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:

f2(a, y) -> f2(y, g1(y))
g1(a) -> b
g1(b) -> b

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

f2(a, y) -> f2(y, g1(y))
g1(a) -> b
g1(b) -> b

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(a, y) -> f2(y, g1(y))
g1(a) -> b
g1(b) -> b

The set Q consists of the following terms:

f2(a, x0)
g1(a)
g1(b)


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

F2(a, y) -> G1(y)
F2(a, y) -> F2(y, g1(y))

The TRS R consists of the following rules:

f2(a, y) -> f2(y, g1(y))
g1(a) -> b
g1(b) -> b

The set Q consists of the following terms:

f2(a, x0)
g1(a)
g1(b)

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(a, y) -> G1(y)
F2(a, y) -> F2(y, g1(y))

The TRS R consists of the following rules:

f2(a, y) -> f2(y, g1(y))
g1(a) -> b
g1(b) -> b

The set Q consists of the following terms:

f2(a, x0)
g1(a)
g1(b)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP

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

F2(a, y) -> F2(y, g1(y))

The TRS R consists of the following rules:

f2(a, y) -> f2(y, g1(y))
g1(a) -> b
g1(b) -> b

The set Q consists of the following terms:

f2(a, x0)
g1(a)
g1(b)

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