Termination w.r.t. Q proof of TRS/secret06jambox3.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:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, 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:

A(0, x) → B(0, b(0, x))
B(0, a(1, a(x, y))) → A(0, a(x, y))
B(0, a(1, a(x, y))) → B(1, a(0, a(x, y)))
A(0, b(0, x)) → B(0, a(0, x))
A(0, a(1, a(x, y))) → A(1, a(0, a(x, y)))
A(0, b(0, x)) → A(0, x)
A(0, a(x, y)) → A(1, a(1, a(x, y)))
A(0, x) → B(0, x)
A(0, a(x, y)) → A(1, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

A(0, x) → B(0, b(0, x))
B(0, a(1, a(x, y))) → A(0, a(x, y))
B(0, a(1, a(x, y))) → B(1, a(0, a(x, y)))
A(0, b(0, x)) → B(0, a(0, x))
A(0, a(1, a(x, y))) → A(1, a(0, a(x, y)))
A(0, b(0, x)) → A(0, x)
A(0, a(x, y)) → A(1, a(1, a(x, y)))
A(0, x) → B(0, x)
A(0, a(x, y)) → A(1, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
B(0, a(1, a(x, y))) → B(1, a(0, a(x, y)))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(1, a(x, y))) → A(1, a(0, a(x, y)))
A(0, b(0, x)) → A(0, x)
A(0, x) → B(0, x)
A(0, a(x, y)) → A(1, a(1, a(x, y)))
A(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(x, y)) → A(1, a(x, y))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, 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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

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

A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, b(0, x)) → A(0, x)
A(0, x) → B(0, x)
A(0, a(1, a(x, y))) → A(0, a(x, y))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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