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(f2(0, x), 1) -> f2(g1(f2(x, x)), x)
f2(g1(x), y) -> g1(f2(x, y))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

f2(f2(0, x), 1) -> f2(g1(f2(x, x)), x)
f2(g1(x), y) -> g1(f2(x, y))

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(f2(0, x), 1) -> f2(g1(f2(x, x)), x)
f2(g1(x), y) -> g1(f2(x, y))

The set Q consists of the following terms:

f2(f2(0, x0), 1)
f2(g1(x0), x1)


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(g1(x), y) -> F2(x, y)
F2(f2(0, x), 1) -> F2(x, x)
F2(f2(0, x), 1) -> F2(g1(f2(x, x)), x)

The TRS R consists of the following rules:

f2(f2(0, x), 1) -> f2(g1(f2(x, x)), x)
f2(g1(x), y) -> g1(f2(x, y))

The set Q consists of the following terms:

f2(f2(0, x0), 1)
f2(g1(x0), x1)

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

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

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

F2(g1(x), y) -> F2(x, y)
F2(f2(0, x), 1) -> F2(x, x)
F2(f2(0, x), 1) -> F2(g1(f2(x, x)), x)

The TRS R consists of the following rules:

f2(f2(0, x), 1) -> f2(g1(f2(x, x)), x)
f2(g1(x), y) -> g1(f2(x, y))

The set Q consists of the following terms:

f2(f2(0, x0), 1)
f2(g1(x0), x1)

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