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:

g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))

Q is empty.

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

G2(x, s1(y)) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(y, 0)
G2(s1(x), y) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(x, 0)

The TRS R consists of the following rules:

g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))

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:

G2(x, s1(y)) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(y, 0)
G2(s1(x), y) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(x, 0)

The TRS R consists of the following rules:

g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))

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:

G2(f2(x, y), 0) -> G2(y, 0)
G2(s1(x), y) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(x, 0)

The TRS R consists of the following rules:

g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))

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