Termination w.r.t. Q proof of /tmp/tmp2c5N8T/while.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)

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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)

Q is empty.

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

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

The TRS R 2 is

f(t, x, y) → f(g(x, y), x, s(y))

The signature Sigma is {f}

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

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

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

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

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(t, x, y) → G(x, y)
G(s(x), s(y)) → G(x, y)
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:

f(t, x0, x1)
g(s(x0), 0)
g(s(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 1 less node.

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

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

  ↳ AAECC Innermost

    ↳ 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:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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 [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), s(y)) → G(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

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

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus [18] 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, x2, x3)≥F(g(x2, x3), x2, s(x3)))

We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2)    (g(x0, x1)=t ⇒ F(t, x0, s(x1))≥F(g(x0, s(x1)), x0, s(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), s(0))≥F(g(s(x4), s(0)), s(x4), s(s(0))))

(4)    (g(x5, x6)=t∧(g(x5, x6)=t ⇒ F(t, x5, s(x6))≥F(g(x5, s(x6)), x5, s(s(x6)))) ⇒ F(t, s(x5), s(s(x6)))≥F(g(s(x5), s(s(x6))), s(x5), s(s(s(x6)))))

We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
(5)    (F(t, s(x4), s(0))≥F(g(s(x4), s(0)), s(x4), s(s(0))))

We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (g(x5, x6)=t ⇒ F(t, x5, s(x6))≥F(g(x5, s(x6)), x5, s(s(x6)))) with σ = [ ] which results in the following new constraint:
(6)    (F(t, x5, s(x6))≥F(g(x5, s(x6)), x5, s(s(x6))) ⇒ F(t, s(x5), s(s(x6)))≥F(g(s(x5), s(s(x6))), s(x5), s(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), s(0))≥F(g(s(x4), s(0)), s(x4), s(s(0))))
- (F(t, x5, s(x6))≥F(g(x5, s(x6)), x5, s(s(x6))) ⇒ F(t, s(x5), s(s(x6)))≥F(g(s(x5), s(s(x6))), s(x5), s(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 [18]:

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

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

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

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