Termination w.r.t. Q proof of TRS/AG01#3.33.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:

p₁(f₁(f₁(x))) -> q₁(f₁(g₁(x)))
p₁(g₁(g₁(x))) -> q₁(g₁(f₁(x)))
q₁(f₁(f₁(x))) -> p₁(f₁(g₁(x)))
q₁(g₁(g₁(x))) -> p₁(g₁(f₁(x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p₁(f₁(f₁(x))) -> q₁(f₁(g₁(x)))
p₁(g₁(g₁(x))) -> q₁(g₁(f₁(x)))
q₁(f₁(f₁(x))) -> p₁(f₁(g₁(x)))
q₁(g₁(g₁(x))) -> p₁(g₁(f₁(x)))

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:

Q₁(g₁(g₁(x))) -> P₁(g₁(f₁(x)))
P₁(f₁(f₁(x))) -> Q₁(f₁(g₁(x)))
P₁(g₁(g₁(x))) -> Q₁(g₁(f₁(x)))
Q₁(f₁(f₁(x))) -> P₁(f₁(g₁(x)))

The TRS R consists of the following rules:

p₁(f₁(f₁(x))) -> q₁(f₁(g₁(x)))
p₁(g₁(g₁(x))) -> q₁(g₁(f₁(x)))
q₁(f₁(f₁(x))) -> p₁(f₁(g₁(x)))
q₁(g₁(g₁(x))) -> p₁(g₁(f₁(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:

Q₁(g₁(g₁(x))) -> P₁(g₁(f₁(x)))
P₁(f₁(f₁(x))) -> Q₁(f₁(g₁(x)))
P₁(g₁(g₁(x))) -> Q₁(g₁(f₁(x)))
Q₁(f₁(f₁(x))) -> P₁(f₁(g₁(x)))

The TRS R consists of the following rules:

p₁(f₁(f₁(x))) -> q₁(f₁(g₁(x)))
p₁(g₁(g₁(x))) -> q₁(g₁(f₁(x)))
q₁(f₁(f₁(x))) -> p₁(f₁(g₁(x)))
q₁(g₁(g₁(x))) -> p₁(g₁(f₁(x)))

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 4 less nodes.