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(x, x) → f(g(x), x)
g(x) → s(x)

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

f(x, x) → f(g(x), x)
g(x) → s(x)

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

g(x) → s(x)

The TRS R 2 is

f(x, x) → f(g(x), x)

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(x, x) → f(g(x), x)
g(x) → s(x)

The set Q consists of the following terms:

f(x0, x0)
g(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:

F(x, x) → F(g(x), x)
F(x, x) → G(x)

The TRS R consists of the following rules:

f(x, x) → f(g(x), x)
g(x) → s(x)

The set Q consists of the following terms:

f(x0, x0)
g(x0)

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(x, x) → F(g(x), x)
F(x, x) → G(x)

The TRS R consists of the following rules:

f(x, x) → f(g(x), x)
g(x) → s(x)

The set Q consists of the following terms:

f(x0, x0)
g(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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof

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

F(x, x) → F(g(x), x)

The TRS R consists of the following rules:

f(x, x) → f(g(x), x)
g(x) → s(x)

The set Q consists of the following terms:

f(x0, x0)
g(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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof

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

F(x, x) → F(g(x), x)

The TRS R consists of the following rules:

g(x) → s(x)

The set Q consists of the following terms:

f(x0, x0)
g(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(x0, x0)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting

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

F(x, x) → F(g(x), x)

The TRS R consists of the following rules:

g(x) → s(x)

The set Q consists of the following terms:

g(x0)

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(x, x) → F(g(x), x) at position [0] we obtained the following new rules:

F(x, x) → F(s(x), x)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ DependencyGraphProof

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

F(x, x) → F(s(x), x)

The TRS R consists of the following rules:

g(x) → s(x)

The set Q consists of the following terms:

g(x0)

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.