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:

f3(f3(x, y, z), u, f3(x, y, v)) -> f3(x, y, f3(z, u, v))
f3(x, y, y) -> y
f3(x, y, g1(y)) -> x
f3(x, x, y) -> x
f3(g1(x), x, y) -> y

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f3(f3(x, y, z), u, f3(x, y, v)) -> f3(x, y, f3(z, u, v))
f3(x, y, y) -> y
f3(x, y, g1(y)) -> x
f3(x, x, y) -> x
f3(g1(x), x, y) -> y

Q is empty.

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

F3(f3(x, y, z), u, f3(x, y, v)) -> F3(x, y, f3(z, u, v))
F3(f3(x, y, z), u, f3(x, y, v)) -> F3(z, u, v)

The TRS R consists of the following rules:

f3(f3(x, y, z), u, f3(x, y, v)) -> f3(x, y, f3(z, u, v))
f3(x, y, y) -> y
f3(x, y, g1(y)) -> x
f3(x, x, y) -> x
f3(g1(x), x, y) -> y

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPAfsSolverProof

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

F3(f3(x, y, z), u, f3(x, y, v)) -> F3(x, y, f3(z, u, v))
F3(f3(x, y, z), u, f3(x, y, v)) -> F3(z, u, v)

The TRS R consists of the following rules:

f3(f3(x, y, z), u, f3(x, y, v)) -> f3(x, y, f3(z, u, v))
f3(x, y, y) -> y
f3(x, y, g1(y)) -> x
f3(x, x, y) -> x
f3(g1(x), x, y) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F3(f3(x, y, z), u, f3(x, y, v)) -> F3(x, y, f3(z, u, v))
F3(f3(x, y, z), u, f3(x, y, v)) -> F3(z, u, v)
Used argument filtering: F3(x1, x2, x3)  =  x1
f3(x1, x2, x3)  =  f2(x1, x3)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
QDP
          ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f3(f3(x, y, z), u, f3(x, y, v)) -> f3(x, y, f3(z, u, v))
f3(x, y, y) -> y
f3(x, y, g1(y)) -> x
f3(x, x, y) -> x
f3(g1(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.