(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
f(c(s(x), y)) → f(c(x, s(y)))
f(f(x)) → f(d(f(x)))
f(x) → x
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:
G(c(x, s(y))) → G(c(s(x), y))
F(c(s(x), y)) → F(c(x, s(y)))
F(f(x)) → F(d(f(x)))
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
f(c(s(x), y)) → f(c(x, s(y)))
f(f(x)) → f(d(f(x)))
f(x) → x
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 with 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(c(s(x), y)) → F(c(x, s(y)))
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
f(c(s(x), y)) → f(c(x, s(y)))
f(f(x)) → f(d(f(x)))
f(x) → x
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(c(s(x), y)) → F(c(x, s(y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(
x1) =
x1
c(
x1,
x2) =
x1
s(
x1) =
s(
x1)
g(
x1) =
g
f(
x1) =
x1
d(
x1) =
d
Lexicographic path order with status [LPO].
Precedence:
s1 > d
g > d
Status:
trivial
The following usable rules [FROCOS05] were oriented:
g(c(x, s(y))) → g(c(s(x), y))
f(c(s(x), y)) → f(c(x, s(y)))
f(f(x)) → f(d(f(x)))
f(x) → x
(7) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
f(c(s(x), y)) → f(c(x, s(y)))
f(f(x)) → f(d(f(x)))
f(x) → x
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:
G(c(x, s(y))) → G(c(s(x), y))
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
f(c(s(x), y)) → f(c(x, s(y)))
f(f(x)) → f(d(f(x)))
f(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.