Termination w.r.t. Q proof of /tmp/tmp9yUebe/#3.8b.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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
LOG(s(s(x))) → QUOT(x, s(s(0)))
MINUS(s(x), y) → LE(s(x), y)
LOG(s(s(x))) → LOG(s(quot(x, s(s(0)))))
IF_MINUS(false, s(x), y) → MINUS(x, y)
MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
LE(s(x), s(y)) → LE(x, y)
QUOT(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
LOG(s(s(x))) → QUOT(x, s(s(0)))
MINUS(s(x), y) → LE(s(x), y)
LOG(s(s(x))) → LOG(s(quot(x, s(s(0)))))
IF_MINUS(false, s(x), y) → MINUS(x, y)
MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
LE(s(x), s(y)) → LE(x, y)
QUOT(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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 4 SCCs with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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

              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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

              ↳ QDP

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

LE(s(x), s(y)) → LE(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:

LE(s(x), s(y)) → LE(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

              ↳ QDP

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

IF_MINUS(false, s(x), y) → MINUS(x, y)
MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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

              ↳ QDP

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

IF_MINUS(false, s(x), y) → MINUS(x, y)
MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
le(0, y) → true

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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.

minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
IF_MINUS(false, s(x), y) → MINUS(x, y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
le(0, y) → true

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(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:

MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
The graph contains the following edges 1 >= 2, 2 >= 3
IF_MINUS(false, s(x), y) → MINUS(x, y)
The graph contains the following edges 2 > 1, 3 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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

              ↳ QDP

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
le(0, y) → true

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

              ↳ QDP

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
le(0, y) → true

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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(minus(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(false) = 0
POL(if_minus(x₁, x₂, x₃)) = x₂
POL(le(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following usable rules [17] were oriented:

if_minus(false, s(x), y) → s(minus(x, y))
if_minus(true, s(x), y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
minus(0, y) → 0

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

              ↳ QDP

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

minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
le(0, y) → true

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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

              ↳ 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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

              ↳ 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(minus(x, y), s(y)))
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
le(0, y) → true

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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

              ↳ 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(minus(x, y), s(y)))
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
le(0, y) → true

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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(false) = 0
POL(if_minus(x₁, x₂, x₃)) = x₂
POL(le(x₁, x₂)) = 1
POL(minus(x₁, x₂)) = x₁
POL(quot(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁
POL(true) = 1

The following usable rules [17] were oriented:

minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
minus(0, y) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
quot(0, s(y)) → 0
if_minus(false, s(x), y) → s(minus(x, y))
if_minus(true, s(x), y) → 0

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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(minus(x, y), s(y)))
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
le(0, y) → true

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), 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.