Termination w.r.t. Q proof of /tmp/tmpGYULGZ/z117.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:

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

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

a(d(x)) → d(b(x))
b(x) → a(a(a(x)))
c(d(c(x))) → d(a(x))
d(d(b(x))) → c(d(d(c(c(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(d(x)) → d(b(x))
b(x) → a(a(a(x)))
c(d(c(x))) → d(a(x))
d(d(b(x))) → c(d(d(c(c(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(d(d(x1))) → C(x1)
C(d(c(x1))) → D(x1)
D(a(x1)) → D(x1)
B(d(d(x1))) → C(d(d(c(x1))))
B(d(d(x1))) → C(c(d(d(c(x1)))))
B(d(d(x1))) → D(c(x1))
B(d(d(x1))) → D(d(c(x1)))
D(a(x1)) → B(d(x1))

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

d(a(x1)) → b(d(x1))
b(x1) → a(a(a(x1)))
c(d(c(x1))) → a(d(x1))
b(d(d(x1))) → c(c(d(d(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 1 SCC with 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

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

B(d(d(x1))) → C(x1)
B(d(d(x1))) → D(c(x1))

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(B(x₁)) = 10 + x₁
POL(C(x₁)) = x₁
POL(D(x₁)) = 3·x₁
POL(a(x₁)) = 4 + x₁
POL(b(x₁)) = 12 + x₁
POL(c(x₁)) = 1 + x₁
POL(d(x₁)) = 3·x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

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

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.