NO proof of Transformed_CSR_04_OvConsOS_nosorts_GM.trs # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 65 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 15 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) MRRProof [EQUIVALENT, 0 ms] (10) QDP (11) MRRProof [EQUIVALENT, 0 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) AND (15) QDP (16) TransformationProof [EQUIVALENT, 0 ms] (17) QDP (18) DependencyGraphProof [EQUIVALENT, 0 ms] (19) QDP (20) TransformationProof [EQUIVALENT, 0 ms] (21) QDP (22) DependencyGraphProof [EQUIVALENT, 0 ms] (23) AND (24) QDP (25) UsableRulesProof [EQUIVALENT, 0 ms] (26) QDP (27) TransformationProof [EQUIVALENT, 0 ms] (28) QDP (29) NonTerminationLoopProof [COMPLETE, 0 ms] (30) NO (31) QDP (32) MRRProof [EQUIVALENT, 14 ms] (33) QDP (34) QDP (35) UsableRulesProof [EQUIVALENT, 0 ms] (36) QDP (37) QDPSizeChangeProof [EQUIVALENT, 0 ms] (38) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__and(tt, X) -> mark(X) a__length(nil) -> 0 a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(0, IL) -> nil a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__and(x_1, x_2)) = 1 + 2*x_1 + x_2 POL(a__length(x_1)) = 2*x_1 POL(a__take(x_1, x_2)) = x_1 + 2*x_2 POL(a__zeros) = 0 POL(and(x_1, x_2)) = 1 + 2*x_1 + x_2 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(length(x_1)) = 2*x_1 POL(mark(x_1)) = x_1 POL(nil) = 0 POL(s(x_1)) = 2*x_1 POL(take(x_1, x_2)) = x_1 + 2*x_2 POL(tt) = 2 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__and(tt, X) -> mark(X) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(nil) -> 0 a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(0, IL) -> nil a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__and(x_1, x_2)) = 1 + x_1 + x_2 POL(a__length(x_1)) = 1 + x_1 POL(a__take(x_1, x_2)) = 2 + x_1 + 2*x_2 POL(a__zeros) = 1 POL(and(x_1, x_2)) = 1 + x_1 + x_2 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(length(x_1)) = 1 + x_1 POL(mark(x_1)) = x_1 POL(nil) = 1 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = 2 + x_1 + 2*x_2 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: a__length(nil) -> 0 a__take(0, IL) -> nil ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) 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: A__LENGTH(cons(N, L)) -> A__LENGTH(mark(L)) A__LENGTH(cons(N, L)) -> MARK(L) A__TAKE(s(M), cons(N, IL)) -> MARK(N) MARK(zeros) -> A__ZEROS MARK(and(X1, X2)) -> A__AND(mark(X1), X2) MARK(and(X1, X2)) -> MARK(X1) MARK(length(X)) -> A__LENGTH(mark(X)) MARK(length(X)) -> MARK(X) MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2)) MARK(take(X1, X2)) -> MARK(X1) MARK(take(X1, X2)) -> MARK(X2) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> MARK(L) MARK(and(X1, X2)) -> MARK(X1) MARK(length(X)) -> A__LENGTH(mark(X)) A__LENGTH(cons(N, L)) -> A__LENGTH(mark(L)) MARK(length(X)) -> MARK(X) MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2)) A__TAKE(s(M), cons(N, IL)) -> MARK(N) MARK(take(X1, X2)) -> MARK(X1) MARK(take(X1, X2)) -> MARK(X2) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: MARK(and(X1, X2)) -> MARK(X1) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(A__LENGTH(x_1)) = 2 + x_1 POL(A__TAKE(x_1, x_2)) = 2 + x_1 + x_2 POL(MARK(x_1)) = 2 + 2*x_1 POL(a__and(x_1, x_2)) = 2 + x_1 + x_2 POL(a__length(x_1)) = x_1 POL(a__take(x_1, x_2)) = x_1 + 2*x_2 POL(a__zeros) = 0 POL(and(x_1, x_2)) = 2 + x_1 + x_2 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(length(x_1)) = x_1 POL(mark(x_1)) = x_1 POL(nil) = 0 POL(s(x_1)) = 2*x_1 POL(take(x_1, x_2)) = x_1 + 2*x_2 POL(tt) = 0 POL(zeros) = 0 ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> MARK(L) MARK(length(X)) -> A__LENGTH(mark(X)) A__LENGTH(cons(N, L)) -> A__LENGTH(mark(L)) MARK(length(X)) -> MARK(X) MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2)) A__TAKE(s(M), cons(N, IL)) -> MARK(N) MARK(take(X1, X2)) -> MARK(X1) MARK(take(X1, X2)) -> MARK(X2) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: A__LENGTH(cons(N, L)) -> MARK(L) MARK(length(X)) -> MARK(X) A__TAKE(s(M), cons(N, IL)) -> MARK(N) MARK(take(X1, X2)) -> MARK(X1) MARK(take(X1, X2)) -> MARK(X2) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(A__LENGTH(x_1)) = 2 + x_1 POL(A__TAKE(x_1, x_2)) = 1 + x_1 + x_2 POL(MARK(x_1)) = x_1 POL(a__and(x_1, x_2)) = 2*x_1 + x_2 POL(a__length(x_1)) = 2 + x_1 POL(a__take(x_1, x_2)) = 1 + x_1 + x_2 POL(a__zeros) = 1 POL(and(x_1, x_2)) = 2*x_1 + x_2 POL(cons(x_1, x_2)) = x_1 + x_2 POL(length(x_1)) = 2 + x_1 POL(mark(x_1)) = x_1 POL(nil) = 1 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = 1 + x_1 + x_2 POL(tt) = 1 POL(zeros) = 1 ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(length(X)) -> A__LENGTH(mark(X)) A__LENGTH(cons(N, L)) -> A__LENGTH(mark(L)) MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2)) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes. ---------------------------------------- (14) Complex