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 Order [17,21] with Interpretation:

POL( A₂(x₁, x₂) ) = x₂ + 1

POL( b₂(x₁, x₂) ) = x₁ + x₂

POL( 0 ) = 1

POL( B₂(x₁, x₂) ) = max{0, x₁ + x₂ - 1}

POL( a₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( 1 ) = 0

The following usable rules [14] were oriented:

a₂(0, x) -> b₂(0, b₂(0, x))
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(x, y)) -> a₂(1, a₂(1, a₂(x, y)))
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 Order [17,21] with Interpretation:

POL( A₂(x₁, x₂) ) = 1

POL( B₂(x₁, x₂) ) = max{0, x₂ - 1}

POL( b₂(x₁, x₂) ) = 0

POL( a₂(x₁, x₂) ) = x₁ + 1

POL( 1 ) = 1

POL( 0 ) = 1

The following usable rules [14] were oriented:

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