Termination w.r.t. Q proof of TRS/nontermin/AG01#4.13.trs

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(x, x, x)
f₃(x, y, z) -> 2
0 -> 2
1 -> 2
g₃(x, x, y) -> y
g₃(x, y, y) -> x

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₃(0, 1, x) -> f₃(x, x, x)
f₃(x, y, z) -> 2
0 -> 2
1 -> 2
g₃(x, x, y) -> y
g₃(x, y, y) -> x

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(x, x, x)

The TRS R consists of the following rules:

f₃(0, 1, x) -> f₃(x, x, x)
f₃(x, y, z) -> 2
0 -> 2
1 -> 2
g₃(x, x, y) -> y
g₃(x, y, y) -> x

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) -> F₃(x, x, x)

The TRS R consists of the following rules:

f₃(0, 1, x) -> f₃(x, x, x)
f₃(x, y, z) -> 2
0 -> 2
1 -> 2
g₃(x, x, y) -> y
g₃(x, y, y) -> x

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