Termination w.r.t. Q proof of /tmp/tmplOBtJP/test77.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:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

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

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

f(0, s(0), x0)
+(x0, s(x1))
double(x0)
+(x0, 0)
g(x0, 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(0, s(0), X) → F(X, double(X), X)
F(0, s(0), X) → DOUBLE(X)
+¹(X, s(Y)) → +¹(X, Y)
DOUBLE(X) → +¹(X, X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

f(0, s(0), x0)
+(x0, s(x1))
double(x0)
+(x0, 0)
g(x0, 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(0, s(0), X) → F(X, double(X), X)
F(0, s(0), X) → DOUBLE(X)
+¹(X, s(Y)) → +¹(X, Y)
DOUBLE(X) → +¹(X, X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

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

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

↳ 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:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

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

+¹(X, s(Y)) → +¹(X, Y)

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

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

↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

F(0, s(0), X) → F(X, double(X), X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → Y

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

F(0, s(0), X) → F(X, double(X), X)

The TRS R consists of the following rules:

double(X) → +(X, X)
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))

The set Q consists of the following terms:

f(0, s(0), x0)
+(x0, s(x1))
double(x0)
+(x0, 0)
g(x0, 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, s(0), x0)
g(x0, 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(0, s(0), X) → F(X, double(X), X)

The TRS R consists of the following rules:

double(X) → +(X, X)
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(0, s(0), X) → F(X, double(X), X) at position [1] we obtained the following new rules:

F(0, s(0), X) → F(X, +(X, X), X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ UsableRulesProof

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

F(0, s(0), X) → F(X, +(X, X), X)

The TRS R consists of the following rules:

double(X) → +(X, X)
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

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

F(0, s(0), X) → F(X, +(X, X), X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))

The set Q consists of the following terms:

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

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.

double(x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

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

F(0, s(0), X) → F(X, +(X, X), X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(0, s(0), X) → F(X, +(X, X), X) at position [1] we obtained the following new rules:

F(0, s(0), 0) → F(0, 0, 0)
F(0, s(0), s(x1)) → F(s(x1), s(+(s(x1), x1)), s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

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

F(0, s(0), 0) → F(0, 0, 0)
F(0, s(0), s(x1)) → F(s(x1), s(+(s(x1), x1)), s(x1))

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.