Termination w.r.t. Q proof of /tmp/tmp88sN

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

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

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

          ↳ 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:

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
f(true, x0, x1)
gt(s(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.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ 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 [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:

GT(s(u), s(v)) → GT(u, v)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

TRUNC(s(s(x))) → TRUNC(x)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

TRUNC(s(s(x))) → TRUNC(x)

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

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
f(true, x0, x1)
gt(s(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.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

TRUNC(s(s(x))) → TRUNC(x)

From the DPs we obtained the following set of size-change graphs:

TRUNC(s(s(x))) → TRUNC(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

F(true, x, y) → F(gt(x, y), trunc(x), 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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
f(true, x0, x1)
gt(s(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.

f(true, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

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

F(true, x, y) → F(gt(x, y), trunc(x), 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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
gt(s(x0), 0)

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

F(true, s(x0), s(x1)) → F(gt(x0, x1), trunc(s(x0)), s(s(x1)))
F(true, s(x0), 0) → F(true, trunc(s(x0)), s(0))
F(true, 0, x0) → F(false, trunc(0), s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                    ↳ NonInfProof

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

F(true, s(x0), 0) → F(true, trunc(s(x0)), s(0))
F(true, s(x0), s(x1)) → F(gt(x0, x1), trunc(s(x0)), s(s(x1)))
F(true, 0, x0) → F(false, trunc(0), s(x0))

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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
gt(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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                    ↳ NonInfProof

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

F(true, s(x0), s(x1)) → F(gt(x0, x1), trunc(s(x0)), s(s(x1)))

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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
gt(s(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule F(true, s(x0), s(x1)) → F(gt(x0, x1), trunc(s(x0)), s(s(x1))) at position [1] we obtained the following new rules:

F(true, s(0), s(y1)) → F(gt(0, y1), 0, s(s(y1)))
F(true, s(s(x0)), s(y1)) → F(gt(s(x0), y1), s(s(trunc(x0))), s(s(y1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                    ↳ NonInfProof

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

F(true, s(0), s(y1)) → F(gt(0, y1), 0, s(s(y1)))
F(true, s(s(x0)), s(y1)) → F(gt(s(x0), y1), s(s(trunc(x0))), s(s(y1)))

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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
gt(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

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                    ↳ NonInfProof

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

F(true, s(s(x0)), s(y1)) → F(gt(s(x0), y1), s(s(trunc(x0))), s(s(y1)))

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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
gt(s(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule F(true, s(s(x0)), s(y1)) → F(gt(s(x0), y1), s(s(trunc(x0))), s(s(y1))) we obtained the following new rules:

F(true, s(s(y_1)), s(s(z1))) → F(gt(s(y_1), s(z1)), s(s(trunc(y_1))), s(s(s(z1))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Rewriting

                    ↳ NonInfProof

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

F(true, s(s(y_1)), s(s(z1))) → F(gt(s(y_1), s(z1)), s(s(trunc(y_1))), s(s(s(z1))))

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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
gt(s(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule F(true, s(s(y_1)), s(s(z1))) → F(gt(s(y_1), s(z1)), s(s(trunc(y_1))), s(s(s(z1)))) at position [0] we obtained the following new rules:

F(true, s(s(y_1)), s(s(z1))) → F(gt(y_1, z1), s(s(trunc(y_1))), s(s(s(z1))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                    ↳ NonInfProof

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

F(true, s(s(y_1)), s(s(z1))) → F(gt(y_1, z1), s(s(trunc(y_1))), s(s(s(z1))))

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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
gt(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(true, x, y) → F(gt(x, y), trunc(x), s(y)) the following chains were created:

We consider the chain F(true, x, y) → F(gt(x, y), trunc(x), s(y)), F(true, x, y) → F(gt(x, y), trunc(x), s(y)) which results in the following constraint:
- (1) (F(gt(x0, x1), trunc(x0), s(x1))=F(true, x2, x3) ⇒ F(true, x0, x1)≥F(gt(x0, x1), trunc(x0), s(x1)))
We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
- (2) (gt(x0, x1)=true ⇒ F(true, x0, x1)≥F(gt(x0, x1), trunc(x0), 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(x5), 0)≥F(gt(s(x5), 0), trunc(s(x5)), s(0)))
- (4) (gt(x6, x7)=true∧(gt(x6, x7)=true ⇒ F(true, x6, x7)≥F(gt(x6, x7), trunc(x6), s(x7))) ⇒ F(true, s(x6), s(x7))≥F(gt(s(x6), s(x7)), trunc(s(x6)), s(s(x7))))
We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
- (5) (F(true, s(x5), 0)≥F(gt(s(x5), 0), trunc(s(x5)), s(0)))
We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gt(x6, x7)=true ⇒ F(true, x6, x7)≥F(gt(x6, x7), trunc(x6), s(x7))) with σ = [ ] which results in the following new constraint:
- (6) (F(true, x6, x7)≥F(gt(x6, x7), trunc(x6), s(x7)) ⇒ F(true, s(x6), s(x7))≥F(gt(s(x6), s(x7)), trunc(s(x6)), s(s(x7))))

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

F(true, x, y) → F(gt(x, y), trunc(x), s(y))
- (F(true, s(x5), 0)≥F(gt(s(x5), 0), trunc(s(x5)), s(0)))
- (F(true, x6, x7)≥F(gt(x6, x7), trunc(x6), s(x7)) ⇒ F(true, s(x6), s(x7))≥F(gt(s(x6), s(x7)), trunc(s(x6)), 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 [21]:

POL(0) = 1
POL(F(x₁, x₂, x₃)) = -1 + 2·x₂ - x₃
POL(c) = -1
POL(false) = 1
POL(gt(x₁, x₂)) = 1 + x₂
POL(s(x₁)) = 1 + x₁
POL(true) = 0
POL(trunc(x₁)) = x₁

The following pairs are in P_>:

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

The following pairs are in P_bound:

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

The following rules are usable:

trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
trunc(0) → 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ 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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

gt(0, x0)
trunc(s(s(x0)))
gt(s(x0), s(x1))
trunc(0)
trunc(s(0))
gt(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.