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
  ↳ Non-Overlap Check

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.

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

The set Q consists of the following terms:

f2(g1(h2(x0, x1)), f2(a, a))


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)))

The set Q consists of the following terms:

f2(g1(h2(x0, x1)), f2(a, a))

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:

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)))

The set Q consists of the following terms:

f2(g1(h2(x0, x1)), f2(a, a))

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.