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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f2(x, y) -> h2(x, y)
f2(x, y) -> h2(y, x)
h2(x, x) -> 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:

F2(x, y) -> H2(y, x)
F2(x, y) -> H2(x, y)

The TRS R consists of the following rules:

f2(x, y) -> h2(x, y)
f2(x, y) -> h2(y, x)
h2(x, x) -> 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:

F2(x, y) -> H2(y, x)
F2(x, y) -> H2(x, y)

The TRS R consists of the following rules:

f2(x, y) -> h2(x, y)
f2(x, y) -> h2(y, x)
h2(x, x) -> 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 2 less nodes.