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:

f(0, 1, x) → f(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(0, 1, x) → f(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, z))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)
F(x, y, g(z)) → F(x, y, z)

The TRS R consists of the following rules:

f(0, 1, x) → f(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(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:

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)
F(x, y, g(z)) → F(x, y, z)

The TRS R consists of the following rules:

f(0, 1, x) → f(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(x, y, g(z)) → F(x, y, z)
The remaining pairs can at least be oriented weakly.

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)
Used ordering: Polynomial interpretation [25,35]:

POL(g(x1)) = 1/4 + x_1   
POL(F(x1, x2, x3)) = (1/4)x_3   
POL(0) = 0   
POL(1) = 0   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP

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

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)

The TRS R consists of the following rules:

f(0, 1, x) → f(g(x), g(x), x)
f(g(x), y, z) → g(f(x, y, z))
f(x, g(y), z) → g(f(x, y, z))
f(x, y, g(z)) → g(f(x, y, z))

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