(0) Obligation:

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

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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(f(x, y), z) → F(x, f(y, z))
F(f(x, y), z) → F(y, z)
F(g(x, y), z) → F(x, z)
F(g(x, y), z) → F(y, z)

The TRS R consists of the following rules:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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(f(x, y), z) → F(x, f(y, z))
F(f(x, y), z) → F(y, z)
F(g(x, y), z) → F(x, z)
F(g(x, y), z) → F(y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
F2 > f2 > [g2, i1]
0 > [g2, i1]
1 > [g2, i1]
2 > [g2, i1]


The following usable rules [FROCOS05] were oriented:

f(x, 0) → x
f(0, y) → y
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(2, g(x, y)) → y

(4) Obligation:

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

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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

(5) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(6) TRUE