Termination w.r.t. Q proof of TRS/secret06jambox2.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:

c₁(c₁(c₁(b₁(x)))) -> a₂(1, b₁(c₁(x)))
b₁(c₁(b₁(c₁(x)))) -> a₂(0, a₂(1, x))
a₂(0, x) -> c₁(c₁(x))
a₂(1, x) -> c₁(b₁(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

c₁(c₁(c₁(b₁(x)))) -> a₂(1, b₁(c₁(x)))
b₁(c₁(b₁(c₁(x)))) -> a₂(0, a₂(1, x))
a₂(0, x) -> c₁(c₁(x))
a₂(1, x) -> c₁(b₁(x))

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(c₁(c₁(b₁(x)))) -> C₁(x)
A₂(1, x) -> C₁(b₁(x))
B₁(c₁(b₁(c₁(x)))) -> A₂(1, x)
A₂(1, x) -> B₁(x)
C₁(c₁(c₁(b₁(x)))) -> A₂(1, b₁(c₁(x)))
B₁(c₁(b₁(c₁(x)))) -> A₂(0, a₂(1, x))
A₂(0, x) -> C₁(x)
A₂(0, x) -> C₁(c₁(x))
C₁(c₁(c₁(b₁(x)))) -> B₁(c₁(x))

The TRS R consists of the following rules:

c₁(c₁(c₁(b₁(x)))) -> a₂(1, b₁(c₁(x)))
b₁(c₁(b₁(c₁(x)))) -> a₂(0, a₂(1, x))
a₂(0, x) -> c₁(c₁(x))
a₂(1, x) -> c₁(b₁(x))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

C₁(c₁(c₁(b₁(x)))) -> C₁(x)
A₂(1, x) -> C₁(b₁(x))
B₁(c₁(b₁(c₁(x)))) -> A₂(1, x)
A₂(1, x) -> B₁(x)
C₁(c₁(c₁(b₁(x)))) -> A₂(1, b₁(c₁(x)))
B₁(c₁(b₁(c₁(x)))) -> A₂(0, a₂(1, x))
A₂(0, x) -> C₁(x)
A₂(0, x) -> C₁(c₁(x))
C₁(c₁(c₁(b₁(x)))) -> B₁(c₁(x))

The TRS R consists of the following rules:

c₁(c₁(c₁(b₁(x)))) -> a₂(1, b₁(c₁(x)))
b₁(c₁(b₁(c₁(x)))) -> a₂(0, a₂(1, x))
a₂(0, x) -> c₁(c₁(x))
a₂(1, x) -> c₁(b₁(x))

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.

C₁(c₁(c₁(b₁(x)))) -> C₁(x)
B₁(c₁(b₁(c₁(x)))) -> A₂(1, x)
C₁(c₁(c₁(b₁(x)))) -> B₁(c₁(x))

The remaining pairs can at least be oriented weakly.

A₂(1, x) -> C₁(b₁(x))
A₂(1, x) -> B₁(x)
C₁(c₁(c₁(b₁(x)))) -> A₂(1, b₁(c₁(x)))
B₁(c₁(b₁(c₁(x)))) -> A₂(0, a₂(1, x))
A₂(0, x) -> C₁(x)
A₂(0, x) -> C₁(c₁(x))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( C₁(x₁) ) = max{0, x₁ - 2}

POL( c₁(x₁) ) = x₁ + 1

POL( b₁(x₁) ) = x₁ + 1

POL( A₂(x₁, x₂) ) = max{0, x₁ + x₂ - 2}

POL( 1 ) = 1

POL( B₁(x₁) ) = max{0, x₁ - 1}

POL( 0 ) = 2

POL( a₂(x₁, x₂) ) = x₂ + 2

The following usable rules [14] were oriented:

a₂(0, x) -> c₁(c₁(x))
b₁(c₁(b₁(c₁(x)))) -> a₂(0, a₂(1, x))
a₂(1, x) -> c₁(b₁(x))
c₁(c₁(c₁(b₁(x)))) -> a₂(1, b₁(c₁(x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

A₂(1, x) -> C₁(b₁(x))
A₂(1, x) -> B₁(x)
C₁(c₁(c₁(b₁(x)))) -> A₂(1, b₁(c₁(x)))
B₁(c₁(b₁(c₁(x)))) -> A₂(0, a₂(1, x))
A₂(0, x) -> C₁(x)
A₂(0, x) -> C₁(c₁(x))

The TRS R consists of the following rules:

c₁(c₁(c₁(b₁(x)))) -> a₂(1, b₁(c₁(x)))
b₁(c₁(b₁(c₁(x)))) -> a₂(0, a₂(1, x))
a₂(0, x) -> c₁(c₁(x))
a₂(1, x) -> c₁(b₁(x))

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(1, x) -> B₁(x)
A₂(0, x) -> C₁(x)

The remaining pairs can at least be oriented weakly.

A₂(1, x) -> C₁(b₁(x))
C₁(c₁(c₁(b₁(x)))) -> A₂(1, b₁(c₁(x)))
B₁(c₁(b₁(c₁(x)))) -> A₂(0, a₂(1, x))
A₂(0, x) -> C₁(c₁(x))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( A₂(x₁, x₂) ) = x₂ + 1

POL( B₁(x₁) ) = max{0, x₁ - 1}

POL( C₁(x₁) ) = max{0, x₁ - 1}

POL( b₁(x₁) ) = x₁ + 2

POL( c₁(x₁) ) = x₁ + 2

POL( a₂(x₁, x₂) ) = x₁ + x₂ + 2

POL( 1 ) = 2

POL( 0 ) = 2

The following usable rules [14] were oriented:

b₁(c₁(b₁(c₁(x)))) -> a₂(0, a₂(1, x))
a₂(0, x) -> c₁(c₁(x))
a₂(1, x) -> c₁(b₁(x))
c₁(c₁(c₁(b₁(x)))) -> a₂(1, b₁(c₁(x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A₂(1, x) -> C₁(b₁(x))
C₁(c₁(c₁(b₁(x)))) -> A₂(1, b₁(c₁(x)))
B₁(c₁(b₁(c₁(x)))) -> A₂(0, a₂(1, x))
A₂(0, x) -> C₁(c₁(x))

The TRS R consists of the following rules:

c₁(c₁(c₁(b₁(x)))) -> a₂(1, b₁(c₁(x)))
b₁(c₁(b₁(c₁(x)))) -> a₂(0, a₂(1, x))
a₂(0, x) -> c₁(c₁(x))
a₂(1, x) -> c₁(b₁(x))

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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

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

A₂(1, x) -> C₁(b₁(x))
C₁(c₁(c₁(b₁(x)))) -> A₂(1, b₁(c₁(x)))

The TRS R consists of the following rules:

c₁(c₁(c₁(b₁(x)))) -> a₂(1, b₁(c₁(x)))
b₁(c₁(b₁(c₁(x)))) -> a₂(0, a₂(1, x))
a₂(0, x) -> c₁(c₁(x))
a₂(1, x) -> c₁(b₁(x))

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