Termination w.r.t. Q proof of /tmp/tmpxTzdGs/1.srs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

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

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

Q is empty.

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

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

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

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

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

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))

Q is empty.

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

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

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

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ Overlay + Local Confluence

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

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

            ↳ QDP

          ↳ Narrowing

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

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

The TRS R consists of the following rules:

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

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

B(c(c(x0))) → B(a(b(b(x0))))
B(c(a(x0))) → B(a(c(b(x0))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

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

B(c(c(x0))) → B(a(b(b(x0))))
B(c(a(x0))) → B(a(c(b(x0))))
B(a(x1)) → B(x1)
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B(c(x1)) → B(x1) we obtained the following new rules:

B(c(c(y_0))) → B(c(y_0))
B(c(c(c(y_0)))) → B(c(c(y_0)))
B(c(a(y_0))) → B(a(y_0))
B(c(c(a(y_0)))) → B(c(a(y_0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

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

B(c(c(y_0))) → B(c(y_0))
B(c(c(c(y_0)))) → B(c(c(y_0)))
B(c(c(x0))) → B(a(b(b(x0))))
B(c(a(y_0))) → B(a(y_0))
B(a(x1)) → B(x1)
B(c(a(x0))) → B(a(c(b(x0))))
B(c(c(a(y_0)))) → B(c(a(y_0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B(a(x1)) → B(x1) we obtained the following new rules:

B(a(c(c(y_0)))) → B(c(c(y_0)))
B(a(a(y_0))) → B(a(y_0))
B(a(c(c(c(y_0))))) → B(c(c(c(y_0))))
B(a(c(a(y_0)))) → B(c(a(y_0)))
B(a(c(c(a(y_0))))) → B(c(c(a(y_0))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

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

B(a(a(y_0))) → B(a(y_0))
B(a(c(c(y_0)))) → B(c(c(y_0)))
B(a(c(c(c(y_0))))) → B(c(c(c(y_0))))
B(c(c(y_0))) → B(c(y_0))
B(c(c(x0))) → B(a(b(b(x0))))
B(c(c(c(y_0)))) → B(c(c(y_0)))
B(a(c(a(y_0)))) → B(c(a(y_0)))
B(c(a(x0))) → B(a(c(b(x0))))
B(c(a(y_0))) → B(a(y_0))
B(a(c(c(a(y_0))))) → B(c(c(a(y_0))))
B(c(c(a(y_0)))) → B(c(a(y_0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.c: 1
B: 0
a: 0
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

B.1(c.0(a.0(x0))) → B.0(a.1(c.0(b.0(x0))))
B.1(c.1(c.0(a.1(y_0)))) → B.1(c.0(a.1(y_0)))
B.0(a.1(c.1(c.0(a.0(y_0))))) → B.1(c.1(c.0(a.0(y_0))))
B.1(c.1(c.1(c.0(y_0)))) → B.1(c.1(c.0(y_0)))
B.0(a.1(c.0(a.1(y_0)))) → B.1(c.0(a.1(y_0)))
B.0(a.1(c.0(a.0(y_0)))) → B.1(c.0(a.0(y_0)))
B.1(c.1(c.1(x0))) → B.0(a.0(b.0(b.1(x0))))
B.1(c.1(c.1(y_0))) → B.1(c.1(y_0))
B.1(c.1(c.1(c.1(y_0)))) → B.1(c.1(c.1(y_0)))
B.0(a.1(c.1(c.0(y_0)))) → B.1(c.1(c.0(y_0)))
B.1(c.0(a.1(y_0))) → B.0(a.1(y_0))
B.1(c.0(a.0(y_0))) → B.0(a.0(y_0))
B.1(c.1(c.0(a.0(y_0)))) → B.1(c.0(a.0(y_0)))
B.0(a.1(c.1(c.1(c.1(y_0))))) → B.1(c.1(c.1(c.1(y_0))))
B.0(a.1(c.1(c.0(a.1(y_0))))) → B.1(c.1(c.0(a.1(y_0))))
B.0(a.1(c.1(c.1(y_0)))) → B.1(c.1(c.1(y_0)))
B.0(a.0(a.1(y_0))) → B.0(a.1(y_0))
B.1(c.1(c.0(x0))) → B.0(a.0(b.0(b.0(x0))))
B.0(a.1(c.1(c.1(c.0(y_0))))) → B.1(c.1(c.1(c.0(y_0))))
B.0(a.0(a.0(y_0))) → B.0(a.0(y_0))
B.1(c.0(a.1(x0))) → B.0(a.1(c.0(b.1(x0))))
B.1(c.1(c.0(y_0))) → B.1(c.0(y_0))

