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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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
      ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F3(x1, x2, x3)  =  F1(x1)
f3(x1, x2, x3)  =  f3(x1, x2, x3)
g1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[F1, f3]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ 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.