Termination w.r.t. Q proof of /tmp/tmpqbesxu/02.srs

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:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

The set Q is empty.
We have obtained the following QTRS:

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(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:

A¹(p(x1)) → A¹(A(x1))
P(A(A(x1))) → P(x1)
A¹(p(x1)) → P(a(A(x1)))
P(A(A(x1))) → A¹(p(x1))
A¹(A(x1)) → A¹(x1)

The TRS R consists of the following rules:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

A¹(p(x1)) → A¹(A(x1))
P(A(A(x1))) → P(x1)
A¹(p(x1)) → P(a(A(x1)))
P(A(A(x1))) → A¹(p(x1))
A¹(A(x1)) → A¹(x1)

The TRS R consists of the following rules:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

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:

A¹(p(x1)) → A¹(A(x1))

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(A¹(x₁)) = 2·x₁
POL(P(x₁)) = 2 + 2·x₁
POL(a(x₁)) = x₁
POL(p(x₁)) = 1 + x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

  ↳ QTRS Reverse

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

P(A(A(x1))) → P(x1)
A¹(p(x1)) → P(a(A(x1)))
A¹(A(x1)) → A¹(x1)
P(A(A(x1))) → A¹(p(x1))

The TRS R consists of the following rules:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

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.

A¹(A(x1)) → A¹(x1)

The remaining pairs can at least be oriented weakly.

P(A(A(x1))) → P(x1)
A¹(p(x1)) → P(a(A(x1)))
P(A(A(x1))) → A¹(p(x1))

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x₁) ) = x₁ + 1

POL( p(x₁) ) = max{0, -1}

POL( P(x₁) ) = 1

POL( A¹(x₁) ) = x₁ + 1

POL( a(x₁) ) = x₁

The following usable rules [17] were oriented:

a(A(x1)) → A(a(x1))
a(p(x1)) → p(a(A(x1)))
p(A(A(x1))) → a(p(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

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

P(A(A(x1))) → P(x1)
A¹(p(x1)) → P(a(A(x1)))
P(A(A(x1))) → A¹(p(x1))

The TRS R consists of the following rules:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A¹(p(x1)) → P(a(A(x1))) at position [0] we obtained the following new rules:

A¹(p(x0)) → P(A(a(x0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

  ↳ QTRS Reverse

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

A¹(p(x0)) → P(A(a(x0)))
P(A(A(x1))) → P(x1)
P(A(A(x1))) → A¹(p(x1))

The TRS R consists of the following rules:

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

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

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))
A¹(p(x0)) → P(A(a(x0)))
P(A(A(x1))) → P(x1)
P(A(A(x1))) → A¹(p(x1))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))
A¹(p(x0)) → P(A(a(x0)))
P(A(A(x1))) → P(x1)
P(A(A(x1))) → A¹(p(x1))

The set Q is empty.
We have obtained the following QTRS:

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))
p(A¹(x)) → a(A(P(x)))
A(A(P(x))) → P(x)
A(A(P(x))) → p(A¹(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ RRRPoloQTRSProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

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

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))
p(A¹(x)) → a(A(P(x)))
A(A(P(x))) → P(x)
A(A(P(x))) → p(A¹(x))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))
p(A¹(x)) → a(A(P(x)))
A(A(P(x))) → P(x)
A(A(P(x))) → p(A¹(x))

The set Q is empty.
We have obtained the following QTRS:

a(p(x)) → p(a(A(x)))
a(A(x)) → A(a(x))
p(A(A(x))) → a(p(x))
A¹(p(x)) → P(A(a(x)))
P(A(A(x))) → P(x)
P(A(A(x))) → A¹(p(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                          ↳ RRRPoloQTRSProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

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

a(p(x)) → p(a(A(x)))
a(A(x)) → A(a(x))
p(A(A(x))) → a(p(x))
A¹(p(x)) → P(A(a(x)))
P(A(A(x))) → P(x)
P(A(A(x))) → A¹(p(x))

Q is empty.

