(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) QDPSizeChangeProof (EQUIVALENT transformation)
We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.
Order:Homeomorphic Embedding Order
AFS:
cons(x1, x2) = cons(x2)
From the DPs we obtained the following set of size-change graphs:
- R1(cons(x, k), a) → R1(k, cons(x, a)) (allowed arguments on rhs = {1, 2})
The graph contains the following edges 1 > 1
We oriented the following set of usable rules [AAECC05,FROCOS05].
none
(6) TRUE