Termination w.r.t. Q proof of /tmp/tmpUBojD7/logarquot.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:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

LOG(s(s(X))) → QUOT(X, s(s(0)))
QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))
LOG(s(s(X))) → LOG(s(quot(X, s(s(0)))))
MIN(s(X), s(Y)) → MIN(X, Y)
QUOT(s(X), s(Y)) → MIN(X, Y)

The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

LOG(s(s(X))) → QUOT(X, s(s(0)))
QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))
LOG(s(s(X))) → LOG(s(quot(X, s(s(0)))))
MIN(s(X), s(Y)) → MIN(X, Y)
QUOT(s(X), s(Y)) → MIN(X, Y)

The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

MIN(s(X), s(Y)) → MIN(X, Y)

The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

MIN(s(X), s(Y)) → MIN(X, Y)

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

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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.

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

MIN(s(X), s(Y)) → MIN(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:

MIN(s(X), s(Y)) → MIN(X, Y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))

The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))

The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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.

quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

              ↳ QDP

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

QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))

The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(QUOT(x₁, x₂)) = x₁
POL(min(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)

The set Q consists of the following terms:

min(x0, 0)
min(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.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

LOG(s(s(X))) → LOG(s(quot(X, s(s(0)))))

The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

LOG(s(s(X))) → LOG(s(quot(X, s(s(0)))))

The TRS R consists of the following rules:

quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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.

log(s(0))
log(s(s(x0)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

LOG(s(s(X))) → LOG(s(quot(X, s(s(0)))))

The TRS R consists of the following rules:

quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

LOG(s(s(X))) → LOG(s(quot(X, s(s(0)))))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(LOG(x₁)) = x₁
POL(min(x₁, x₂)) = x₁
POL(quot(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(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.