Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a2(f, a2(f, x)) -> a2(x, g)
a2(x, g) -> a2(f, a2(g, a2(f, x)))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a2(f, a2(f, x)) -> a2(x, g)
a2(x, g) -> a2(f, a2(g, a2(f, x)))

Q is empty.

Q DP problem:
The TRS P consists of the following rules:

A2(f, a2(f, x)) -> A2(x, g)
A2(x, g) -> A2(g, a2(f, x))
A2(x, g) -> A2(f, x)
A2(x, g) -> A2(f, a2(g, a2(f, x)))

The TRS R consists of the following rules:

a2(f, a2(f, x)) -> a2(x, g)
a2(x, g) -> a2(f, a2(g, a2(f, 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:

A2(f, a2(f, x)) -> A2(x, g)
A2(x, g) -> A2(g, a2(f, x))
A2(x, g) -> A2(f, x)
A2(x, g) -> A2(f, a2(g, a2(f, x)))

The TRS R consists of the following rules:

a2(f, a2(f, x)) -> a2(x, g)
a2(x, g) -> a2(f, a2(g, a2(f, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

Q DP problem:
The TRS P consists of the following rules:

A2(f, a2(f, x)) -> A2(x, g)
A2(x, g) -> A2(f, x)
A2(x, g) -> A2(f, a2(g, a2(f, x)))

The TRS R consists of the following rules:

a2(f, a2(f, x)) -> a2(x, g)
a2(x, g) -> a2(f, a2(g, a2(f, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.