Termination w.r.t. Q proof of TRS/AProVEfac.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(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 3 less nodes.

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

Q is empty.
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)

Lexicographic path order with status [19].
Precedence:

s₁ > P₁

Status:

P₁: [1]
s₁: [1]

The following usable rules [14] were oriented: none

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

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

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

Q is empty.
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(x, s(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) = PLUS(x1, x2)
s(x1) = s(x1)

Lexicographic path order with status [19].
Precedence:

s₁ > PLUS₂

Status:

s₁: multiset
PLUS₂: [1,2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

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

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

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

Q is empty.
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.

TIMES(s(x), y) → TIMES(x, y)

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

Lexicographic path order with status [19].
Precedence:

s₁ > TIMES₁

Status:

TIMES₁: [1]
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

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

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

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

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