Termination w.r.t. Q proof of TRS/secret06jambox5.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(a₂(y, 0), 0) -> y
c₁(c₁(y)) -> y
c₁(a₂(c₁(c₁(y)), x)) -> a₂(c₁(c₁(c₁(a₂(x, 0)))), y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a₂(a₂(y, 0), 0) -> y
c₁(c₁(y)) -> y
c₁(a₂(c₁(c₁(y)), x)) -> a₂(c₁(c₁(c₁(a₂(x, 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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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