Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/secret2005SLASHtpa2.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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

Q is empty.

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

F₂(x, s₁(y)) -> -¹₂(s₁(y), x)
F₂(x, s₁(y)) -> -¹₂(x, s₁(y))
F₂(s₁(x), y) -> P₁(-₂(y, s₁(x)))
F₂(s₁(x), y) -> P₁(-₂(s₁(x), y))
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
F₂(s₁(x), y) -> F₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
F₂(x, s₁(y)) -> F₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))
F₂(x, s₁(y)) -> P₁(-₂(s₁(y), x))
F₂(x, s₁(y)) -> P₁(-₂(x, s₁(y)))
F₂(s₁(x), y) -> -¹₂(y, s₁(x))
F₂(s₁(x), y) -> -¹₂(s₁(x), y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

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:

F₂(x, s₁(y)) -> -¹₂(s₁(y), x)
F₂(x, s₁(y)) -> -¹₂(x, s₁(y))
F₂(s₁(x), y) -> P₁(-₂(y, s₁(x)))
F₂(s₁(x), y) -> P₁(-₂(s₁(x), y))
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
F₂(s₁(x), y) -> F₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
F₂(x, s₁(y)) -> F₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))
F₂(x, s₁(y)) -> P₁(-₂(s₁(y), x))
F₂(x, s₁(y)) -> P₁(-₂(x, s₁(y)))
F₂(s₁(x), y) -> -¹₂(y, s₁(x))
F₂(s₁(x), y) -> -¹₂(s₁(x), y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

Used argument filtering: -¹₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), 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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

F₂(s₁(x), y) -> F₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
F₂(x, s₁(y)) -> F₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
p₁(s₁(x)) -> x
f₂(s₁(x), y) -> f₂(p₁(-₂(s₁(x), y)), p₁(-₂(y, s₁(x))))
f₂(x, s₁(y)) -> f₂(p₁(-₂(x, s₁(y))), p₁(-₂(s₁(y), x)))

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