(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(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:
REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)
REV(++(x, x)) → REV(x)
The TRS R consists of the following rules:
rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)
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.
REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)
REV(++(x, x)) → REV(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REV(
x1) =
x1
++(
x1,
x2) =
++(
x1,
x2)
Recursive Path Order [RPO].
Precedence:
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:
rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)
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