Termination w.r.t. Q proof of /tmp/tmp6hUfYC/fac.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:

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 [1,15] 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)
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))
P(s(s(x))) → P(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

TIMES(s(x), y) → PLUS(times(x, y), y)
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))
P(s(s(x))) → P(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 [15,17,22] contains 4 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

          ↳ 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 modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

          ↳ 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))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

P(s(s(x))) → P(s(x))

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

P(s(s(x))) → P(s(x))

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:

P(s(s(x))) → P(s(x))
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

          ↳ 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 modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

          ↳ 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))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

PLUS(x, s(y)) → PLUS(x, y)

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

PLUS(x, s(y)) → PLUS(x, y)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

          ↳ 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 modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

          ↳ 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))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

          ↳ QDP

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

TIMES(s(x), y) → TIMES(x, y)

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

          ↳ QDP

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

TIMES(s(x), y) → TIMES(x, y)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

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.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

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))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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:

p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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:

p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0

The set Q consists of the following terms:

p(s(s(x0)))
p(s(0))

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 rules of the TRS R:

p(s(0)) → 0

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 2
POL(FAC(x₁)) = x₁
POL(p(x₁)) = x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

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:

p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(s(x0)))
p(s(0))

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.

FAC(s(x)) → FAC(p(s(x)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( p(x₁) ) =

/	1	\
\	0	/

/	1	1	\
\	1	0	/

x₁

M( s(x₁) ) =

/	0	\
\	1	/

/	0	1	\
\	0	1	/

x₁

Tuple symbols:

M( FAC(x₁) ) =

[

]

x₁

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

p(s(s(x))) → s(p(s(x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

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

p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(s(x0)))
p(s(0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.