(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)
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:
FROM(X) → FROM(s(X))
2ND(cons(X, XS)) → HEAD(XS)
TAKE(s(N), cons(X, XS)) → TAKE(N, XS)
SEL(s(N), cons(X, XS)) → SEL(N, XS)
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)
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 3 SCCs with 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SEL(s(N), cons(X, XS)) → SEL(N, XS)
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- SEL(s(N), cons(X, XS)) → SEL(N, XS)
The graph contains the following edges 1 > 1, 2 > 2
(7) YES
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TAKE(s(N), cons(X, XS)) → TAKE(N, XS)
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- TAKE(s(N), cons(X, XS)) → TAKE(N, XS)
The graph contains the following edges 1 > 1, 2 > 2
(10) YES
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(X) → FROM(s(X))
The TRS R consists of the following rules:
from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) NonLoopProof (EQUIVALENT transformation)
By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x0 / s(x0)] on the rule
FROM(s(x0))[ ]n[ ] → FROM(s(x0))[ ]n[x0 / s(x0)]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Simplify mu) - Instantiate mu - Instantiation
FROM(X)[ ]n[ ] → FROM(s(X))[ ]n[ ]
by OriginalRule from TRS P
(13) NO