The TRS R consists of the following rules:

b.1(c.1(x1)) → a.0(b.0(b.1(x1)))
b.1(c.0(x1)) → a.0(b.0(b.0(x1)))
b.0(a.1(x1)) → a.1(c.0(b.1(x1)))
b.0(a.0(x1)) → a.1(c.0(b.0(x1)))

The set Q consists of the following terms:

b.1(c.0(x0))
b.1(c.1(x0))
b.0(a.0(x0))
b.0(a.1(x0))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                      ↳ SemLabProof2

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

B.1(c.0(a.0(x0))) → B.0(a.1(c.0(b.0(x0))))
B.1(c.1(c.0(a.1(y_0)))) → B.1(c.0(a.1(y_0)))
B.0(a.1(c.1(c.0(a.0(y_0))))) → B.1(c.1(c.0(a.0(y_0))))
B.1(c.1(c.1(c.0(y_0)))) → B.1(c.1(c.0(y_0)))
B.0(a.1(c.0(a.1(y_0)))) → B.1(c.0(a.1(y_0)))
B.0(a.1(c.0(a.0(y_0)))) → B.1(c.0(a.0(y_0)))
B.1(c.1(c.1(x0))) → B.0(a.0(b.0(b.1(x0))))
B.1(c.1(c.1(y_0))) → B.1(c.1(y_0))
B.1(c.1(c.1(c.1(y_0)))) → B.1(c.1(c.1(y_0)))
B.0(a.1(c.1(c.0(y_0)))) → B.1(c.1(c.0(y_0)))
B.1(c.0(a.1(y_0))) → B.0(a.1(y_0))
B.1(c.0(a.0(y_0))) → B.0(a.0(y_0))
B.1(c.1(c.0(a.0(y_0)))) → B.1(c.0(a.0(y_0)))
B.0(a.1(c.1(c.1(c.1(y_0))))) → B.1(c.1(c.1(c.1(y_0))))
B.0(a.1(c.1(c.0(a.1(y_0))))) → B.1(c.1(c.0(a.1(y_0))))
B.0(a.1(c.1(c.1(y_0)))) → B.1(c.1(c.1(y_0)))
B.0(a.0(a.1(y_0))) → B.0(a.1(y_0))
B.1(c.1(c.0(x0))) → B.0(a.0(b.0(b.0(x0))))
B.0(a.1(c.1(c.1(c.0(y_0))))) → B.1(c.1(c.1(c.0(y_0))))
B.0(a.0(a.0(y_0))) → B.0(a.0(y_0))
B.1(c.0(a.1(x0))) → B.0(a.1(c.0(b.1(x0))))
B.1(c.1(c.0(y_0))) → B.1(c.0(y_0))

The TRS R consists of the following rules:

b.1(c.1(x1)) → a.0(b.0(b.1(x1)))
b.1(c.0(x1)) → a.0(b.0(b.0(x1)))
b.0(a.1(x1)) → a.1(c.0(b.1(x1)))
b.0(a.0(x1)) → a.1(c.0(b.0(x1)))

