ACTIVATE(N)
U121(tt, M, N) → ACTIVATE(N)
U121(tt, M, N) → PLUS(activate(N), activate(M))
U121(tt, M, N) → ACTIVATE(M)
U111(tt, M, 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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

PLUS(N, s(M)) → U111(tt, M, N)
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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


PLUS(N, s(M)) → U111(tt, M, N)
U111(tt, M, N) → U121(tt, activate(M), activate(N))
The remaining pairs can at least be oriented weakly.

U121(tt, M, N) → PLUS(activate(N), activate(M))
Used ordering: Polynomial interpretation [25,35]:

POL(PLUS(x1, x2)) = 1 + (4)x_2   
POL(U121(x1, x2, x3)) = 1 + (4)x_2   
POL(tt) = 4   
POL(U111(x1, x2, x3)) = x_1 + (4)x_2   
POL(s(x1)) = 4 + (4)x_1   
POL(activate(x1)) = x_1   
The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented:

activate(X) → X



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

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

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.