Termination w.r.t. Q proof of /tmp/tmpofaAMA/dup15.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(a(d(d(x1)))) → d(d(b(b(x1))))
a(a(x1)) → b(b(b(b(b(b(x1))))))
b(b(d(d(b(b(x1)))))) → a(a(c(c(x1))))
c(c(x1)) → d(d(x1))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(a(d(d(x1)))) → d(d(b(b(x1))))
a(a(x1)) → b(b(b(b(b(b(x1))))))
b(b(d(d(b(b(x1)))))) → a(a(c(c(x1))))
c(c(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 2 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(B(x₁)) = x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 1 + 2·x₁
POL(d(x₁)) = 1 + 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

B(b(d(d(b(b(x0)))))) → A(d(d(x0)))
B(b(d(d(b(b(c(x0))))))) → A(c(d(d(x0))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(a(d(d(x1)))) → d(d(b(b(x1))))
a(a(x1)) → b(b(b(b(b(b(x1))))))
b(b(d(d(b(b(x1)))))) → a(a(c(c(x1))))
c(c(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 2 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = 8 + 6·x₁
POL(B(x₁)) = 4·x₁
POL(a(x₁)) = 4 + x₁
POL(b(x₁)) = 8
POL(c(x₁)) = 0
POL(d(x₁)) = 0

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ QDPOrderProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

A(a(x1)) → B(b(b(b(b(b(x1))))))

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = 14 + 2·x₁
POL(B(x₁)) = 12 + 2·x₁
POL(a(x₁)) = 9 + x₁
POL(b(x₁)) = 2 + x₁
POL(c(x₁)) = 2·x₁
POL(d(x₁)) = 2·x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

A(a(x1)) → B(b(b(b(b(b(x1))))))

The TRS R consists of the following rules:

a(a(d(d(x1)))) → d(d(b(b(x1))))
a(a(x1)) → b(b(b(b(b(b(x1))))))
b(b(d(d(b(b(x1)))))) → a(a(c(c(x1))))
c(c(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 0 SCCs with 1 less node.
We have reversed the following QTRS:
The set of rules R is

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

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

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

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

Q is empty.