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

Q is empty.


QTRS
  ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

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

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

f(f(x, y, z), u, f(x, y, v)) → f(x, y, f(z, u, v))
f(x, y, y) → y
f(x, y, g(y)) → x
f(x, x, y) → x
f(g(x), x, y) → y
Used ordering:
Polynomial interpretation [25]:

POL(f(x1, x2, x3)) = 2 + 2·x1 + x2 + x3   
POL(g(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
QTRS
      ↳ RisEmptyProof

Q restricted rewrite system:
R is empty.
Q is empty.

The TRS R is empty. Hence, termination is trivially proven.