Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.

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

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(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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.