Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x3)
0  =  0
1  =  1
g(x1)  =  g(x1)

Recursive path order with status [RPO].
Precedence:

0 > F₁
1 > F₁
g₁ > F₁

Status:

F₁: [1]
0: multiset
1: multiset
g₁: multiset

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

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