Termination w.r.t. Q proof of TRS/secret06jambox6.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(x, y) -> c₁(a₂(c₁(y), a₂(0, x)))
a₂(y, x) -> y
a₂(y, c₁(b₂(a₂(0, x), 0))) -> b₂(a₂(c₁(b₂(0, y)), x), 0)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

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

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

The TRS R consists of the following rules:

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

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