Termination w.r.t. Q proof of TRS/Thiemannquot.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:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

QUOT(x, y) → IF_QUOT(minus(x, y), y, le(y, 0), le(y, x))
QUOT(x, y) → LE(y, x)
QUOT(x, y) → LE(y, 0)
LE(s(x), s(y)) → LE(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
IF_QUOT(x, y, false, true) → QUOT(x, y)
QUOT(x, y) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

QUOT(x, y) → IF_QUOT(minus(x, y), y, le(y, 0), le(y, x))
QUOT(x, y) → LE(y, x)
QUOT(x, y) → LE(y, 0)
LE(s(x), s(y)) → LE(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
IF_QUOT(x, y, false, true) → QUOT(x, y)
QUOT(x, y) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

QUOT(x, y) → IF_QUOT(minus(x, y), y, le(y, 0), le(y, x))
QUOT(x, y) → LE(y, 0)
QUOT(x, y) → LE(y, x)
LE(s(x), s(y)) → LE(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
IF_QUOT(x, y, false, true) → QUOT(x, y)
QUOT(x, y) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ 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:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LE(x1, x2) = LE(x2)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

s₁ > LE₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

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

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = MINUS(x2)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

s₁ > MINUS₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

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

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

QUOT(x, y) → IF_QUOT(minus(x, y), y, le(y, 0), le(y, x))
IF_QUOT(x, y, false, true) → QUOT(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

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