(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
and(tt, X) → activate(X)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(activate(x1)) = 1 + x1
POL(and(x1, x2)) = 2 + x1 + x2
POL(cons(x1, x2)) = 1 + x1 + x2
POL(length(x1)) = 1 + x1
POL(n__take(x1, x2)) = x1 + x2
POL(n__zeros) = 0
POL(nil) = 0
POL(s(x1)) = x1
POL(take(x1, x2)) = 1 + x1 + x2
POL(tt) = 0
POL(zeros) = 1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
and(tt, X) → activate(X)
length(nil) → 0
take(0, IL) → nil
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(X) → X
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
Q is empty.
(3) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))
activate(n__zeros)
activate(n__take(x0, x1))
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
LENGTH(cons(N, L)) → ACTIVATE(L)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
ACTIVATE(n__zeros) → ZEROS
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))
activate(n__zeros)
activate(n__take(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))
activate(n__zeros)
activate(n__take(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(10) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
R is empty.
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))
activate(n__zeros)
activate(n__take(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(12) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))
activate(n__zeros)
activate(n__take(x0, x1))
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) 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(M), cons(N, IL)) → ACTIVATE(IL)
The graph contains the following edges 2 > 1
- ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
The graph contains the following edges 1 > 1, 1 > 2
(15) TRUE
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))
activate(n__zeros)
activate(n__take(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(17) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))
activate(n__zeros)
activate(n__take(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(19) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
length(cons(x0, x1))
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
take(s(x0), cons(x1, x2))
activate(n__zeros)
activate(n__take(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(21) UsableRulesReductionPairsProof (EQUIVALENT transformation)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
No dependency pairs are removed.
The following rules are removed from R:
activate(n__take(X1, X2)) → take(X1, X2)
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(0) = 0
POL(LENGTH(x1)) = x1
POL(activate(x1)) = x1
POL(cons(x1, x2)) = x1 + 2·x2
POL(n__take(x1, x2)) = 1 + 2·x1 + x2
POL(n__zeros) = 0
POL(s(x1)) = 2 + 2·x1
POL(take(x1, x2)) = 2·x1 + x2
POL(zeros) = 0
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
take(s(x0), cons(x1, x2))
activate(n__zeros)
activate(n__take(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(23) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
take(s(x0), cons(x1, x2))
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
activate(n__zeros)
activate(n__take(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(25) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
LENGTH(
cons(
N,
L)) →
LENGTH(
activate(
L)) at position [0] we obtained the following new rules [LPAR04]:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
The TRS R consists of the following rules:
activate(n__zeros) → zeros
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
activate(n__zeros)
activate(n__take(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(27) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
activate(n__zeros)
activate(n__take(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(29) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
activate(n__zeros)
activate(n__take(x0, x1))
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
(31) Rewriting (EQUIVALENT transformation)
By rewriting [LPAR04] the rule
LENGTH(
cons(
y0,
n__zeros)) →
LENGTH(
zeros) at position [0] we obtained the following new rules [LPAR04]:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
(33) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
R is empty.
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
(35) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
zeros
(36) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(37) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
LENGTH(
cons(
y0,
n__zeros)) →
LENGTH(
cons(
0,
n__zeros)) we obtained the following new rules [LPAR04]:
LENGTH(cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(39) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
s =
LENGTH(
cons(
0,
n__zeros)) evaluates to t =
LENGTH(
cons(
0,
n__zeros))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from LENGTH(cons(0, n__zeros)) to LENGTH(cons(0, n__zeros)).
(40) FALSE