Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/nonterminSLASHcaribooSLASHex3.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:

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(x))

Q is empty.

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

G₁(x) -> F₁(g₁(x))
F₁(g₁(a)) -> G₁(b)
G₁(x) -> G₁(x)
F₁(g₁(a)) -> F₁(s₁(g₁(b)))

The TRS R consists of the following rules:

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(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:

G₁(x) -> F₁(g₁(x))
F₁(g₁(a)) -> G₁(b)
G₁(x) -> G₁(x)
F₁(g₁(a)) -> F₁(s₁(g₁(b)))

The TRS R consists of the following rules:

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

G₁(x) -> F₁(g₁(x))
F₁(g₁(a)) -> G₁(b)
G₁(x) -> G₁(x)

The TRS R consists of the following rules:

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(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.

G₁(x) -> F₁(g₁(x))
F₁(g₁(a)) -> G₁(b)

Used argument filtering: G₁(x1) = G₁(x1)
F₁(x1) = x1
g₁(x1) = g₁(x1)
a = a
b = b
f₁(x1) = f
Used ordering: Precedence:

a > G₁ > g₁ > f > b

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

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

G₁(x) -> G₁(x)

The TRS R consists of the following rules:

f₁(g₁(a)) -> f₁(s₁(g₁(b)))
f₁(f₁(x)) -> b
g₁(x) -> f₁(g₁(x))

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