Obligation (AND) ---------------------------------------- (15) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__LENGTH(mark(L)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (16) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule A__LENGTH(cons(N, L)) -> A__LENGTH(mark(L)) at position [0] we obtained the following new rules [LPAR04]: (A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros),A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros)) (A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1)),A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1))) (A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0))),A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0)))) (A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1))),A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1)))) (A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1)),A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1))) (A__LENGTH(cons(y0, 0)) -> A__LENGTH(0),A__LENGTH(cons(y0, 0)) -> A__LENGTH(0)) (A__LENGTH(cons(y0, tt)) -> A__LENGTH(tt),A__LENGTH(cons(y0, tt)) -> A__LENGTH(tt)) (A__LENGTH(cons(y0, nil)) -> A__LENGTH(nil),A__LENGTH(cons(y0, nil)) -> A__LENGTH(nil)) (A__LENGTH(cons(y0, s(x0))) -> A__LENGTH(s(mark(x0))),A__LENGTH(cons(y0, s(x0))) -> A__LENGTH(s(mark(x0)))) ---------------------------------------- (17) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros) A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1)) A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0))) A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1)) A__LENGTH(cons(y0, 0)) -> A__LENGTH(0) A__LENGTH(cons(y0, tt)) -> A__LENGTH(tt) A__LENGTH(cons(y0, nil)) -> A__LENGTH(nil) A__LENGTH(cons(y0, s(x0))) -> A__LENGTH(s(mark(x0))) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (18) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros) A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1)) A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0))) A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros) at position [0] we obtained the following new rules [LPAR04]: (A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros)),A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros))) (A__LENGTH(cons(y0, zeros)) -> A__LENGTH(zeros),A__LENGTH(cons(y0, zeros)) -> A__LENGTH(zeros)) ---------------------------------------- (21) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1)) A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0))) A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1)) A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros)) A__LENGTH(cons(y0, zeros)) -> A__LENGTH(zeros) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. ---------------------------------------- (23) Complex Obligation (AND) ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros)) we obtained the following new rules [LPAR04]: (A__LENGTH(cons(0, zeros)) -> A__LENGTH(cons(0, zeros)),A__LENGTH(cons(0, zeros)) -> A__LENGTH(cons(0, zeros))) ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(0, zeros)) -> A__LENGTH(cons(0, zeros)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) NonTerminationLoopProof (COMPLETE) 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 = A__LENGTH(cons(0, zeros)) evaluates to t =A__LENGTH(cons(0, zeros)) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from A__LENGTH(cons(0, zeros)) to A__LENGTH(cons(0, zeros)). ---------------------------------------- (30) NO ---------------------------------------- (31) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1)) A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0))) A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (32) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1)) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(A__LENGTH(x_1)) = x_1 POL(a__and(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 POL(a__length(x_1)) = x_1 POL(a__take(x_1, x_2)) = 2*x_1 + x_2 POL(a__zeros) = 0 POL(and(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(length(x_1)) = x_1 POL(mark(x_1)) = x_1 POL(nil) = 2 POL(s(x_1)) = 2*x_1 POL(take(x_1, x_2)) = 2*x_1 + x_2 POL(tt) = 0 POL(zeros) = 0 ---------------------------------------- (33) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0))) A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__length(cons(N, L)) -> s(a__length(mark(L))) a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) a__length(X) -> length(X) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (37) QDPSizeChangeProof (EQUIVALENT) 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: *MARK(s(X)) -> MARK(X) The graph contains the following edges 1 > 1 *MARK(cons(X1, X2)) -> MARK(X1) The graph contains the following edges 1 > 1 ---------------------------------------- (38) YES