Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/SK90SLASH2.58.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:

f₂(x, y) -> x
g₁(a) -> h₃(a, b, a)
i₁(x) -> f₂(x, x)
h₃(x, x, y) -> g₁(x)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₂(x, y) -> x
g₁(a) -> h₃(a, b, a)
i₁(x) -> f₂(x, x)
h₃(x, x, y) -> g₁(x)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

f₂(x, y) -> x
g₁(a) -> h₃(a, b, a)
i₁(x) -> f₂(x, x)
h₃(x, x, y) -> g₁(x)

The set Q consists of the following terms:

f₂(x0, x1)
g₁(a)
i₁(x0)
h₃(x0, x0, x1)

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

G₁(a) -> H₃(a, b, a)
H₃(x, x, y) -> G₁(x)
I₁(x) -> F₂(x, x)

The TRS R consists of the following rules:

f₂(x, y) -> x
g₁(a) -> h₃(a, b, a)
i₁(x) -> f₂(x, x)
h₃(x, x, y) -> g₁(x)

The set Q consists of the following terms:

f₂(x0, x1)
g₁(a)
i₁(x0)
h₃(x0, x0, x1)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

G₁(a) -> H₃(a, b, a)
H₃(x, x, y) -> G₁(x)
I₁(x) -> F₂(x, x)

The TRS R consists of the following rules:

f₂(x, y) -> x
g₁(a) -> h₃(a, b, a)
i₁(x) -> f₂(x, x)
h₃(x, x, y) -> g₁(x)

The set Q consists of the following terms:

f₂(x0, x1)
g₁(a)
i₁(x0)
h₃(x0, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 3 less nodes.