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(a, x) -> g2(a, x)
g2(a, x) -> f2(b, x)
f2(a, x) -> f2(b, x)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f2(a, x) -> g2(a, x)
g2(a, x) -> f2(b, x)
f2(a, x) -> f2(b, 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(a, x) -> F2(b, x)
F2(a, x) -> G2(a, x)
G2(a, x) -> F2(b, x)

The TRS R consists of the following rules:

f2(a, x) -> g2(a, x)
g2(a, x) -> f2(b, x)
f2(a, x) -> f2(b, 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(a, x) -> F2(b, x)
F2(a, x) -> G2(a, x)
G2(a, x) -> F2(b, x)

The TRS R consists of the following rules:

f2(a, x) -> g2(a, x)
g2(a, x) -> f2(b, x)
f2(a, x) -> f2(b, 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 3 less nodes.