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

The set Q consists of the following terms:

g(0)
g(s(x0))
f(g(x0), s(0), x1)



↳ QTRS
  ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

g(0)
g(s(x0))
f(g(x0), s(0), x1)


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

g(0)
g(s(x0))
f(g(x0), s(0), x1)

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:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

g(0)
g(s(x0))
f(g(x0), s(0), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

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

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

The set Q consists of the following terms:

g(0)
g(s(x0))
f(g(x0), s(0), x1)

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
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
          ↳ QDP

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:

g(0)
g(s(x0))
f(g(x0), s(0), x1)

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.

g(0)
g(s(x0))
f(g(x0), s(0), x1)



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

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: 



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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

g(0)
g(s(x0))
f(g(x0), s(0), x1)

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
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof

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

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

The TRS R consists of the following rules:

g(s(x)) → s(g(x))
g(0) → 0

The set Q consists of the following terms:

g(0)
g(s(x0))
f(g(x0), s(0), x1)

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(g(x0), s(0), x1)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ Rewriting

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

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

The TRS R consists of the following rules:

g(s(x)) → s(g(x))
g(0) → 0

The set Q consists of the following terms:

g(0)
g(s(x0))

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

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ Rewriting
                      ↳ QDP
                        ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

g(s(x)) → s(g(x))
g(0) → 0

The set Q consists of the following terms:

g(0)
g(s(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.