Termination w.r.t. Q proof of /tmp/tmphscT8y/aprove5.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(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
e(x1) → a(x1)
b(d(x1)) → d(d(x1))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

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

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
e(x1) → a(x1)
b(d(x1)) → d(d(x1))

Q is empty.

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

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
e(x1) → a(x1)
b(d(x1)) → d(d(x1))

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

b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
c(a(x)) → b(a(c(x)))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
f(f(e(e(x)))) → e(e(f(f(f(x)))))
e(x) → a(x)
d(b(x)) → d(d(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

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

b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
c(a(x)) → b(a(c(x)))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
f(f(e(e(x)))) → e(e(f(f(f(x)))))
e(x) → a(x)
d(b(x)) → d(d(x))

Q is empty.

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

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
e(x1) → a(x1)
b(d(x1)) → d(d(x1))

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

b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
c(a(x)) → b(a(c(x)))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
f(f(e(e(x)))) → e(e(f(f(f(x)))))
e(x) → a(x)
d(b(x)) → d(d(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ RRRPoloQTRSProof

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

b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
c(a(x)) → b(a(c(x)))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
f(f(e(e(x)))) → e(e(f(f(f(x)))))
e(x) → a(x)
d(b(x)) → d(d(x))

Q is empty.

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

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
e(x1) → a(x1)
b(d(x1)) → d(d(x1))

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

e(x1) → a(x1)

Used ordering:
Polynomial interpretation [25]:

POL(a(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = x₁
POL(d(x₁)) = x₁
POL(e(x₁)) = 2 + x₁
POL(f(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
b(d(x1)) → d(d(x1))

Q is empty.

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

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
e(e(f(f(x1)))) → f(f(f(e(e(x1)))))
b(d(x1)) → d(d(x1))

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

e(e(f(f(x1)))) → f(f(f(e(e(x1)))))

Used ordering:
Polynomial interpretation [25]:

POL(a(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = x₁
POL(d(x₁)) = x₁
POL(e(x₁)) = 2 + 2·x₁
POL(f(x₁)) = 1 + x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

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

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

D(a(x1)) → D(d(c(x1)))
A(a(x1)) → A(d(d(d(x1))))
A(b(x1)) → B(c(a(x1)))
B(d(x1)) → D(d(x1))
A(d(d(c(x1)))) → D(x1)
A(c(x1)) → B(x1)
A(b(x1)) → A(x1)
B(c(x1)) → B(b(x1))
D(a(x1)) → D(c(x1))
A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))
A(a(x1)) → D(d(x1))
A(d(d(c(x1)))) → A(a(a(d(x1))))
A(a(x1)) → D(d(d(x1)))
A(a(x1)) → D(x1)
A(d(d(c(x1)))) → A(a(d(x1)))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
b(d(x1)) → d(d(x1))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

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

D(a(x1)) → D(d(c(x1)))
A(a(x1)) → A(d(d(d(x1))))
A(b(x1)) → B(c(a(x1)))
B(d(x1)) → D(d(x1))
A(d(d(c(x1)))) → D(x1)
A(c(x1)) → B(x1)
A(b(x1)) → A(x1)
B(c(x1)) → B(b(x1))
D(a(x1)) → D(c(x1))
A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))
A(a(x1)) → D(d(x1))
A(d(d(c(x1)))) → A(a(a(d(x1))))
A(a(x1)) → D(d(d(x1)))
A(a(x1)) → D(x1)
A(d(d(c(x1)))) → A(a(d(x1)))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
b(d(x1)) → d(d(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 9 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                  ↳ QDP

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

B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
b(d(x1)) → d(d(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ MNOCProof

                    ↳ UsableRulesProof

                  ↳ QDP

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

B(c(x1)) → B(x1)
B(c(x1)) → B(b(x1))

The TRS R consists of the following rules:

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))
d(a(x1)) → d(d(c(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ MNOCProof

                          ↳ QDP

                            ↳ UsableRulesProof

                    ↳ UsableRulesProof

                  ↳ QDP

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

B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))
d(a(x1)) → d(d(c(x1)))

The set Q consists of the following terms:

b(c(x0))
b(d(x0))
d(a(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ MNOCProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                    ↳ UsableRulesProof

