Termination w.r.t. Q proof of /tmp/tmpHsli2e/thiemann09.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:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), 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:

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

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), 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:

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

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

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 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), 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

                ↳ RuleRemovalProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

PLUS(s(x), y) → PLUS(y, x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(PLUS(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s(x₁)) = 1 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(x, s(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, s(y))
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

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:

ACK(s(x), 0) → ACK(x, s(0))
The graph contains the following edges 1 > 1
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
The graph contains the following edges 1 > 1
ACK(s(x), s(y)) → ACK(s(x), y)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))

From the DPs we obtained the following set of size-change graphs:

MINUS(s(x), s(y)) → MINUS(x, y)
The graph contains the following edges 1 > 1, 2 > 2
MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ Instantiation

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

QUOT(s(x), s(y), z) → QUOT(minus(p(ack(0, x)), y), s(y), s(z))

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule QUOT(s(x), s(y), z) → QUOT(minus(p(ack(0, x)), y), s(y), s(z)) we obtained the following new rules:

QUOT(s(x0), s(y_5), s(y_6)) → QUOT(minus(p(ack(0, x0)), y_5), s(y_5), s(s(y_6)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ Instantiation

              ↳ QDP

                ↳ Instantiation

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

QUOT(s(x0), s(y_5), s(y_6)) → QUOT(minus(p(ack(0, x0)), y_5), s(y_5), s(s(y_6)))

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule QUOT(s(x0), s(y_5), s(y_6)) → QUOT(minus(p(ack(0, x0)), y_5), s(y_5), s(s(y_6))) we obtained the following new rules:

QUOT(s(x0), s(y_5), s(s(y_6))) → QUOT(minus(p(ack(0, x0)), y_5), s(y_5), s(s(s(y_6))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ Instantiation

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ QDPOrderProof

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

QUOT(s(x0), s(y_5), s(s(y_6))) → QUOT(minus(p(ack(0, x0)), y_5), s(y_5), s(s(s(y_6))))

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), 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(x0), s(y_5), s(s(y_6))) → QUOT(minus(p(ack(0, x0)), y_5), s(y_5), s(s(s(y_6))))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( plus(x₁, x₂) ) = x₁ + x₂

POL( minus(x₁, x₂) ) = x₁

POL( QUOT(x₁, ..., x₃) ) = x₁ + 1

POL( s(x₁) ) = x₁ + 1

POL( p(x₁) ) = max{0, x₁ - 1}

POL( 0 ) = max{0, -1}

POL( ack(x₁, x₂) ) = x₂ + 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ Instantiation

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), 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.