(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
rev(ls) → r1(ls, empty)
r1(empty, a) → a
r1(cons(x, k), a) → r1(k, cons(x, a))
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(ls) → R1(ls, empty)
R1(cons(x, k), a) → R1(k, cons(x, a))
The TRS R consists of the following rules:
rev(ls) → r1(ls, empty)
r1(empty, a) → a
r1(cons(x, k), a) → r1(k, cons(x, a))
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 1 less node.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
R1(cons(x, k), a) → R1(k, cons(x, a))
The TRS R consists of the following rules:
rev(ls) → r1(ls, empty)
r1(empty, a) → a
r1(cons(x, k), a) → r1(k, cons(x, a))
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.
R1(cons(x, k), a) → R1(k, cons(x, a))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(R1(x1, x2)) = x1
POL(cons(x1, x2)) = 1 + x2
The following usable rules [FROCOS05] were oriented:
none
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rev(ls) → r1(ls, empty)
r1(empty, a) → a
r1(cons(x, k), a) → r1(k, cons(x, a))
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