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) -> f3(g2(X, X), 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) -> f3(g2(X, X), 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) -> F3(g2(X, X), X, X)
F3(0, 1, X) -> G2(X, X)

The TRS R consists of the following rules:

f3(0, 1, X) -> f3(g2(X, X), 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
      ↳ DependencyGraphProof

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

F3(0, 1, X) -> F3(g2(X, X), X, X)
F3(0, 1, X) -> G2(X, X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

F3(0, 1, X) -> F3(g2(X, X), X, X)

The TRS R consists of the following rules:

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

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