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
  ↳ DependencyPairsProof

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.

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

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

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.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F(f(x, y, z), u, f(x, y, v)) → F(x, y, f(z, u, v))
F(f(x, y, z), u, f(x, y, v)) → F(z, u, v)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x1, x3)
f(x1, x2, x3)  =  f(x1, x3)
g(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
F2 > f2

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ PisEmptyProof

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.