Termination w.r.t. Q proof of TRS/Thiemanndiv

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:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

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:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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:

PLUS(0, s(x)) → PLUS(0, x)
GE(s(x), 0) → GE(x, 0)
IFY(true, x, y) → GE(x, y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
DIV(x, y) → GE(y, s(0))
GE(0, s(s(x))) → GE(0, s(x))
IF(true, x, y) → DIV(minus(x, y), y)
PLUS(s(x), y) → PLUS(x, y)
MINUS(0, s(x)) → MINUS(0, x)
MINUS(s(x), 0) → MINUS(x, 0)
IF(true, x, y) → MINUS(x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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:

PLUS(0, s(x)) → PLUS(0, x)
GE(s(x), 0) → GE(x, 0)
IFY(true, x, y) → GE(x, y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
DIV(x, y) → GE(y, s(0))
GE(0, s(s(x))) → GE(0, s(x))
IF(true, x, y) → DIV(minus(x, y), y)
PLUS(s(x), y) → PLUS(x, y)
MINUS(0, s(x)) → MINUS(0, x)
MINUS(s(x), 0) → MINUS(x, 0)
IF(true, x, y) → MINUS(x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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:

PLUS(0, s(x)) → PLUS(0, x)
IFY(true, x, y) → GE(x, y)
GE(s(x), 0) → GE(x, 0)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
DIV(x, y) → GE(y, s(0))
GE(0, s(s(x))) → GE(0, s(x))
IF(true, x, y) → DIV(minus(x, y), y)
PLUS(s(x), y) → PLUS(x, y)
MINUS(s(x), 0) → MINUS(x, 0)
MINUS(0, s(x)) → MINUS(0, x)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → MINUS(x, y)
GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 9 SCCs with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

PLUS(0, s(x)) → PLUS(0, x)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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.

PLUS(0, s(x)) → PLUS(0, x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = PLUS(x2)
0 = 0
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

0 > [PLUS₁, s_1]

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

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ 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:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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.

PLUS(s(x), y) → PLUS(x, y)

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

Lexicographic Path Order [19].
Precedence:

trivial

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

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

MINUS(s(x), 0) → MINUS(x, 0)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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), 0) → MINUS(x, 0)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = MINUS(x1)
s(x1) = s(x1)
0 = 0

Lexicographic Path Order [19].
Precedence:

s₁ > 0 > MINUS₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

MINUS(0, s(x)) → MINUS(0, x)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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(0, s(x)) → MINUS(0, x)

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)
0 = 0
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

0 > [MINUS₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ 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:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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(x1)
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

GE(0, s(s(x))) → GE(0, s(x))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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.

GE(0, s(s(x))) → GE(0, s(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
GE(x1, x2) = GE(x2)
0 = 0
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

0 > GE₁ > s₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

GE(s(x), 0) → GE(x, 0)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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.

GE(s(x), 0) → GE(x, 0)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
GE(x1, x2) = GE(x1)
s(x1) = s(x1)
0 = 0

Lexicographic Path Order [19].
Precedence:

s₁ > 0 > GE₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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.

GE(s(x), s(y)) → GE(x, y)

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

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

IF(true, x, y) → DIV(minus(x, y), y)
IFY(true, x, y) → IF(ge(x, y), x, y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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