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(g(x, y), x, z) → f(z, z, z)
g(x, y) → x
g(x, y) → y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(g(x, y), x, z) → f(z, z, z)
g(x, y) → x
g(x, y) → y

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(g(x, y), x, z) → F(z, z, z)

The TRS R consists of the following rules:

f(g(x, y), x, z) → f(z, z, z)
g(x, y) → x
g(x, y) → y

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(g(x, y), x, z) → F(z, z, z)

The TRS R consists of the following rules:

f(g(x, y), x, z) → f(z, z, z)
g(x, y) → x
g(x, y) → y

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(g(x, y), x, z) → F(z, z, z)

The TRS R consists of the following rules:

f(g(x, y), x, z) → f(z, z, z)
g(x, y) → x
g(x, y) → y


s = F(g(y', y), g(x', y'), z) evaluates to t =F(z, z, z)

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



Rewriting sequence

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

F(g(y', y'), y', g(y', y')) → F(g(y', y'), g(y', y'), g(y', y'))
with rule F(g(x, y), x, z) → F(z, z, z)

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.