Termination w.r.t. Q proof of TRS/secret06jambox4.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:

c₁(c₁(b₁(c₁(x)))) -> b₁(a₂(0, c₁(x)))
c₁(c₁(x)) -> b₁(c₁(b₁(c₁(x))))
a₂(0, x) -> c₁(c₁(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

c₁(c₁(b₁(c₁(x)))) -> b₁(a₂(0, c₁(x)))
c₁(c₁(x)) -> b₁(c₁(b₁(c₁(x))))
a₂(0, x) -> c₁(c₁(x))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

C₁(c₁(b₁(c₁(x)))) -> A₂(0, c₁(x))
C₁(c₁(x)) -> C₁(b₁(c₁(x)))
A₂(0, x) -> C₁(x)
A₂(0, x) -> C₁(c₁(x))

The TRS R consists of the following rules:

c₁(c₁(b₁(c₁(x)))) -> b₁(a₂(0, c₁(x)))
c₁(c₁(x)) -> b₁(c₁(b₁(c₁(x))))
a₂(0, x) -> c₁(c₁(x))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

C₁(c₁(b₁(c₁(x)))) -> A₂(0, c₁(x))
C₁(c₁(x)) -> C₁(b₁(c₁(x)))
A₂(0, x) -> C₁(x)
A₂(0, x) -> C₁(c₁(x))

The TRS R consists of the following rules:

c₁(c₁(b₁(c₁(x)))) -> b₁(a₂(0, c₁(x)))
c₁(c₁(x)) -> b₁(c₁(b₁(c₁(x))))
a₂(0, x) -> c₁(c₁(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

C₁(c₁(b₁(c₁(x)))) -> A₂(0, c₁(x))
A₂(0, x) -> C₁(x)
A₂(0, x) -> C₁(c₁(x))

The TRS R consists of the following rules:

c₁(c₁(b₁(c₁(x)))) -> b₁(a₂(0, c₁(x)))
c₁(c₁(x)) -> b₁(c₁(b₁(c₁(x))))
a₂(0, x) -> c₁(c₁(x))

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