Termination w.r.t. Q proof of /tmp/tmp7YBpk6/aprove5.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, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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:

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(s(x), s(y)) → P(s(x))
DIV(plus(x, y), z) → DIV(x, z)
MINUS(x, plus(y, z)) → MINUS(x, y)
PLUS(s(x), y) → MINUS(s(x), s(0))
DIV(plus(x, y), z) → PLUS(div(x, z), div(y, z))
DIV(s(x), s(y)) → MINUS(x, y)
MINUS(s(x), s(y)) → P(s(y))
MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
DIV(plus(x, y), z) → DIV(y, z)
P(s(s(x))) → P(s(x))
PLUS(s(x), y) → PLUS(y, minus(s(x), s(0)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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:

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(s(x), s(y)) → P(s(x))
DIV(plus(x, y), z) → DIV(x, z)
MINUS(x, plus(y, z)) → MINUS(x, y)
PLUS(s(x), y) → MINUS(s(x), s(0))
DIV(plus(x, y), z) → PLUS(div(x, z), div(y, z))
DIV(s(x), s(y)) → MINUS(x, y)
MINUS(s(x), s(y)) → P(s(y))
MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
DIV(plus(x, y), z) → DIV(y, z)
P(s(s(x))) → P(s(x))
PLUS(s(x), y) → PLUS(y, minus(s(x), s(0)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

P(s(s(x))) → P(s(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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.

P(s(s(x))) → P(s(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(P(x₁)) = (4)x_1
POL(s(x₁)) = 4 + (4)x_1

The value of delta used in the strict ordering is 64.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
MINUS(x, plus(y, z)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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.

MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
MINUS(x, plus(y, z)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = 1 + x_1 + (4)x_2
POL(minus(x₁, x₂)) = 0
POL(MINUS(x₁, x₂)) = (1/4)x_2
POL(s(x₁)) = 0
POL(p(x₁)) = 0
POL(0) = 0

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

p(s(s(x))) → s(p(s(x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(y, minus(s(x), s(0)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
DIV(plus(x, y), z) → DIV(x, z)
DIV(plus(x, y), z) → DIV(y, z)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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.

DIV(plus(x, y), z) → DIV(x, z)
DIV(plus(x, y), z) → DIV(y, z)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = 1 + (2)x_1 + (2)x_2
POL(minus(x₁, x₂)) = x_1
POL(DIV(x₁, x₂)) = (1/2)x_1
POL(s(x₁)) = x_1
POL(p(x₁)) = 0
POL(0) = 0

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = x_1
POL(DIV(x₁, x₂)) = (1/4)x_1
POL(s(x₁)) = 1/4 + x_1
POL(p(x₁)) = x_1
POL(0) = 0

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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.