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

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

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

a(a(x1)) → b(b(b(x1)))
a(x1) → c(d(x1))
b(b(x1)) → c(c(c(x1)))
c(c(x1)) → d(d(d(x1)))
e(d(x1)) → a(b(c(d(e(x1)))))
b(x1) → d(d(x1))
e(c(x1)) → b(a(a(e(x1))))
c(d(d(x1))) → a(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 6 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPOrderProof

            ↳ UsableRulesProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = 8 + x₁
POL(B(x₁)) = 4 + x₁
POL(C(x₁)) = 1 + x₁
POL(a(x₁)) = 14 + x₁
POL(b(x₁)) = 9 + x₁
POL(c(x₁)) = 6 + x₁
POL(d(x₁)) = 4 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

            ↳ UsableRulesProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

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

E(c(x1)) → E(x1)
E(d(x1)) → E(x1)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

            ↳ UsableRulesProof

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

E(c(x1)) → E(x1)
E(d(x1)) → E(x1)

R is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

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

E(c(x1)) → E(x1)
E(d(x1)) → E(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:

E(c(x1)) → E(x1)
E(d(x1)) → E(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(E(x₁)) = 2·x₁
POL(c(x₁)) = 2·x₁
POL(d(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ 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.