Termination w.r.t. Q proof of TRS/SK902.57.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(g₁(h₂(x, y)), f₂(a, a)) -> f₂(h₂(x, x), g₁(f₂(y, a)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(g₁(h₂(x, y)), f₂(a, a)) -> f₂(h₂(x, x), g₁(f₂(y, a)))

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:

F₂(g₁(h₂(x, y)), f₂(a, a)) -> F₂(h₂(x, x), g₁(f₂(y, a)))
F₂(g₁(h₂(x, y)), f₂(a, a)) -> F₂(y, a)

The TRS R consists of the following rules:

f₂(g₁(h₂(x, y)), f₂(a, a)) -> f₂(h₂(x, x), g₁(f₂(y, a)))

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(g₁(h₂(x, y)), f₂(a, a)) -> F₂(h₂(x, x), g₁(f₂(y, a)))
F₂(g₁(h₂(x, y)), f₂(a, a)) -> F₂(y, a)

The TRS R consists of the following rules:

f₂(g₁(h₂(x, y)), f₂(a, a)) -> f₂(h₂(x, x), g₁(f₂(y, a)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.