The set Q consists of the following terms:

b.1(c.0(x0))
b.1(c.1(x0))
b.0(a.0(x0))
b.0(a.1(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.1(c.1(c.0(a.1(y_0)))) → B.1(c.0(a.1(y_0)))
B.1(c.1(c.1(c.0(y_0)))) → B.1(c.1(c.0(y_0)))
B.1(c.1(c.1(x0))) → B.0(a.0(b.0(b.1(x0))))
B.1(c.1(c.1(y_0))) → B.1(c.1(y_0))
B.1(c.1(c.1(c.1(y_0)))) → B.1(c.1(c.1(y_0)))
B.1(c.1(c.0(a.0(y_0)))) → B.1(c.0(a.0(y_0)))
B.1(c.1(c.0(x0))) → B.0(a.0(b.0(b.0(x0))))
B.1(c.1(c.0(y_0))) → B.1(c.0(y_0))

Strictly oriented rules of the TRS R:

b.1(c.1(x1)) → a.0(b.0(b.1(x1)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(B.0(x₁)) = 1 + x₁
POL(B.1(x₁)) = 1 + x₁
POL(a.0(x₁)) = x₁
POL(a.1(x₁)) = x₁
POL(b.0(x₁)) = x₁
POL(b.1(x₁)) = x₁
POL(c.0(x₁)) = x₁
POL(c.1(x₁)) = 1 + x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                      ↳ SemLabProof2

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

B.1(c.0(a.0(x0))) → B.0(a.1(c.0(b.0(x0))))
B.0(a.1(c.1(c.0(a.0(y_0))))) → B.1(c.1(c.0(a.0(y_0))))
B.0(a.1(c.0(a.1(y_0)))) → B.1(c.0(a.1(y_0)))
B.0(a.1(c.0(a.0(y_0)))) → B.1(c.0(a.0(y_0)))
B.0(a.1(c.1(c.0(y_0)))) → B.1(c.1(c.0(y_0)))
B.1(c.0(a.1(y_0))) → B.0(a.1(y_0))
B.1(c.0(a.0(y_0))) → B.0(a.0(y_0))
B.0(a.1(c.1(c.1(c.1(y_0))))) → B.1(c.1(c.1(c.1(y_0))))
B.0(a.1(c.1(c.0(a.1(y_0))))) → B.1(c.1(c.0(a.1(y_0))))
B.0(a.1(c.1(c.1(y_0)))) → B.1(c.1(c.1(y_0)))
B.0(a.0(a.1(y_0))) → B.0(a.1(y_0))
B.0(a.1(c.1(c.1(c.0(y_0))))) → B.1(c.1(c.1(c.0(y_0))))
B.0(a.0(a.0(y_0))) → B.0(a.0(y_0))
B.1(c.0(a.1(x0))) → B.0(a.1(c.0(b.1(x0))))

The TRS R consists of the following rules:

b.1(c.0(x1)) → a.0(b.0(b.0(x1)))
b.0(a.1(x1)) → a.1(c.0(b.1(x1)))
b.0(a.0(x1)) → a.1(c.0(b.0(x1)))

The set Q consists of the following terms:

b.1(c.0(x0))
b.1(c.1(x0))
b.0(a.0(x0))
b.0(a.1(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 6 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                      ↳ SemLabProof2

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

B.1(c.0(a.0(y_0))) → B.0(a.0(y_0))
B.1(c.0(a.1(y_0))) → B.0(a.1(y_0))
B.1(c.0(a.0(x0))) → B.0(a.1(c.0(b.0(x0))))
B.0(a.1(c.0(a.1(y_0)))) → B.1(c.0(a.1(y_0)))
B.0(a.1(c.0(a.0(y_0)))) → B.1(c.0(a.0(y_0)))
B.0(a.0(a.1(y_0))) → B.0(a.1(y_0))
B.0(a.0(a.0(y_0))) → B.0(a.0(y_0))
B.1(c.0(a.1(x0))) → B.0(a.1(c.0(b.1(x0))))

The TRS R consists of the following rules:

b.1(c.0(x1)) → a.0(b.0(b.0(x1)))
b.0(a.1(x1)) → a.1(c.0(b.1(x1)))
b.0(a.0(x1)) → a.1(c.0(b.0(x1)))