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

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))
p(A¹(x)) → a(A(P(x)))
A(A(P(x))) → P(x)
A(A(P(x))) → p(A¹(x))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

A(A(P(x))) → P(x)

Used ordering:
Polynomial interpretation [25]:

POL(A(x₁)) = 2 + x₁
POL(A¹(x₁)) = 2 + x₁
POL(P(x₁)) = 2·x₁
POL(a(x₁)) = 2 + x₁
POL(p(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ RRRPoloQTRSProof

                            ↳ QTRS

                              ↳ DependencyPairsProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

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

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))
p(A¹(x)) → a(A(P(x)))
A(A(P(x))) → p(A¹(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:

P¹(A¹(x)) → A²(P(x))
A²(a(x)) → A²(x)
P¹(a(x)) → P¹(x)
A²(A(P(x))) → P¹(A¹(x))
A²(A(p(x))) → P¹(a(x))
P¹(a(x)) → A²(a(p(x)))

The TRS R consists of the following rules:

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))
p(A¹(x)) → a(A(P(x)))
A(A(P(x))) → p(A¹(x))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ RRRPoloQTRSProof

                            ↳ QTRS

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

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

P¹(A¹(x)) → A²(P(x))
A²(a(x)) → A²(x)
P¹(a(x)) → P¹(x)
A²(A(P(x))) → P¹(A¹(x))
A²(A(p(x))) → P¹(a(x))
P¹(a(x)) → A²(a(p(x)))

The TRS R consists of the following rules:

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))
p(A¹(x)) → a(A(P(x)))
A(A(P(x))) → p(A¹(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ RRRPoloQTRSProof

                            ↳ QTRS

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A²(a(x)) → A²(x)
P¹(a(x)) → P¹(x)
P¹(a(x)) → A²(a(p(x)))
A²(A(p(x))) → P¹(a(x))

The TRS R consists of the following rules:

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))
p(A¹(x)) → a(A(P(x)))
A(A(P(x))) → p(A¹(x))

A²(a(x)) → A²(x)
P¹(a(x)) → P¹(x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 1 + x₁
POL(A¹(x₁)) = 1 + x₁
POL(A²(x₁)) = 2 + x₁
POL(P(x₁)) = 2·x₁
POL(P¹(x₁)) = 1 + 2·x₁
POL(a(x₁)) = 1 + x₁
POL(p(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ RRRPoloQTRSProof

                            ↳ QTRS

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A²(A(p(x))) → P¹(a(x))
P¹(a(x)) → A²(a(p(x)))

The TRS R consists of the following rules:

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))
p(A¹(x)) → a(A(P(x)))
A(A(P(x))) → p(A¹(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.
We have reversed the following QTRS:
The set of rules R is

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))
p(A¹(x)) → a(A(P(x)))
A(A(P(x))) → P(x)
A(A(P(x))) → p(A¹(x))

The set Q is empty.
We have obtained the following QTRS:

a(p(x)) → p(a(A(x)))
a(A(x)) → A(a(x))
p(A(A(x))) → a(p(x))
A¹(p(x)) → P(A(a(x)))
P(A(A(x))) → P(x)
P(A(A(x))) → A¹(p(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ RRRPoloQTRSProof

                          ↳ QTRS Reverse

                            ↳ QTRS

  ↳ QTRS Reverse

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

a(p(x)) → p(a(A(x)))
a(A(x)) → A(a(x))
p(A(A(x))) → a(p(x))
A¹(p(x)) → P(A(a(x)))
P(A(A(x))) → P(x)
P(A(A(x))) → A¹(p(x))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(p(x1)) → p(a(A(x1)))
a(A(x1)) → A(a(x1))
p(A(A(x1))) → a(p(x1))

The set Q is empty.
We have obtained the following QTRS:

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

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

p(a(x)) → A(a(p(x)))
A(a(x)) → a(A(x))
A(A(p(x))) → p(a(x))

Q is empty.