Termination w.r.t. Q proof of /tmp/tmpJnkDVV/ex10.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f₃(true, x, y) -> f₃(gt₂(x, y), trunc₁(x), s₁(y))
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(x0), 0)

↳ QTRS

  ↳ DependencyPairsProof

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

f₃(true, x, y) -> f₃(gt₂(x, y), trunc₁(x), s₁(y))
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(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:

TRUNC₁(s₁(s₁(x))) -> TRUNC₁(x)
F₃(true, x, y) -> GT₂(x, y)
F₃(true, x, y) -> F₃(gt₂(x, y), trunc₁(x), s₁(y))
GT₂(s₁(u), s₁(v)) -> GT₂(u, v)
F₃(true, x, y) -> TRUNC₁(x)

The TRS R consists of the following rules:

f₃(true, x, y) -> f₃(gt₂(x, y), trunc₁(x), s₁(y))
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(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:

TRUNC₁(s₁(s₁(x))) -> TRUNC₁(x)
F₃(true, x, y) -> GT₂(x, y)
F₃(true, x, y) -> F₃(gt₂(x, y), trunc₁(x), s₁(y))
GT₂(s₁(u), s₁(v)) -> GT₂(u, v)
F₃(true, x, y) -> TRUNC₁(x)

The TRS R consists of the following rules:

f₃(true, x, y) -> f₃(gt₂(x, y), trunc₁(x), s₁(y))
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(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:

GT₂(s₁(u), s₁(v)) -> GT₂(u, v)

The TRS R consists of the following rules:

f₃(true, x, y) -> f₃(gt₂(x, y), trunc₁(x), s₁(y))
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(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:

GT₂(s₁(u), s₁(v)) -> GT₂(u, v)

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

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(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.

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(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:

GT₂(s₁(u), s₁(v)) -> GT₂(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:

GT₂(s₁(u), s₁(v)) -> GT₂(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:

TRUNC₁(s₁(s₁(x))) -> TRUNC₁(x)

The TRS R consists of the following rules:

f₃(true, x, y) -> f₃(gt₂(x, y), trunc₁(x), s₁(y))
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(x0), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

TRUNC₁(s₁(s₁(x))) -> TRUNC₁(x)

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

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(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.

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(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:

TRUNC₁(s₁(s₁(x))) -> TRUNC₁(x)

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

TRUNC₁(s₁(s₁(x))) -> TRUNC₁(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

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

F₃(true, x, y) -> F₃(gt₂(x, y), trunc₁(x), s₁(y))

The TRS R consists of the following rules:

f₃(true, x, y) -> f₃(gt₂(x, y), trunc₁(x), s₁(y))
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))
gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(x0), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

            ↳ UsableRulesProof

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

F₃(true, x, y) -> F₃(gt₂(x, y), trunc₁(x), s₁(y))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(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.

f₃(true, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

            ↳ UsableRulesProof

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

F₃(true, x, y) -> F₃(gt₂(x, y), trunc₁(x), s₁(y))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule F₃(true, x, y) -> F₃(gt₂(x, y), trunc₁(x), s₁(y)) at position [0] we obtained the following new rules:

F₃(true, s₁(x0), s₁(x1)) -> F₃(gt₂(x0, x1), trunc₁(s₁(x0)), s₁(s₁(x1)))
F₃(true, s₁(x0), 0) -> F₃(true, trunc₁(s₁(x0)), s₁(0))
F₃(true, 0, x0) -> F₃(false, trunc₁(0), s₁(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

F₃(true, s₁(x0), 0) -> F₃(true, trunc₁(s₁(x0)), s₁(0))
F₃(true, s₁(x0), s₁(x1)) -> F₃(gt₂(x0, x1), trunc₁(s₁(x0)), s₁(s₁(x1)))
F₃(true, 0, x0) -> F₃(false, trunc₁(0), s₁(x0))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
gt₂(s₁(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

            ↳ UsableRulesProof

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

F₃(true, s₁(x0), s₁(x1)) -> F₃(gt₂(x0, x1), trunc₁(s₁(x0)), s₁(s₁(x1)))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule F₃(true, s₁(x0), s₁(x1)) -> F₃(gt₂(x0, x1), trunc₁(s₁(x0)), s₁(s₁(x1))) at position [1] we obtained the following new rules:

F₃(true, s₁(0), s₁(y1)) -> F₃(gt₂(0, y1), 0, s₁(s₁(y1)))
F₃(true, s₁(s₁(x0)), s₁(y1)) -> F₃(gt₂(s₁(x0), y1), s₁(s₁(trunc₁(x0))), s₁(s₁(y1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

F₃(true, s₁(0), s₁(y1)) -> F₃(gt₂(0, y1), 0, s₁(s₁(y1)))
F₃(true, s₁(s₁(x0)), s₁(y1)) -> F₃(gt₂(s₁(x0), y1), s₁(s₁(trunc₁(x0))), s₁(s₁(y1)))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
gt₂(s₁(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

            ↳ UsableRulesProof

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

F₃(true, s₁(s₁(x0)), s₁(y1)) -> F₃(gt₂(s₁(x0), y1), s₁(s₁(trunc₁(x0))), s₁(s₁(y1)))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule F₃(true, s₁(s₁(x0)), s₁(y1)) -> F₃(gt₂(s₁(x0), y1), s₁(s₁(trunc₁(x0))), s₁(s₁(y1))) we obtained the following new rules:

F₃(true, s₁(s₁(y_1)), s₁(s₁(z1))) -> F₃(gt₂(s₁(y_1), s₁(z1)), s₁(s₁(trunc₁(y_1))), s₁(s₁(s₁(z1))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Rewriting

            ↳ UsableRulesProof

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

F₃(true, s₁(s₁(y_1)), s₁(s₁(z1))) -> F₃(gt₂(s₁(y_1), s₁(z1)), s₁(s₁(trunc₁(y_1))), s₁(s₁(s₁(z1))))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule F₃(true, s₁(s₁(y_1)), s₁(s₁(z1))) -> F₃(gt₂(s₁(y_1), s₁(z1)), s₁(s₁(trunc₁(y_1))), s₁(s₁(s₁(z1)))) at position [0] we obtained the following new rules:

F₃(true, s₁(s₁(y_1)), s₁(s₁(z1))) -> F₃(gt₂(y_1, z1), s₁(s₁(trunc₁(y_1))), s₁(s₁(s₁(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

            ↳ UsableRulesProof

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

F₃(true, s₁(s₁(y_1)), s₁(s₁(z1))) -> F₃(gt₂(y_1, z1), s₁(s₁(trunc₁(y_1))), s₁(s₁(s₁(z1))))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
gt₂(s₁(x0), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

F₃(true, x, y) -> F₃(gt₂(x, y), trunc₁(x), s₁(y))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
f₃(true, x0, x1)
gt₂(s₁(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.

f₃(true, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

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

F₃(true, x, y) -> F₃(gt₂(x, y), trunc₁(x), s₁(y))

The TRS R consists of the following rules:

gt₂(0, v) -> false
gt₂(s₁(u), 0) -> true
gt₂(s₁(u), s₁(v)) -> gt₂(u, v)
trunc₁(0) -> 0
trunc₁(s₁(0)) -> 0
trunc₁(s₁(s₁(x))) -> s₁(s₁(trunc₁(x)))

The set Q consists of the following terms:

gt₂(0, x0)
trunc₁(s₁(s₁(x0)))
gt₂(s₁(x0), s₁(x1))
trunc₁(0)
trunc₁(s₁(0))
gt₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.