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:

a₂(0, b₂(0, x)) -> b₂(0, a₂(0, x))
a₂(0, x) -> b₂(0, b₂(0, x))
a₂(0, a₂(1, a₂(x, y))) -> a₂(1, a₂(0, a₂(x, y)))
b₂(0, a₂(1, a₂(x, y))) -> b₂(1, a₂(0, a₂(x, y)))
a₂(0, a₂(x, y)) -> a₂(1, a₂(1, a₂(x, y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a₂(0, b₂(0, x)) -> b₂(0, a₂(0, x))
a₂(0, x) -> b₂(0, b₂(0, x))
a₂(0, a₂(1, a₂(x, y))) -> a₂(1, a₂(0, a₂(x, y)))
b₂(0, a₂(1, a₂(x, y))) -> b₂(1, a₂(0, a₂(x, y)))
a₂(0, a₂(x, y)) -> a₂(1, a₂(1, a₂(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:

A₂(0, x) -> B₂(0, b₂(0, x))
B₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))
B₂(0, a₂(1, a₂(x, y))) -> B₂(1, a₂(0, a₂(x, y)))
A₂(0, b₂(0, x)) -> B₂(0, a₂(0, x))
A₂(0, a₂(1, a₂(x, y))) -> A₂(1, a₂(0, a₂(x, y)))
A₂(0, b₂(0, x)) -> A₂(0, x)
A₂(0, a₂(x, y)) -> A₂(1, a₂(1, a₂(x, y)))
A₂(0, x) -> B₂(0, x)
A₂(0, a₂(x, y)) -> A₂(1, a₂(x, y))
A₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))

The TRS R consists of the following rules:

a₂(0, b₂(0, x)) -> b₂(0, a₂(0, x))
a₂(0, x) -> b₂(0, b₂(0, x))
a₂(0, a₂(1, a₂(x, y))) -> a₂(1, a₂(0, a₂(x, y)))
b₂(0, a₂(1, a₂(x, y))) -> b₂(1, a₂(0, a₂(x, y)))
a₂(0, a₂(x, y)) -> a₂(1, a₂(1, a₂(x, 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:

A₂(0, x) -> B₂(0, b₂(0, x))
B₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))
B₂(0, a₂(1, a₂(x, y))) -> B₂(1, a₂(0, a₂(x, y)))
A₂(0, b₂(0, x)) -> B₂(0, a₂(0, x))
A₂(0, a₂(1, a₂(x, y))) -> A₂(1, a₂(0, a₂(x, y)))
A₂(0, b₂(0, x)) -> A₂(0, x)
A₂(0, a₂(x, y)) -> A₂(1, a₂(1, a₂(x, y)))
A₂(0, x) -> B₂(0, x)
A₂(0, a₂(x, y)) -> A₂(1, a₂(x, y))
A₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))

The TRS R consists of the following rules:

a₂(0, b₂(0, x)) -> b₂(0, a₂(0, x))
a₂(0, x) -> b₂(0, b₂(0, x))
a₂(0, a₂(1, a₂(x, y))) -> a₂(1, a₂(0, a₂(x, y)))
b₂(0, a₂(1, a₂(x, y))) -> b₂(1, a₂(0, a₂(x, y)))
a₂(0, a₂(x, y)) -> a₂(1, a₂(1, a₂(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

A₂(0, x) -> B₂(0, b₂(0, x))
A₂(0, b₂(0, x)) -> B₂(0, a₂(0, x))
B₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))
A₂(0, b₂(0, x)) -> A₂(0, x)
A₂(0, x) -> B₂(0, x)
A₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))

The TRS R consists of the following rules:

a₂(0, b₂(0, x)) -> b₂(0, a₂(0, x))
a₂(0, x) -> b₂(0, b₂(0, x))
a₂(0, a₂(1, a₂(x, y))) -> a₂(1, a₂(0, a₂(x, y)))
b₂(0, a₂(1, a₂(x, y))) -> b₂(1, a₂(0, a₂(x, y)))
a₂(0, a₂(x, y)) -> a₂(1, a₂(1, a₂(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

A₂(0, b₂(0, x)) -> A₂(0, x)
A₂(0, x) -> B₂(0, x)
A₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))

The remaining pairs can at least be oriented weakly.

A₂(0, x) -> B₂(0, b₂(0, x))
A₂(0, b₂(0, x)) -> B₂(0, a₂(0, x))
B₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))

Used ordering: Polynomial interpretation [21]:

POL(0) = 2
POL(1) = 0
POL(A₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(B₂(x₁, x₂)) = 2·x₂
POL(a₂(x₁, x₂)) = 2 + x₁ + x₂
POL(b₂(x₁, x₂)) = x₁ + x₂

The following usable rules [14] were oriented:

a₂(0, x) -> b₂(0, b₂(0, x))
a₂(0, a₂(x, y)) -> a₂(1, a₂(1, a₂(x, y)))
b₂(0, a₂(1, a₂(x, y))) -> b₂(1, a₂(0, a₂(x, y)))
a₂(0, b₂(0, x)) -> b₂(0, a₂(0, x))
a₂(0, a₂(1, a₂(x, y))) -> a₂(1, a₂(0, a₂(x, y)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

A₂(0, x) -> B₂(0, b₂(0, x))
B₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))
A₂(0, b₂(0, x)) -> B₂(0, a₂(0, x))

The TRS R consists of the following rules:

a₂(0, b₂(0, x)) -> b₂(0, a₂(0, x))
a₂(0, x) -> b₂(0, b₂(0, x))
a₂(0, a₂(1, a₂(x, y))) -> a₂(1, a₂(0, a₂(x, y)))
b₂(0, a₂(1, a₂(x, y))) -> b₂(1, a₂(0, a₂(x, y)))
a₂(0, a₂(x, y)) -> a₂(1, a₂(1, a₂(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

A₂(0, x) -> B₂(0, b₂(0, x))

The remaining pairs can at least be oriented weakly.

B₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))
A₂(0, b₂(0, x)) -> B₂(0, a₂(0, x))

Used ordering: Polynomial interpretation [21]:

POL(0) = 2
POL(1) = 2
POL(A₂(x₁, x₂)) = 2 + 3·x₁
POL(B₂(x₁, x₂)) = 2·x₂
POL(a₂(x₁, x₂)) = 2·x₁
POL(b₂(x₁, x₂)) = 0

The following usable rules [14] were oriented:

a₂(0, x) -> b₂(0, b₂(0, x))
a₂(0, a₂(x, y)) -> a₂(1, a₂(1, a₂(x, y)))
b₂(0, a₂(1, a₂(x, y))) -> b₂(1, a₂(0, a₂(x, y)))
a₂(0, b₂(0, x)) -> b₂(0, a₂(0, x))
a₂(0, a₂(1, a₂(x, y))) -> a₂(1, a₂(0, a₂(x, y)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

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

A₂(0, b₂(0, x)) -> B₂(0, a₂(0, x))
B₂(0, a₂(1, a₂(x, y))) -> A₂(0, a₂(x, y))

The TRS R consists of the following rules:

a₂(0, b₂(0, x)) -> b₂(0, a₂(0, x))
a₂(0, x) -> b₂(0, b₂(0, x))
a₂(0, a₂(1, a₂(x, y))) -> a₂(1, a₂(0, a₂(x, y)))
b₂(0, a₂(1, a₂(x, y))) -> b₂(1, a₂(0, a₂(x, y)))
a₂(0, a₂(x, y)) -> a₂(1, a₂(1, a₂(x, y)))

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