Termination w.r.t. Q proof of /tmp/tmp4Q

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:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

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:

C(R(x1)) → B(a(R(x1)))
L¹(a(a(x1))) → C(x1)
C(a(x1)) → C(x1)
L¹(a(a(x1))) → L¹(a(b(c(x1))))
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
L¹(a(a(x1))) → B(c(x1))
C(b(x1)) → B(a(x1))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(R(x1)) → B(a(R(x1)))
L¹(a(a(x1))) → C(x1)
C(a(x1)) → C(x1)
L¹(a(a(x1))) → L¹(a(b(c(x1))))
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
L¹(a(a(x1))) → B(c(x1))
C(b(x1)) → B(a(x1))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(x1)) → C(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
C(b(x1)) → B(a(x1))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

            ↳ UsableRulesProof

          ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(x1)) → C(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
C(b(x1)) → B(a(x1))

The TRS R consists of the following rules:

c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
c(R(x1)) → b(a(R(x1)))
b(a(a(x1))) → a(b(c(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:

C(a(x1)) → C(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(B(x₁)) = 2·x₁
POL(C(x₁)) = 2 + 2·x₁
POL(R(x₁)) = x₁
POL(a(x₁)) = 1 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 1 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ DependencyGraphProof

            ↳ UsableRulesProof

          ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(b(x1)) → B(a(x1))

The TRS R consists of the following rules:

c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
c(R(x1)) → b(a(R(x1)))
b(a(a(x1))) → a(b(c(x1)))

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 1 less node.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

          ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(x1)) → C(x1)
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
C(b(x1)) → B(a(x1))

The TRS R consists of the following rules:

c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
c(R(x1)) → b(a(R(x1)))
b(a(a(x1))) → a(b(c(x1)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

L¹(a(a(x1))) → L¹(a(b(c(x1))))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

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

            ↳ MNOCProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

L¹(a(a(x1))) → L¹(a(b(c(x1))))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

The set Q consists of the following terms:

b(a(a(x0)))
c(a(x0))
c(b(x0))
L(a(a(x0)))
c(R(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

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

            ↳ MNOCProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

L¹(a(a(x1))) → L¹(a(b(c(x1))))

The TRS R consists of the following rules:

c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
c(R(x1)) → b(a(R(x1)))
b(a(a(x1))) → a(b(c(x1)))

The set Q consists of the following terms:

b(a(a(x0)))
c(a(x0))
c(b(x0))
L(a(a(x0)))
c(R(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.

L(a(a(x0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

            ↳ MNOCProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

L¹(a(a(x1))) → L¹(a(b(c(x1))))

The TRS R consists of the following rules:

c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
c(R(x1)) → b(a(R(x1)))
b(a(a(x1))) → a(b(c(x1)))

The set Q consists of the following terms:

b(a(a(x0)))
c(a(x0))
c(b(x0))
c(R(x0))

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

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

L¹(a(a(x1))) → L¹(a(b(c(x1))))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

The set Q consists of the following terms:

b(a(a(x0)))
c(a(x0))
c(b(x0))
L(a(a(x0)))
c(R(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

L¹(a(a(x1))) → L¹(a(b(c(x1))))

The TRS R consists of the following rules:

c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
c(R(x1)) → b(a(R(x1)))
b(a(a(x1))) → a(b(c(x1)))

The set Q consists of the following terms:

b(a(a(x0)))
c(a(x0))
c(b(x0))
L(a(a(x0)))
c(R(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.

L(a(a(x0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

L¹(a(a(x1))) → L¹(a(b(c(x1))))

The TRS R consists of the following rules:

c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
c(R(x1)) → b(a(R(x1)))
b(a(a(x1))) → a(b(c(x1)))

The set Q consists of the following terms:

b(a(a(x0)))
c(a(x0))
c(b(x0))
c(R(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.

L¹(a(a(x1))) → L¹(a(b(c(x1))))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( c(x₁) ) = x₁ + 1

POL( b(x₁) ) = max{0, x₁ - 1}

POL( L¹(x₁) ) = max{0, x₁ - 1}

POL( R(x₁) ) = 1

POL( a(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

c(a(x1)) → a(c(x1))
b(a(a(x1))) → a(b(c(x1)))
c(R(x1)) → b(a(R(x1)))
c(b(x1)) → b(a(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
c(R(x1)) → b(a(R(x1)))
b(a(a(x1))) → a(b(c(x1)))

The set Q consists of the following terms:

b(a(a(x0)))
c(a(x0))
c(b(x0))
c(R(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We have reversed the following QTRS:
The set of rules R is

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

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

a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
a(a(L(x))) → c(b(a(L(x))))
R(c(x)) → R(a(b(x)))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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

a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
a(a(L(x))) → c(b(a(L(x))))
R(c(x)) → R(a(b(x)))

Q is empty.

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

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

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

a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
a(a(L(x))) → c(b(a(L(x))))
R(c(x)) → R(a(b(x)))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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

a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
a(a(L(x))) → c(b(a(L(x))))
R(c(x)) → R(a(b(x)))

Q is empty.