NO Termination w.r.t. Q proof of ../tpdb/qualif/toyama2.trs
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(a, b, x) → f(x, x, x)
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(a, b, x) → f(x, x, x)
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(a, b, x) → F(x, x, x)

The TRS R consists of the following rules:

f(a, b, x) → f(x, x, x)
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(a, b, x) → F(x, x, x)

The TRS R consists of the following rules:

f(a, b, x) → f(x, x, x)
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(a, b, x) → F(x, x, x)

The TRS R consists of the following rules:

f(a, b, x) → f(x, x, x)
g(x, y) → x
g(x, y) → y


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

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

F(a, b, g(a, b)) → F(g(a, b), g(a, b), g(a, b))
with rule F(a, b, 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.