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

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
g(x, x, y) → y
g(x, y, y) → x

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
g(x, x, y) → y
g(x, y, y) → x

Q is empty.

Using Dependency Pairs [1,15] 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
g(x, x, y) → y
g(x, y, y) → x

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ NonTerminationProof

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
g(x, x, y) → y
g(x, y, y) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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
g(x, x, y) → y
g(x, y, y) → x


s = F(g(1, 0, 0), g(1, y, y), x) evaluates to t =F(x, x, x)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

F(g(1, 0, 0), g(1, 0, 0), g(1, 0, 0)) → F(g(1, 0, 0), 1, g(1, 0, 0))
with rule g(x', y, y) → x' at position [1] and matcher [y / 0, x' / 1]

F(g(1, 0, 0), 1, g(1, 0, 0)) → F(g(1, 2, 0), 1, g(1, 0, 0))
with rule 0 → 2 at position [0,1] and matcher [ ]

F(g(1, 2, 0), 1, g(1, 0, 0)) → F(g(2, 2, 0), 1, g(1, 0, 0))
with rule 1 → 2 at position [0,0] and matcher [ ]

F(g(2, 2, 0), 1, g(1, 0, 0)) → F(0, 1, g(1, 0, 0))
with rule g(x', x', y) → y at position [0] and matcher [y / 0, x' / 2]

F(0, 1, g(1, 0, 0)) → F(g(1, 0, 0), g(1, 0, 0), g(1, 0, 0))
with rule F(0, 1, x) → F(x, x, x)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.