                  ↳ QDP

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

B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))

The set Q consists of the following terms:

b(c(x0))
b(d(x0))
d(a(x0))

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:

B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(B(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 2 + 2·x₁
POL(d(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ MNOCProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ PisEmptyProof

                    ↳ UsableRulesProof

                  ↳ QDP

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

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))

The set Q consists of the following terms:

b(c(x0))
b(d(x0))
d(a(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 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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ MNOCProof

                  ↳ QDP

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

B(c(x1)) → B(x1)
B(c(x1)) → B(b(x1))

The TRS R consists of the following rules:

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))
d(a(x1)) → d(d(c(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ MNOCProof

                          ↳ QDP

                            ↳ UsableRulesProof

                  ↳ QDP

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

B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))
d(a(x1)) → d(d(c(x1)))

The set Q consists of the following terms:

b(c(x0))
b(d(x0))
d(a(x0))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ MNOCProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                  ↳ QDP

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

B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))

The set Q consists of the following terms:

b(c(x0))
b(d(x0))
d(a(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as they contain symbols which do neither occur in P nor in R.

d(a(x0))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ MNOCProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ MNOCProof

                  ↳ QDP

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

B(c(x1)) → B(x1)
B(c(x1)) → B(b(x1))

The TRS R consists of the following rules:

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))

The set Q consists of the following terms:

b(c(x0))
b(d(x0))

We have to consider all (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ MNOCProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ MNOCProof

                                      ↳ QDP

                  ↳ QDP

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

B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

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

A(a(x1)) → A(d(d(d(x1))))
A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))
A(d(d(c(x1)))) → A(a(a(d(x1))))
A(d(d(c(x1)))) → A(a(d(x1)))
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
b(d(x1)) → d(d(x1))

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

A(a(a(x0))) → A(d(d(d(d(c(x0))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

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

A(a(a(x0))) → A(d(d(d(d(c(x0))))))
A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))
A(d(d(c(x1)))) → A(a(a(d(x1))))
A(b(x1)) → A(x1)
A(d(d(c(x1)))) → A(a(d(x1)))

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
b(d(x1)) → d(d(x1))

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 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))
A(d(d(c(x1)))) → A(a(a(d(x1))))
A(d(d(c(x1)))) → A(a(d(x1)))
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
b(d(x1)) → d(d(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.

A(d(d(c(x1)))) → A(a(a(d(x1))))
A(d(d(c(x1)))) → A(a(d(x1)))

The remaining pairs can at least be oriented weakly.

A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))
A(b(x1)) → A(x1)

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

POL( A(x₁) ) = x₁

POL( c(x₁) ) = x₁

POL( d(x₁) ) = 1

POL( b(x₁) ) = x₁

POL( a(x₁) ) = max{0, -1}

The following usable rules [17] were oriented:

d(a(x1)) → d(d(c(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(b(x1)) → b(c(a(x1)))
a(a(x1)) → a(d(d(d(x1))))
a(d(d(c(x1)))) → a(a(a(d(x1))))
b(d(x1)) → d(d(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ UsableRulesProof

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

A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
a(c(x1)) → c(a(b(x1)))
a(a(x1)) → a(d(d(d(x1))))
d(a(x1)) → d(d(c(x1)))
a(d(d(c(x1)))) → a(a(a(d(x1))))
b(d(x1)) → d(d(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ UsableRulesReductionPairsProof

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

A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))
d(a(x1)) → d(d(c(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

A(c(x1)) → A(b(x1))
A(d(d(c(x1)))) → A(d(x1))

The following rules are removed from R:

d(a(x1)) → d(d(c(x1)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 2·x₁
POL(a(x₁)) = 2 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 2 + x₁
POL(d(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ UsableRulesReductionPairsProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

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

A(b(x1)) → A(x1)

The TRS R consists of the following rules:

b(c(x1)) → c(b(b(x1)))
b(d(x1)) → d(d(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ UsableRulesReductionPairsProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ UsableRulesReductionPairsProof

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

A(b(x1)) → A(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

A(b(x1)) → A(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ UsableRulesReductionPairsProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ UsableRulesReductionPairsProof

                                              ↳ QDP

                                                ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
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.