(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
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(x, c(x), c(y)) → F(y, y, f(y, x, y))
F(x, c(x), c(y)) → F(y, x, y)
F(s(x), y, z) → F(x, s(c(y)), c(z))
The TRS R consists of the following rules:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y, z) → F(x, s(c(y)), c(z))
The TRS R consists of the following rules:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F(s(x), y, z) → F(x, s(c(y)), c(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(
x1,
x2,
x3)
s(
x1) =
s(
x1)
c(
x1) =
x1
Recursive path order with status [RPO].
Quasi-Precedence:
F3 > s1
Status:
s1: multiset
F3: [3,1,2]
The following usable rules [FROCOS05] were oriented:
none
(7) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, c(x), c(y)) → F(y, x, y)
F(x, c(x), c(y)) → F(y, y, f(y, x, y))
The TRS R consists of the following rules:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.