The set Q consists of the following terms:

b.1(c.0(x0))
b.1(c.1(x0))
b.0(a.0(x0))
b.0(a.1(x0))

We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used. Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

                        ↳ QDP

                          ↳ QDPToSRSProof

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

B(a(a(y_0))) → B(a(y_0))
B(a(c(a(y_0)))) → B(c(a(y_0)))
B(c(a(x0))) → B(a(c(b(x0))))
B(c(a(y_0))) → B(a(y_0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

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

b(c(x1)) → a(b(b(x1)))
b(a(x1)) → a(c(b(x1)))
B(a(a(y_0))) → B(a(y_0))
B(a(c(a(y_0)))) → B(c(a(y_0)))
B(c(a(x0))) → B(a(c(b(x0))))
B(c(a(y_0))) → B(a(y_0))

The set Q consists of the following terms:

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

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

b(c(x1)) → a(b(b(x1)))
b(a(x1)) → a(c(b(x1)))
B(a(a(y_0))) → B(a(y_0))
B(a(c(a(y_0)))) → B(c(a(y_0)))
B(c(a(x0))) → B(a(c(b(x0))))
B(c(a(y_0))) → B(a(y_0))

The set Q is {b(c(x0)), b(a(x0))}.
We have obtained the following QTRS:

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

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

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(x))

Q is empty.

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

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(x))

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

b(c(x)) → a(b(b(x)))
b(a(x)) → a(c(b(x)))
B(a(a(x))) → B(a(x))
B(a(c(a(x)))) → B(c(a(x)))
B(c(a(x))) → B(a(c(b(x))))
B(c(a(x))) → B(a(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

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

b(c(x)) → a(b(b(x)))
b(a(x)) → a(c(b(x)))
B(a(a(x))) → B(a(x))
B(a(c(a(x)))) → B(c(a(x)))
B(c(a(x))) → B(a(c(b(x))))
B(c(a(x))) → B(a(x))

Q is empty.

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

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(x))

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

b(c(x)) → a(b(b(x)))
b(a(x)) → a(c(b(x)))
B(a(a(x))) → B(a(x))
B(a(c(a(x)))) → B(c(a(x)))
B(c(a(x))) → B(a(c(b(x))))
B(c(a(x))) → B(a(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                  ↳ DependencyPairsProof

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

b(c(x)) → a(b(b(x)))
b(a(x)) → a(c(b(x)))
B(a(a(x))) → B(a(x))
B(a(c(a(x)))) → B(c(a(x)))
B(c(a(x))) → B(a(c(b(x))))
B(c(a(x))) → B(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:

C(b(x)) → A(x)
A(c(B(x))) → A(B(x))
A(c(B(x))) → C(a(B(x)))
A(c(a(B(x)))) → C(B(x))
A(c(a(B(x)))) → A(c(B(x)))
A(b(x)) → C(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(x))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

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

C(b(x)) → A(x)
A(c(B(x))) → A(B(x))
A(c(B(x))) → C(a(B(x)))
A(c(a(B(x)))) → C(B(x))
A(c(a(B(x)))) → A(c(B(x)))
A(b(x)) → C(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(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 4 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                          ↳ NonTerminationProof

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

C(b(x)) → A(x)
A(b(x)) → C(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(x))

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

A(b(b(x0))) → C(b(c(a(x0))))
A(b(c(a(B(x0))))) → C(a(c(B(x0))))
A(b(c(B(x0)))) → C(a(B(x0)))
A(b(c(B(x0)))) → C(b(c(a(B(x0)))))
A(b(a(B(x0)))) → C(a(B(x0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                          ↳ NonTerminationProof

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

C(b(x)) → A(x)
A(b(c(B(x0)))) → C(b(c(a(B(x0)))))
A(b(b(x0))) → C(b(c(a(x0))))
A(b(c(a(B(x0))))) → C(a(c(B(x0))))
A(b(c(B(x0)))) → C(a(B(x0)))
A(b(a(B(x0)))) → C(a(B(x0)))
A(b(x)) → A(x)

