Termination w.r.t. Q proof of /tmp/tmp2FhguF/ex02.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(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

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

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] 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), s(y)) → G(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

g(s(x0), s(x1))
g(s(x0), 0)
f(t, 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 [13] we can delete all non-usable rules [14] 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), s(y)) → G(x, y)

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

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

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

G(s(x), s(y)) → G(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [16] together with the size-change analysis [27] 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), s(y)) → G(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(t, x, y) → F(g(x, y), x, s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

F(t, x, y) → F(g(x, y), x, s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

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

F(t, x, y) → F(g(x, y), x, s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule F(t, x, y) → F(g(x, y), x, s(y)) at position [0] we obtained the following new rules:

F(t, s(x0), 0) → F(t, s(x0), s(0))
F(t, s(x0), s(x1)) → F(g(x0, x1), s(x0), s(s(x1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                    ↳ NonInfProof

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

F(t, s(x0), 0) → F(t, s(x0), s(0))
F(t, s(x0), s(x1)) → F(g(x0, x1), s(x0), s(s(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

                    ↳ NonInfProof

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

F(t, s(x0), s(x1)) → F(g(x0, x1), s(x0), s(s(x1)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule F(t, s(x0), s(x1)) → F(g(x0, x1), s(x0), s(s(x1))) we obtained the following new rules:

F(t, s(z0), s(s(z1))) → F(g(z0, s(z1)), s(z0), s(s(s(z1))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                    ↳ NonInfProof

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

F(t, s(z0), s(s(z1))) → F(g(z0, s(z1)), s(z0), s(s(s(z1))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus with the following steps:
Note that final constraints are written in bold face.

For Pair F(t, x, y) → F(g(x, y), x, s(y)) the following chains were created:

We consider the chain F(t, x, y) → F(g(x, y), x, s(y)), F(t, x, y) → F(g(x, y), x, s(y)) which results in the following constraint:
- (1) (F(g(x0, x1), x0, s(x1))=F(t, x2, x3) ⇒ F(t, x0, x1)≥F(g(x0, x1), x0, s(x1)))
We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
- (2) (g(x0, x1)=t ⇒ F(t, x0, x1)≥F(g(x0, x1), x0, s(x1)))
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on g(x0, x1)=t which results in the following new constraints:
- (3) (t=t ⇒ F(t, s(x4), 0)≥F(g(s(x4), 0), s(x4), s(0)))
- (4) (g(x5, x6)=t∧(g(x5, x6)=t ⇒ F(t, x5, x6)≥F(g(x5, x6), x5, s(x6))) ⇒ F(t, s(x5), s(x6))≥F(g(s(x5), s(x6)), s(x5), s(s(x6))))
We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
- (5) (F(t, s(x4), 0)≥F(g(s(x4), 0), s(x4), s(0)))
We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (g(x5, x6)=t ⇒ F(t, x5, x6)≥F(g(x5, x6), x5, s(x6))) with σ = [ ] which results in the following new constraint:
- (6) (F(t, x5, x6)≥F(g(x5, x6), x5, s(x6)) ⇒ F(t, s(x5), s(x6))≥F(g(s(x5), s(x6)), s(x5), s(s(x6))))

To summarize, we get the following constraints P_≥ for the following pairs.

F(t, x, y) → F(g(x, y), x, s(y))
- (F(t, s(x4), 0)≥F(g(s(x4), 0), s(x4), s(0)))
- (F(t, x5, x6)≥F(g(x5, x6), x5, s(x6)) ⇒ F(t, s(x5), s(x6))≥F(g(s(x5), s(x6)), s(x5), s(s(x6))))

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved.
Polynomial interpretation [21]:

POL(0) = 0
POL(F(x₁, x₂, x₃)) = -1 + x₁ + 2·x₂ - x₃
POL(c) = -1
POL(g(x₁, x₂)) = 0
POL(s(x₁)) = 2 + x₁
POL(t) = 0

The following pairs are in P_>:

F(t, x, y) → F(g(x, y), x, s(y))

The following pairs are in P_bound:

F(t, x, y) → F(g(x, y), x, s(y))

The following rules are usable:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ QDP

                        ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.