Termination w.r.t. Q proof of /tmp/tmpVVcAXn/ex03.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:

minus₂(x, y) -> cond₃(ge₂(x, s₁(y)), x, y)
cond₃(false, x, y) -> 0
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
ge₂(u, 0) -> true
ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)

The set Q consists of the following terms:

ge₂(x0, 0)
minus₂(x0, x1)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
cond₃(true, x0, x1)
cond₃(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(x, y) -> cond₃(ge₂(x, s₁(y)), x, y)
cond₃(false, x, y) -> 0
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
ge₂(u, 0) -> true
ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)

The set Q consists of the following terms:

ge₂(x0, 0)
minus₂(x0, x1)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
cond₃(true, x0, x1)
cond₃(false, 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:

MINUS₂(x, y) -> COND₃(ge₂(x, s₁(y)), x, y)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(x, y) -> GE₂(x, s₁(y))
GE₂(s₁(u), s₁(v)) -> GE₂(u, v)

The TRS R consists of the following rules:

minus₂(x, y) -> cond₃(ge₂(x, s₁(y)), x, y)
cond₃(false, x, y) -> 0
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
ge₂(u, 0) -> true
ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)

The set Q consists of the following terms:

ge₂(x0, 0)
minus₂(x0, x1)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
cond₃(true, x0, x1)
cond₃(false, 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:

MINUS₂(x, y) -> COND₃(ge₂(x, s₁(y)), x, y)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(x, y) -> GE₂(x, s₁(y))
GE₂(s₁(u), s₁(v)) -> GE₂(u, v)

The TRS R consists of the following rules:

minus₂(x, y) -> cond₃(ge₂(x, s₁(y)), x, y)
cond₃(false, x, y) -> 0
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
ge₂(u, 0) -> true
ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)

The set Q consists of the following terms:

ge₂(x0, 0)
minus₂(x0, x1)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
cond₃(true, x0, x1)
cond₃(false, 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:

GE₂(s₁(u), s₁(v)) -> GE₂(u, v)

The TRS R consists of the following rules:

minus₂(x, y) -> cond₃(ge₂(x, s₁(y)), x, y)
cond₃(false, x, y) -> 0
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
ge₂(u, 0) -> true
ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)

The set Q consists of the following terms:

ge₂(x0, 0)
minus₂(x0, x1)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
cond₃(true, x0, x1)
cond₃(false, 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:

GE₂(s₁(u), s₁(v)) -> GE₂(u, v)

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

ge₂(x0, 0)
minus₂(x0, x1)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
cond₃(true, x0, x1)
cond₃(false, 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.

ge₂(x0, 0)
minus₂(x0, x1)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
cond₃(true, x0, x1)
cond₃(false, 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:

GE₂(s₁(u), s₁(v)) -> GE₂(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:

GE₂(s₁(u), s₁(v)) -> GE₂(u, v)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

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

MINUS₂(x, y) -> COND₃(ge₂(x, s₁(y)), x, y)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))

The TRS R consists of the following rules:

minus₂(x, y) -> cond₃(ge₂(x, s₁(y)), x, y)
cond₃(false, x, y) -> 0
cond₃(true, x, y) -> s₁(minus₂(x, s₁(y)))
ge₂(u, 0) -> true
ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)

The set Q consists of the following terms:

ge₂(x0, 0)
minus₂(x0, x1)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
cond₃(true, x0, x1)
cond₃(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

            ↳ UsableRulesProof

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

MINUS₂(x, y) -> COND₃(ge₂(x, s₁(y)), x, y)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))

The TRS R consists of the following rules:

ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)
ge₂(u, 0) -> true

The set Q consists of the following terms:

ge₂(x0, 0)
minus₂(x0, x1)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
cond₃(true, x0, x1)
cond₃(false, 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.

minus₂(x0, x1)
cond₃(true, x0, x1)
cond₃(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

            ↳ UsableRulesProof

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

MINUS₂(x, y) -> COND₃(ge₂(x, s₁(y)), x, y)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))

The TRS R consists of the following rules:

ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)
ge₂(u, 0) -> true

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule MINUS₂(x, y) -> COND₃(ge₂(x, s₁(y)), x, y) at position [0] we obtained the following new rules:

MINUS₂(0, x0) -> COND₃(false, 0, x0)
MINUS₂(s₁(x0), x1) -> COND₃(ge₂(x0, x1), s₁(x0), x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

MINUS₂(0, x0) -> COND₃(false, 0, x0)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(s₁(x0), x1) -> COND₃(ge₂(x0, x1), s₁(x0), x1)

The TRS R consists of the following rules:

ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)
ge₂(u, 0) -> true

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))

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

            ↳ UsableRulesProof

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

COND₃(true, x, y) -> MINUS₂(x, s₁(y))
MINUS₂(s₁(x0), x1) -> COND₃(ge₂(x0, x1), s₁(x0), x1)

The TRS R consists of the following rules:

ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)
ge₂(u, 0) -> true

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule MINUS₂(s₁(x0), x1) -> COND₃(ge₂(x0, x1), s₁(x0), x1) we obtained the following new rules:

MINUS₂(s₁(x0), s₁(z1)) -> COND₃(ge₂(x0, s₁(z1)), s₁(x0), s₁(z1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

            ↳ UsableRulesProof

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

MINUS₂(s₁(x0), s₁(z1)) -> COND₃(ge₂(x0, s₁(z1)), s₁(x0), s₁(z1))
COND₃(true, x, y) -> MINUS₂(x, s₁(y))

The TRS R consists of the following rules:

ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)
ge₂(u, 0) -> true

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND₃(true, x, y) -> MINUS₂(x, s₁(y)) we obtained the following new rules:

COND₃(true, s₁(z0), s₁(z1)) -> MINUS₂(s₁(z0), s₁(s₁(z1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

            ↳ UsableRulesProof

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

MINUS₂(s₁(x0), s₁(z1)) -> COND₃(ge₂(x0, s₁(z1)), s₁(x0), s₁(z1))
COND₃(true, s₁(z0), s₁(z1)) -> MINUS₂(s₁(z0), s₁(s₁(z1)))

The TRS R consists of the following rules:

ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)
ge₂(u, 0) -> true

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

MINUS₂(x, y) -> COND₃(ge₂(x, s₁(y)), x, y)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))

The TRS R consists of the following rules:

ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)
ge₂(u, 0) -> true

The set Q consists of the following terms:

ge₂(x0, 0)
minus₂(x0, x1)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))
cond₃(true, x0, x1)
cond₃(false, 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.

minus₂(x0, x1)
cond₃(true, x0, x1)
cond₃(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

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

MINUS₂(x, y) -> COND₃(ge₂(x, s₁(y)), x, y)
COND₃(true, x, y) -> MINUS₂(x, s₁(y))

The TRS R consists of the following rules:

ge₂(0, s₁(v)) -> false
ge₂(s₁(u), s₁(v)) -> ge₂(u, v)
ge₂(u, 0) -> true

The set Q consists of the following terms:

ge₂(x0, 0)
ge₂(0, s₁(x0))
ge₂(s₁(x0), s₁(x1))

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