Termination w.r.t. Q proof of /tmp/tmpzygC7N/quotminus.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

Q is empty.

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

MIN(min(X, Y), Z) → PLUS(Y, Z)
QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))
MIN(s(X), s(Y)) → MIN(X, Y)
QUOT(s(X), s(Y)) → MIN(X, Y)
MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))
PLUS(s(X), Y) → PLUS(X, Y)

The TRS R consists of the following rules:

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

Q is empty.
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:

MIN(min(X, Y), Z) → PLUS(Y, Z)
QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))
MIN(s(X), s(Y)) → MIN(X, Y)
QUOT(s(X), s(Y)) → MIN(X, Y)
MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))
PLUS(s(X), Y) → PLUS(X, Y)

The TRS R consists of the following rules:

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

Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

PLUS(s(X), Y) → PLUS(X, Y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

PLUS(s(X), Y) → PLUS(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:

PLUS(s(X), Y) → PLUS(X, Y)
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:

MIN(s(X), s(Y)) → MIN(X, Y)
MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

          ↳ QDP

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

MIN(s(X), s(Y)) → MIN(X, Y)
MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

MIN(s(X), s(Y)) → MIN(X, Y)
MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)

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
MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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:

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

Q is empty.
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 with max and min functions [25]:

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

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.