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(x, x) -> f2(i1(x), g1(g1(x)))
f2(x, y) -> x
g1(x) -> i1(x)
f2(x, i1(x)) -> f2(x, x)
f2(i1(x), i1(g1(x))) -> a

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f2(x, x) -> f2(i1(x), g1(g1(x)))
f2(x, y) -> x
g1(x) -> i1(x)
f2(x, i1(x)) -> f2(x, x)
f2(i1(x), i1(g1(x))) -> a

Q is empty.

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

F2(x, x) -> F2(i1(x), g1(g1(x)))
F2(x, x) -> G1(x)
F2(x, i1(x)) -> F2(x, x)
F2(x, x) -> G1(g1(x))

The TRS R consists of the following rules:

f2(x, x) -> f2(i1(x), g1(g1(x)))
f2(x, y) -> x
g1(x) -> i1(x)
f2(x, i1(x)) -> f2(x, x)
f2(i1(x), i1(g1(x))) -> a

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:

F2(x, x) -> F2(i1(x), g1(g1(x)))
F2(x, x) -> G1(x)
F2(x, i1(x)) -> F2(x, x)
F2(x, x) -> G1(g1(x))

The TRS R consists of the following rules:

f2(x, x) -> f2(i1(x), g1(g1(x)))
f2(x, y) -> x
g1(x) -> i1(x)
f2(x, i1(x)) -> f2(x, x)
f2(i1(x), i1(g1(x))) -> a

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

F2(x, x) -> F2(i1(x), g1(g1(x)))
F2(x, i1(x)) -> F2(x, x)

The TRS R consists of the following rules:

f2(x, x) -> f2(i1(x), g1(g1(x)))
f2(x, y) -> x
g1(x) -> i1(x)
f2(x, i1(x)) -> f2(x, x)
f2(i1(x), i1(g1(x))) -> a

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