Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/SK90SLASH2.46.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

a₁(a₁(x)) -> b₁(b₁(x))
b₁(b₁(a₁(x))) -> a₁(b₁(b₁(x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a₁(a₁(x)) -> b₁(b₁(x))
b₁(b₁(a₁(x))) -> a₁(b₁(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:

A₁(a₁(x)) -> B₁(x)
B₁(b₁(a₁(x))) -> A₁(b₁(b₁(x)))
B₁(b₁(a₁(x))) -> B₁(x)
B₁(b₁(a₁(x))) -> B₁(b₁(x))
A₁(a₁(x)) -> B₁(b₁(x))

The TRS R consists of the following rules:

a₁(a₁(x)) -> b₁(b₁(x))
b₁(b₁(a₁(x))) -> a₁(b₁(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:

A₁(a₁(x)) -> B₁(x)
B₁(b₁(a₁(x))) -> A₁(b₁(b₁(x)))
B₁(b₁(a₁(x))) -> B₁(x)
B₁(b₁(a₁(x))) -> B₁(b₁(x))
A₁(a₁(x)) -> B₁(b₁(x))

The TRS R consists of the following rules:

a₁(a₁(x)) -> b₁(b₁(x))
b₁(b₁(a₁(x))) -> a₁(b₁(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 strictly oriented and are deleted.

A₁(a₁(x)) -> B₁(x)
B₁(b₁(a₁(x))) -> B₁(x)
B₁(b₁(a₁(x))) -> B₁(b₁(x))

The remaining pairs can at least by weakly be oriented.

B₁(b₁(a₁(x))) -> A₁(b₁(b₁(x)))
A₁(a₁(x)) -> B₁(b₁(x))

Used ordering: Combined order from the following AFS and order.
A₁(x1) = A₁(x1)
a₁(x1) = a₁(x1)
B₁(x1) = B₁(x1)
b₁(x1) = b₁(x1)

Lexicographic Path Order [19].
Precedence:

[A₁, B_1] > [a₁, b_1]

The following usable rules [14] were oriented:

a₁(a₁(x)) -> b₁(b₁(x))
b₁(b₁(a₁(x))) -> a₁(b₁(b₁(x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

B₁(b₁(a₁(x))) -> A₁(b₁(b₁(x)))
A₁(a₁(x)) -> B₁(b₁(x))

The TRS R consists of the following rules:

a₁(a₁(x)) -> b₁(b₁(x))
b₁(b₁(a₁(x))) -> a₁(b₁(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 strictly oriented and are deleted.

B₁(b₁(a₁(x))) -> A₁(b₁(b₁(x)))
A₁(a₁(x)) -> B₁(b₁(x))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
B₁(x1) = B₁(x1)
b₁(x1) = x1
a₁(x1) = a₁(x1)
A₁(x1) = A₁(x1)

Lexicographic Path Order [19].
Precedence:

[B₁, a₁, A_1]

The following usable rules [14] were oriented:

a₁(a₁(x)) -> b₁(b₁(x))
b₁(b₁(a₁(x))) -> a₁(b₁(b₁(x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a₁(a₁(x)) -> b₁(b₁(x))
b₁(b₁(a₁(x))) -> a₁(b₁(b₁(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.