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:

f1(s1(x)) -> f1(g2(x, x))
g2(0, 1) -> s1(0)
0 -> 1

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(s1(x)) -> f1(g2(x, x))
g2(0, 1) -> s1(0)
0 -> 1

Q is empty.

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

F1(s1(x)) -> F1(g2(x, x))
F1(s1(x)) -> G2(x, x)

The TRS R consists of the following rules:

f1(s1(x)) -> f1(g2(x, x))
g2(0, 1) -> s1(0)
0 -> 1

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:

F1(s1(x)) -> F1(g2(x, x))
F1(s1(x)) -> G2(x, x)

The TRS R consists of the following rules:

f1(s1(x)) -> f1(g2(x, x))
g2(0, 1) -> s1(0)
0 -> 1

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:

F1(s1(x)) -> F1(g2(x, x))

The TRS R consists of the following rules:

f1(s1(x)) -> f1(g2(x, x))
g2(0, 1) -> s1(0)
0 -> 1

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