(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: Combined order from the following AFS and order.
F(
x1,
x2) =
x1
f(
x1,
x2) =
f(
x1,
x2)
g(
x1,
x2) =
g(
x1,
x2)
Lexicographic path order with status [LPO].
Precedence:
trivial
Status:
trivial
The following usable rules [FROCOS05] were oriented:
none
(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