Termination w.r.t. Q proof of /tmp/tmpO

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(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

Q is empty.

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

gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The TRS R 2 is

f(true, x, y) → f(gt(x, y), s(x), s(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(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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:

GT(s(u), s(v)) → GT(u, v)
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
F(true, x, y) → GT(x, y)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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:

GT(s(u), s(v)) → GT(u, v)
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
F(true, x, y) → GT(x, y)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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:

GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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:

GT(s(u), s(v)) → GT(u, v)

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

f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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:

GT(s(u), s(v)) → GT(u, v)

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:

GT(s(u), s(v)) → GT(u, v)
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(true, x, y) → F(gt(x, y), s(x), s(s(y)))

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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(true, x, y) → F(gt(x, y), s(x), s(s(y)))

The TRS R consists of the following rules:

gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(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(true, 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(true, x, y) → F(gt(x, y), s(x), s(s(y)))

The TRS R consists of the following rules:

gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(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(true, x, y) → F(gt(x, y), s(x), s(s(y))) the following chains were created:

We consider the chain F(true, x, y) → F(gt(x, y), s(x), s(s(y))), F(true, x, y) → F(gt(x, y), s(x), s(s(y))) which results in the following constraint:
(1)    (F(gt(x0, x1), s(x0), s(s(x1)))=F(true, x2, x3) ⇒ F(true, x2, x3)≥F(gt(x2, x3), s(x2), s(s(x3))))

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

We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x0, x1)=true which results in the following new constraints:
(3)    (true=true ⇒ F(true, s(s(x5)), s(s(0)))≥F(gt(s(s(x5)), s(s(0))), s(s(s(x5))), s(s(s(s(0))))))

(4)    (gt(x6, x7)=true∧(gt(x6, x7)=true ⇒ F(true, s(x6), s(s(x7)))≥F(gt(s(x6), s(s(x7))), s(s(x6)), s(s(s(s(x7)))))) ⇒ F(true, s(s(x6)), s(s(s(x7))))≥F(gt(s(s(x6)), s(s(s(x7)))), s(s(s(x6))), s(s(s(s(s(x7)))))))

We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
(5)    (F(true, s(s(x5)), s(s(0)))≥F(gt(s(s(x5)), s(s(0))), s(s(s(x5))), s(s(s(s(0))))))

We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gt(x6, x7)=true ⇒ F(true, s(x6), s(s(x7)))≥F(gt(s(x6), s(s(x7))), s(s(x6)), s(s(s(s(x7)))))) with σ = [ ] which results in the following new constraint:
(6)    (F(true, s(x6), s(s(x7)))≥F(gt(s(x6), s(s(x7))), s(s(x6)), s(s(s(s(x7))))) ⇒ F(true, s(s(x6)), s(s(s(x7))))≥F(gt(s(s(x6)), s(s(s(x7)))), s(s(s(x6))), s(s(s(s(s(x7)))))))

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

F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
- (F(true, s(s(x5)), s(s(0)))≥F(gt(s(s(x5)), s(s(0))), s(s(s(x5))), s(s(s(s(0))))))
- (F(true, s(x6), s(s(x7)))≥F(gt(s(x6), s(s(x7))), s(s(x6)), s(s(s(s(x7))))) ⇒ F(true, s(s(x6)), s(s(s(x7))))≥F(gt(s(s(x6)), s(s(s(x7)))), s(s(s(x6))), s(s(s(s(s(x7)))))))

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(false) = 1
POL(gt(x₁, x₂)) = 1
POL(s(x₁)) = 1 + x₁
POL(true) = 1

The following pairs are in P_>:

F(true, x, y) → F(gt(x, y), s(x), s(s(y)))

The following pairs are in P_bound:

F(true, x, y) → F(gt(x, y), s(x), s(s(y)))

The following rules are usable:

gt(s(u), 0) → true
gt(0, v) → false
gt(s(u), s(v)) → gt(u, v)

↳ 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:

gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(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.