(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(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(a) → F(c(a))
F(c(a)) → F(d(b))
F(a) → F(d(a))
F(c(b)) → F(d(a))
E(g(X)) → E(X)
The TRS R consists of the following rules:
f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(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:
E(g(X)) → E(X)
The TRS R consists of the following rules:
f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(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.
E(g(X)) → E(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
E(
x1) =
x1
g(
x1) =
g(
x1)
f(
x1) =
f(
x1)
a =
a
c(
x1) =
x1
d(
x1) =
x1
b =
b
e(
x1) =
e(
x1)
Recursive Path Order [RPO].
Precedence:
f1 > [a, b]
The following usable rules [FROCOS05] were oriented:
f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(X)
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(a) → f(c(a))
f(c(X)) → X
f(c(a)) → f(d(b))
f(a) → f(d(a))
f(d(X)) → X
f(c(b)) → f(d(a))
e(g(X)) → e(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