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:

f3(0, 1, X) -> h2(X, X)
h2(0, X) -> f3(0, X, X)
g2(X, Y) -> X
g2(X, Y) -> Y

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f3(0, 1, X) -> h2(X, X)
h2(0, X) -> f3(0, X, X)
g2(X, Y) -> X
g2(X, Y) -> Y

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F3(0, 1, X) -> H2(X, X)
H2(0, X) -> F3(0, X, X)

The TRS R consists of the following rules:

f3(0, 1, X) -> h2(X, X)
h2(0, X) -> f3(0, X, X)
g2(X, Y) -> X
g2(X, Y) -> Y

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

↳ QTRS
  ↳ DependencyPairsProof
QDP

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

F3(0, 1, X) -> H2(X, X)
H2(0, X) -> F3(0, X, X)

The TRS R consists of the following rules:

f3(0, 1, X) -> h2(X, X)
h2(0, X) -> f3(0, X, X)
g2(X, Y) -> X
g2(X, Y) -> Y

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