Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f3(0, 1, x) -> f3(g1(x), g1(x), x)
f3(g1(x), y, z) -> g1(f3(x, y, z))
f3(x, g1(y), z) -> g1(f3(x, y, z))
f3(x, y, g1(z)) -> g1(f3(x, y, z))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f3(0, 1, x) -> f3(g1(x), g1(x), x)
f3(g1(x), y, z) -> g1(f3(x, y, z))
f3(x, g1(y), z) -> g1(f3(x, y, z))
f3(x, y, g1(z)) -> g1(f3(x, y, z))

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(g1(x), y, z) -> F3(x, y, z)
F3(0, 1, x) -> F3(g1(x), g1(x), x)
F3(x, y, g1(z)) -> F3(x, y, z)
F3(x, g1(y), z) -> F3(x, y, z)

The TRS R consists of the following rules:

f3(0, 1, x) -> f3(g1(x), g1(x), x)
f3(g1(x), y, z) -> g1(f3(x, y, z))
f3(x, g1(y), z) -> g1(f3(x, y, z))
f3(x, y, g1(z)) -> g1(f3(x, y, z))

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(g1(x), y, z) -> F3(x, y, z)
F3(0, 1, x) -> F3(g1(x), g1(x), x)
F3(x, y, g1(z)) -> F3(x, y, z)
F3(x, g1(y), z) -> F3(x, y, z)

The TRS R consists of the following rules:

f3(0, 1, x) -> f3(g1(x), g1(x), x)
f3(g1(x), y, z) -> g1(f3(x, y, z))
f3(x, g1(y), z) -> g1(f3(x, y, z))
f3(x, y, g1(z)) -> g1(f3(x, y, z))

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.


F3(x, y, g1(z)) -> F3(x, y, z)
The remaining pairs can at least be oriented weakly.

F3(g1(x), y, z) -> F3(x, y, z)
F3(0, 1, x) -> F3(g1(x), g1(x), x)
F3(x, g1(y), z) -> F3(x, y, z)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(1) = 0   
POL(F3(x1, x2, x3)) = x3   
POL(g1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP

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

F3(g1(x), y, z) -> F3(x, y, z)
F3(0, 1, x) -> F3(g1(x), g1(x), x)
F3(x, g1(y), z) -> F3(x, y, z)

The TRS R consists of the following rules:

f3(0, 1, x) -> f3(g1(x), g1(x), x)
f3(g1(x), y, z) -> g1(f3(x, y, z))
f3(x, g1(y), z) -> g1(f3(x, y, z))
f3(x, y, g1(z)) -> g1(f3(x, y, z))

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