Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/AG01SLASHHASH3.33.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

  ↳ Non-Overlap Check

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.

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:

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)))

The set Q consists of the following terms:

p₁(f₁(f₁(x0)))
p₁(g₁(g₁(x0)))
q₁(f₁(f₁(x0)))
q₁(g₁(g₁(x0)))

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)))

The set Q consists of the following terms:

p₁(f₁(f₁(x0)))
p₁(g₁(g₁(x0)))
q₁(f₁(f₁(x0)))
q₁(g₁(g₁(x0)))

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:

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)))

The set Q consists of the following terms:

p₁(f₁(f₁(x0)))
p₁(g₁(g₁(x0)))
q₁(f₁(f₁(x0)))
q₁(g₁(g₁(x0)))

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.