The TRS R consists of the following rules:

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(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

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ QDPOrderProof

                                          ↳ NonTerminationProof

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

C(b(x)) → A(x)
A(b(c(B(x0)))) → C(b(c(a(B(x0)))))
A(b(b(x0))) → C(b(c(a(x0))))
A(b(c(a(B(x0))))) → C(a(c(B(x0))))
A(b(x)) → A(x)

The TRS R consists of the following rules:

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(x))

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(b(c(B(x0)))) → C(b(c(a(B(x0)))))

The remaining pairs can at least be oriented weakly.

C(b(x)) → A(x)
A(b(b(x0))) → C(b(c(a(x0))))
A(b(c(a(B(x0))))) → C(a(c(B(x0))))
A(b(x)) → A(x)

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

POL( A(x₁) ) = x₁

POL( C(x₁) ) = x₁

POL( c(x₁) ) = x₁

POL( b(x₁) ) = x₁

POL( B(x₁) ) = x₁ + 1

POL( a(x₁) ) = max{0, -1}

The following usable rules [17] were oriented:

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(x))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ MNOCProof

          ↳ Narrowing

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ ForwardInstantiation

                    ↳ QDP

                      ↳ SemLabProof

                      ↳ SemLabProof2

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ QDPOrderProof

                                                    ↳ QDP

                                          ↳ NonTerminationProof

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

C(b(x)) → A(x)
A(b(b(x0))) → C(b(c(a(x0))))
A(b(c(a(B(x0))))) → C(a(c(B(x0))))
A(b(x)) → A(x)

The TRS R consists of the following rules:

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

C(b(x)) → A(x)
A(b(x)) → C(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

c(b(x)) → b(b(a(x)))
a(b(x)) → b(c(a(x)))
a(a(B(x))) → a(B(x))
a(c(a(B(x)))) → a(c(B(x)))
a(c(B(x))) → b(c(a(B(x))))
a(c(B(x))) → a(B(x))

s = A(c(a(c(a(B(x')))))) evaluates to t =A(c(a(c(a(B(x'))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

A(c(a(c(a(B(x')))))) → A(c(a(c(B(x')))))
with rule a(c(a(B(x'')))) → a(c(B(x''))) at position [0,0] and matcher [x'' / x']

A(c(a(c(B(x'))))) → A(c(b(c(a(B(x'))))))
with rule a(c(B(x))) → b(c(a(B(x)))) at position [0,0] and matcher [x / x']

A(c(b(c(a(B(x')))))) → A(b(b(a(c(a(B(x')))))))
with rule c(b(x)) → b(b(a(x))) at position [0] and matcher [x / c(a(B(x')))]

A(b(b(a(c(a(B(x'))))))) → A(b(b(a(c(B(x'))))))
with rule a(c(a(B(x'')))) → a(c(B(x''))) at position [0,0,0] and matcher [x'' / x']

A(b(b(a(c(B(x')))))) → A(b(b(b(c(a(B(x')))))))
with rule a(c(B(x))) → b(c(a(B(x)))) at position [0,0,0] and matcher [x / x']

A(b(b(b(c(a(B(x'))))))) → A(b(b(c(a(B(x'))))))
with rule A(b(x)) → A(x) at position [] and matcher [x / b(b(c(a(B(x')))))]

A(b(b(c(a(B(x')))))) → C(a(b(c(a(B(x'))))))
with rule A(b(x)) → C(a(x)) at position [] and matcher [x / b(c(a(B(x'))))]

C(a(b(c(a(B(x')))))) → C(b(c(a(c(a(B(x')))))))
with rule a(b(x'')) → b(c(a(x''))) at position [0] and matcher [x'' / c(a(B(x')))]

C(b(c(a(c(a(B(x'))))))) → A(c(a(c(a(B(x'))))))
with rule C(b(x)) → A(x)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.