Termination w.r.t. Q proof of TRS/secret06jambox9.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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