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(g1(X), Y) -> f2(X, f2(g1(X), Y))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

f2(g1(X), Y) -> f2(X, f2(g1(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(g1(X), Y) -> f2(X, f2(g1(X), Y))

The set Q consists of the following terms:

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, f2(g1(X), Y))
F2(g1(X), Y) -> F2(g1(X), Y)

The TRS R consists of the following rules:

f2(g1(X), Y) -> f2(X, f2(g1(X), Y))

The set Q consists of the following terms:

f2(g1(x0), x1)

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

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

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

F2(g1(X), Y) -> F2(X, f2(g1(X), Y))
F2(g1(X), Y) -> F2(g1(X), Y)

The TRS R consists of the following rules:

f2(g1(X), Y) -> f2(X, f2(g1(X), Y))

The set Q consists of the following terms:

f2(g1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


F2(g1(X), Y) -> F2(X, f2(g1(X), Y))
The remaining pairs can at least by weakly be oriented.

F2(g1(X), Y) -> F2(g1(X), Y)
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  x1
g1(x1)  =  g1(x1)
f2(x1, x2)  =  f

Lexicographic Path Order [19].
Precedence:
f > g1

The following usable rules [14] were oriented:

f2(g1(X), Y) -> f2(X, f2(g1(X), Y))



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

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

F2(g1(X), Y) -> F2(g1(X), Y)

The TRS R consists of the following rules:

f2(g1(X), Y) -> f2(X, f2(g1(X), Y))

The set Q consists of the following terms:

f2(g1(x0), x1)

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