Termination w.r.t. Q proof of /tmp/tmpuMGTzc/jambox2.trs

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:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(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(c(c(b(x)))) → B(c(x))
B(c(b(c(x)))) → A(1, x)
C(c(c(b(x)))) → A(1, b(c(x)))
A(1, x) → B(x)
A(0, x) → C(c(x))
B(c(b(c(x)))) → A(0, a(1, x))
A(1, x) → C(b(x))
C(c(c(b(x)))) → C(x)
A(0, x) → C(x)

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

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

C(c(c(b(x)))) → B(c(x))
B(c(b(c(x)))) → A(1, x)
C(c(c(b(x)))) → A(1, b(c(x)))
A(1, x) → B(x)
A(0, x) → C(c(x))
B(c(b(c(x)))) → A(0, a(1, x))
A(1, x) → C(b(x))
C(c(c(b(x)))) → C(x)
A(0, x) → C(x)

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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:

C(c(c(b(x)))) → B(c(x))
B(c(b(c(x)))) → A(1, x)
A(1, x) → B(x)
B(c(b(c(x)))) → A(0, a(1, x))
C(c(c(b(x)))) → C(x)
A(0, x) → C(x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 2
POL(1) = 2
POL(A(x₁, x₂)) = 2·x₁ + 2·x₂
POL(B(x₁)) = 1 + 2·x₁
POL(C(x₁)) = 2·x₁
POL(a(x₁, x₂)) = 2·x₁ + x₂
POL(b(x₁)) = 2 + x₁
POL(c(x₁)) = 2 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

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

C(c(c(b(x)))) → A(1, b(c(x)))
A(0, x) → C(c(x))
A(1, x) → C(b(x))

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

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

C(c(c(b(x)))) → A(1, b(c(x)))
A(1, x) → C(b(x))

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(1, x) → C(b(x)) at position [0] we obtained the following new rules:

A(1, c(b(c(x0)))) → C(a(0, a(1, x0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

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

C(c(c(b(x)))) → A(1, b(c(x)))
A(1, c(b(c(x0)))) → C(a(0, a(1, x0)))

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(c(b(x)))) → A(1, b(c(x))) at position [1] we obtained the following new rules:

C(c(c(b(b(c(x0)))))) → A(1, a(0, a(1, x0)))
C(c(c(b(c(c(b(x0))))))) → A(1, b(a(1, b(c(x0)))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

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

C(c(c(b(b(c(x0)))))) → A(1, a(0, a(1, x0)))
A(1, c(b(c(x0)))) → C(a(0, a(1, x0)))
C(c(c(b(c(c(b(x0))))))) → A(1, b(a(1, b(c(x0)))))

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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