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:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

The set Q consists of the following terms:

f(s(0), g(x0))
g(s(x0))



↳ QTRS
  ↳ DependencyPairsProof

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

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

The set Q consists of the following terms:

f(s(0), g(x0))
g(s(x0))


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

G(s(x)) → G(x)
F(s(0), g(x)) → F(x, g(x))

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

The set Q consists of the following terms:

f(s(0), g(x0))
g(s(x0))

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:

G(s(x)) → G(x)
F(s(0), g(x)) → F(x, g(x))

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

The set Q consists of the following terms:

f(s(0), g(x0))
g(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof

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

G(s(x)) → G(x)

The TRS R consists of the following rules:

f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

The set Q consists of the following terms:

f(s(0), g(x0))
g(s(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

G(s(x)) → G(x)

R is empty.
The set Q consists of the following terms:

f(s(0), g(x0))
g(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(s(0), g(x0))
g(s(x0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPSizeChangeProof

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

G(s(x)) → G(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: