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:

f(0, 1, x) → f(x, x, x)
f(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:

f(0, 1, x) → f(x, x, x)
f(x, y, z) → 2
0 → 2
1 → 2

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:

F(0, 1, x) → F(x, x, x)

The TRS R consists of the following rules:

f(0, 1, x) → f(x, x, x)
f(x, y, z) → 2
0 → 2
1 → 2

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

F(0, 1, x) → F(x, x, x)

The TRS R consists of the following rules:

f(0, 1, x) → f(x, x, x)
f(x, y, z) → 2
0 → 2
1 → 2

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(0, 1, x) → F(x, x, x)

The TRS R consists of the following rules:

f(0, 1, x) → f(x, x, x)
f(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 [13,14,18] contains 0 SCCs with 1 less node.