Termination w.r.t. Q proof of /tmp/tmpqPL2PR/2.16.trs

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(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

The set Q consists of the following terms:

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

G(s(x), y) → +¹(y, s(x))
G(s(x), y) → +¹(y, x)
G(s(x), y) → G(x, s(+(y, x)))
+¹(x, s(y)) → +¹(x, y)
G(s(x), y) → G(x, +(y, s(x)))
F(s(x)) → G(x, s(x))

The TRS R consists of the following rules:

f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

G(s(x), y) → +¹(y, s(x))
G(s(x), y) → +¹(y, x)
G(s(x), y) → G(x, s(+(y, x)))
+¹(x, s(y)) → +¹(x, y)
G(s(x), y) → G(x, +(y, s(x)))
F(s(x)) → G(x, s(x))

The TRS R consists of the following rules:

f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

+¹(x, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

The set Q consists of the following terms:

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

+¹(x, s(y)) → +¹(x, y)

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

+¹(x, s(y)) → +¹(x, y)

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:

+¹(x, s(y)) → +¹(x, y)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

G(s(x), y) → G(x, s(+(y, x)))
G(s(x), y) → G(x, +(y, s(x)))

The TRS R consists of the following rules:

f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

G(s(x), y) → G(x, s(+(y, x)))
G(s(x), y) → G(x, +(y, s(x)))

The TRS R consists of the following rules:

+(x, s(y)) → s(+(x, y))
+(x, 0) → x

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

G(s(x), y) → G(x, s(+(y, x)))
G(s(x), y) → G(x, +(y, s(x)))

The TRS R consists of the following rules:

+(x, s(y)) → s(+(x, y))
+(x, 0) → x

The set Q consists of the following terms:

+(x0, 0)
+(x0, s(x1))

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:

G(s(x), y) → G(x, s(+(y, x)))
The graph contains the following edges 1 > 1
G(s(x), y) → G(x, +(y, s(x)))
The graph contains the following edges 1 > 1