Termination w.r.t. Q proof of TRS/secret06jambox7.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(c₁(a₂(x, y)))) -> b₂(c₁(c₁(c₁(c₁(y)))), x)
c₁(c₁(b₂(c₁(y), 0))) -> a₂(0, c₁(c₁(a₂(y, 0))))
c₁(c₁(a₂(a₂(y, 0), x))) -> c₁(y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

c₁(c₁(c₁(a₂(x, y)))) -> b₂(c₁(c₁(c₁(c₁(y)))), x)
c₁(c₁(b₂(c₁(y), 0))) -> a₂(0, c₁(c₁(a₂(y, 0))))
c₁(c₁(a₂(a₂(y, 0), x))) -> c₁(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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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