Termination w.r.t. Q proof of TRS/SK902.60.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:

f₁(g₂(f₁(a), h₂(a, f₁(a)))) -> f₁(h₂(g₂(f₁(a), a), g₂(f₁(a), f₁(a))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(g₂(f₁(a), h₂(a, f₁(a)))) -> f₁(h₂(g₂(f₁(a), a), g₂(f₁(a), f₁(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:

F₁(g₂(f₁(a), h₂(a, f₁(a)))) -> F₁(h₂(g₂(f₁(a), a), g₂(f₁(a), f₁(a))))

The TRS R consists of the following rules:

f₁(g₂(f₁(a), h₂(a, f₁(a)))) -> f₁(h₂(g₂(f₁(a), a), g₂(f₁(a), f₁(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:

F₁(g₂(f₁(a), h₂(a, f₁(a)))) -> F₁(h₂(g₂(f₁(a), a), g₂(f₁(a), f₁(a))))

The TRS R consists of the following rules:

f₁(g₂(f₁(a), h₂(a, f₁(a)))) -> f₁(h₂(g₂(f₁(a), a), g₂(f₁(a), f₁(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 1 less node.