Termination w.r.t. Q proof of /tmp/tmp2GswIM/tpa2.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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The TRS R 2 is

f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The signature Sigma is {f}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

-¹(s(x), s(y)) → -¹(x, y)
F(x, s(y)) → P(-(s(y), x))
F(x, s(y)) → P(-(x, s(y)))
F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → -¹(y, s(x))
F(s(x), y) → -¹(s(x), y)
F(x, s(y)) → -¹(s(y), x)
F(x, s(y)) → -¹(x, s(y))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x), y) → P(-(y, s(x)))
F(s(x), y) → P(-(s(x), y))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

-¹(s(x), s(y)) → -¹(x, y)
F(x, s(y)) → P(-(s(y), x))
F(x, s(y)) → P(-(x, s(y)))
F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → -¹(y, s(x))
F(s(x), y) → -¹(s(x), y)
F(x, s(y)) → -¹(s(y), x)
F(x, s(y)) → -¹(x, s(y))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x), y) → P(-(y, s(x)))
F(s(x), y) → P(-(s(x), y))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 8 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, y)

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

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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.

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, 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:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x
f(s(x), y) → f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) → f(p(-(x, s(y))), p(-(s(y), x)))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))
f(s(x0), x1)
f(x0, s(x1))

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.

f(s(x0), x1)
f(x0, s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))

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.

F(s(x), y) → F(p(-(s(x), y)), p(-(y, s(x))))

The remaining pairs can at least be oriented weakly.

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))

Used ordering: Polynomial interpretation [25,35]:

POL(-(x₁, x₂)) = (2)x_1
POL(s(x₁)) = 1/4 + (4)x_1
POL(p(x₁)) = (1/4)x_1
POL(F(x₁, x₂)) = (1/2)x_1
POL(0) = 0

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

p(s(x)) → x
-(s(x), s(y)) → -(x, y)
-(x, 0) → x

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))

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.

F(x, s(y)) → F(p(-(x, s(y))), p(-(s(y), x)))

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

M( -(x₁, x₂) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

/	0	0	\
\	0	0	/

x₂

M( p(x₁) ) =

/	0	\
\	0	/

/	0	1	\
\	1	0	/

x₁

M( s(x₁) ) =

/	1	\
\	0	/

/	1	1	\
\	1	1	/

x₁

M( 0 ) =

/	0	\
\	0	/

Tuple symbols:

M( F(x₁, x₂) ) =

[

]

x₁

[

]

x₂

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
p(s(x)) → x

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
p(s(x0))

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