NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 73 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 16 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 8 ms] (8) QTRS (9) DependencyPairsProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) AND (13) QDP (14) UsableRulesProof [EQUIVALENT, 0 ms] (15) QDP (16) QReductionProof [EQUIVALENT, 0 ms] (17) QDP (18) QDPSizeChangeProof [EQUIVALENT, 0 ms] (19) YES (20) QDP (21) UsableRulesProof [EQUIVALENT, 0 ms] (22) QDP (23) QReductionProof [EQUIVALENT, 0 ms] (24) QDP (25) MRRProof [EQUIVALENT, 0 ms] (26) QDP (27) TransformationProof [EQUIVALENT, 0 ms] (28) QDP (29) UsableRulesProof [EQUIVALENT, 0 ms] (30) QDP (31) QReductionProof [EQUIVALENT, 2 ms] (32) QDP (33) TransformationProof [EQUIVALENT, 0 ms] (34) QDP (35) UsableRulesProof [EQUIVALENT, 0 ms] (36) QDP (37) QReductionProof [EQUIVALENT, 0 ms] (38) QDP (39) TransformationProof [EQUIVALENT, 0 ms] (40) QDP (41) NonTerminationLoopProof [COMPLETE, 0 ms] (42) NO ---------------------------------------- (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))) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) 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) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__and(x_1, x_2)) = 2*x_1 + 2*x_2 POL(a__length(x_1)) = 2 + x_1 POL(a__zeros) = 0 POL(and(x_1, x_2)) = 2*x_1 + 2*x_2 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(length(x_1)) = 1 + x_1 POL(mark(x_1)) = 2*x_1 POL(nil) = 2 POL(s(x_1)) = x_1 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) a__length(nil) -> 0 mark(tt) -> tt mark(nil) -> nil a__length(X) -> length(X) ---------------------------------------- (2) 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))) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(s(X)) -> s(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__and(x_1, x_2)) = 2 + x_1 + 2*x_2 POL(a__length(x_1)) = 2*x_1 POL(a__zeros) = 2 POL(and(x_1, x_2)) = 1 + x_1 + x_2 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(length(x_1)) = 2 + 2*x_1 POL(mark(x_1)) = 2*x_1 POL(s(x_1)) = x_1 POL(zeros) = 1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: mark(length(X)) -> a__length(mark(X)) a__zeros -> zeros a__and(X1, X2) -> and(X1, X2) ---------------------------------------- (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))) mark(zeros) -> a__zeros mark(and(X1, X2)) -> a__and(mark(X1), X2) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(s(X)) -> s(mark(X)) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__and(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(a__length(x_1)) = 2*x_1 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(mark(x_1)) = x_1 POL(s(x_1)) = 2*x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: mark(and(X1, X2)) -> a__and(mark(X1), X2) ---------------------------------------- (6) 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))) mark(zeros) -> a__zeros mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(s(X)) -> s(mark(X)) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) ---------------------------------------- (7) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__length(x_1)) = 2*x_1 POL(a__zeros) = 2 POL(cons(x_1, x_2)) = 2 + x_1 + 2*x_2 POL(mark(x_1)) = 2 + 2*x_1 POL(s(x_1)) = x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 ---------------------------------------- (8) 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))) mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) ---------------------------------------- (9) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (10) 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) MARK(zeros) -> A__ZEROS 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))) mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes. ---------------------------------------- (12) Complex Obligation (AND) ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: 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))) mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (15) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) R is empty. The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (16) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) ---------------------------------------- (17) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (18) 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 ---------------------------------------- (19) YES ---------------------------------------- (20) 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))) mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (22) 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: mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) a__and(x0, x1) a__length(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. a__and(x0, x1) a__length(x0) ---------------------------------------- (24) 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: mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) 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 rules of the TRS R: mark(s(X)) -> s(mark(X)) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(A__LENGTH(x_1)) = 2*x_1 POL(a__zeros) = 0 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(mark(x_1)) = 2*x_1 POL(s(x_1)) = 1 + 2*x_1 POL(zeros) = 0 ---------------------------------------- (26) 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: mark(zeros) -> a__zeros a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) 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)) ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros) The TRS R consists of the following rules: mark(zeros) -> a__zeros a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. mark(zeros) mark(and(x0, x1)) mark(length(x0)) mark(cons(x0, x1)) mark(0) mark(tt) mark(nil) mark(s(x0)) ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) TransformationProof (EQUIVALENT) By rewriting [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))) ---------------------------------------- (34) 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) The set Q consists of the following terms: a__zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (36) 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. The set Q consists of the following terms: a__zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (37) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. a__zeros ---------------------------------------- (38) 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. ---------------------------------------- (39) 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))) ---------------------------------------- (40) 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. ---------------------------------------- (41) 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)). ---------------------------------------- (42) NO