Termination w.r.t. Q proof of TRS/Thiemannfactorial1.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:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(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:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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

FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → P(s(x))
P(s(s(x))) → P(s(x))
TIMES(s(x), y) → TIMES(p(s(x)), y)
TIMES(s(x), y) → P(s(x))
FACTORIAL(x) → FAC(x, s(0))
FAC(s(x), y) → P(s(x))
FAC(s(x), y) → TIMES(s(x), y)
TIMES(s(x), y) → PLUS(y, times(p(s(x)), y))

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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:

FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → P(s(x))
P(s(s(x))) → P(s(x))
TIMES(s(x), y) → TIMES(p(s(x)), y)
TIMES(s(x), y) → P(s(x))
FACTORIAL(x) → FAC(x, s(0))
FAC(s(x), y) → P(s(x))
FAC(s(x), y) → TIMES(s(x), y)
TIMES(s(x), y) → PLUS(y, times(p(s(x)), y))

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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(s(x), y) → PLUS(p(s(x)), y)
FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
PLUS(s(x), y) → P(s(x))
P(s(s(x))) → P(s(x))
TIMES(s(x), y) → TIMES(p(s(x)), y)
TIMES(s(x), y) → P(s(x))
FAC(s(x), y) → P(s(x))
FACTORIAL(x) → FAC(x, s(0))
FAC(s(x), y) → TIMES(s(x), y)
TIMES(s(x), y) → PLUS(y, times(p(s(x)), y))

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 6 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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.

P(s(s(x))) → P(s(x))

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

Recursive Path Order [2].
Precedence:

s₁ > P₁

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

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

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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

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

PLUS(s(x), y) → PLUS(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

TIMES(s(x), y) → TIMES(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

The set Q consists of the following terms:

plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)

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