Termination w.r.t. Q proof of TRS/secret06jambox8.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:

b₂(b₂(0, y), x) -> y
c₁(c₁(c₁(y))) -> c₁(c₁(a₂(a₂(c₁(b₂(0, y)), 0), 0)))
a₂(y, 0) -> b₂(y, 0)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

b₂(b₂(0, y), x) -> y
c₁(c₁(c₁(y))) -> c₁(c₁(a₂(a₂(c₁(b₂(0, y)), 0), 0)))
a₂(y, 0) -> b₂(y, 0)

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(c₁(y))) -> A₂(c₁(b₂(0, y)), 0)
C₁(c₁(c₁(y))) -> C₁(b₂(0, y))
A₂(y, 0) -> B₂(y, 0)
C₁(c₁(c₁(y))) -> C₁(a₂(a₂(c₁(b₂(0, y)), 0), 0))
C₁(c₁(c₁(y))) -> C₁(c₁(a₂(a₂(c₁(b₂(0, y)), 0), 0)))
C₁(c₁(c₁(y))) -> B₂(0, y)
C₁(c₁(c₁(y))) -> A₂(a₂(c₁(b₂(0, y)), 0), 0)

The TRS R consists of the following rules:

b₂(b₂(0, y), x) -> y
c₁(c₁(c₁(y))) -> c₁(c₁(a₂(a₂(c₁(b₂(0, y)), 0), 0)))
a₂(y, 0) -> b₂(y, 0)

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(c₁(y))) -> A₂(c₁(b₂(0, y)), 0)
C₁(c₁(c₁(y))) -> C₁(b₂(0, y))
A₂(y, 0) -> B₂(y, 0)
C₁(c₁(c₁(y))) -> C₁(a₂(a₂(c₁(b₂(0, y)), 0), 0))
C₁(c₁(c₁(y))) -> C₁(c₁(a₂(a₂(c₁(b₂(0, y)), 0), 0)))
C₁(c₁(c₁(y))) -> B₂(0, y)
C₁(c₁(c₁(y))) -> A₂(a₂(c₁(b₂(0, y)), 0), 0)

The TRS R consists of the following rules:

b₂(b₂(0, y), x) -> y
c₁(c₁(c₁(y))) -> c₁(c₁(a₂(a₂(c₁(b₂(0, y)), 0), 0)))
a₂(y, 0) -> b₂(y, 0)

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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

C₁(c₁(c₁(y))) -> C₁(a₂(a₂(c₁(b₂(0, y)), 0), 0))
C₁(c₁(c₁(y))) -> C₁(c₁(a₂(a₂(c₁(b₂(0, y)), 0), 0)))

The TRS R consists of the following rules:

b₂(b₂(0, y), x) -> y
c₁(c₁(c₁(y))) -> c₁(c₁(a₂(a₂(c₁(b₂(0, y)), 0), 0)))
a₂(y, 0) -> b₂(y, 0)

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