(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(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:
F(f(x)) → F(c(f(x)))
F(f(x)) → F(d(f(x)))
G(c(h(0))) → G(d(1))
G(c(1)) → G(d(h(0)))
G(h(x)) → G(x)
The TRS R consists of the following rules:
f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(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 1 SCC with 4 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(h(x)) → G(x)
The TRS R consists of the following rules:
f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
G(h(x)) → G(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(
x1) =
x1
h(
x1) =
h(
x1)
f(
x1) =
f
c(
x1) =
c(
x1)
d(
x1) =
x1
g(
x1) =
x1
0 =
0
1 =
1
Lexicographic path order with status [LPO].
Precedence:
c1 > h1
c1 > 0
c1 > 1
Status:
trivial
The following usable rules [FROCOS05] were oriented:
f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(x)
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(h(0))) → g(d(1))
g(c(1)) → g(d(h(0)))
g(h(x)) → g(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(8) TRUE