NO proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 58 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 19 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) DependencyPairsProof [EQUIVALENT, 0 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) MRRProof [EQUIVALENT, 13 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 52 ms] (14) QDP (15) TransformationProof [EQUIVALENT, 0 ms] (16) QDP (17) NonTerminationLoopProof [COMPLETE, 46 ms] (18) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) adx(nil) -> nil adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) nats -> adx(zeros) zeros -> cons(0, n__zeros) head(cons(X, L)) -> X tail(cons(X, L)) -> activate(L) incr(X) -> n__incr(X) adx(X) -> n__adx(X) zeros -> n__zeros activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(n__zeros) -> zeros activate(X) -> X Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = x_1 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(head(x_1)) = 2*x_1 POL(incr(x_1)) = x_1 POL(n__adx(x_1)) = x_1 POL(n__incr(x_1)) = x_1 POL(n__zeros) = 0 POL(nats) = 2 POL(nil) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = 2 + 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: nats -> adx(zeros) tail(cons(X, L)) -> activate(L) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) adx(nil) -> nil adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) zeros -> cons(0, n__zeros) head(cons(X, L)) -> X incr(X) -> n__incr(X) adx(X) -> n__adx(X) zeros -> n__zeros activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(n__zeros) -> zeros activate(X) -> X Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = x_1 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(head(x_1)) = 1 + 2*x_1 POL(incr(x_1)) = x_1 POL(n__adx(x_1)) = x_1 POL(n__incr(x_1)) = x_1 POL(n__zeros) = 0 POL(nil) = 0 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: head(cons(X, L)) -> X ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) adx(nil) -> nil adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) zeros -> cons(0, n__zeros) incr(X) -> n__incr(X) adx(X) -> n__adx(X) zeros -> n__zeros activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(n__zeros) -> zeros activate(X) -> X Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = x_1 POL(adx(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(incr(x_1)) = x_1 POL(n__adx(x_1)) = 2*x_1 POL(n__incr(x_1)) = x_1 POL(n__zeros) = 0 POL(nil) = 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: adx(nil) -> nil ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) zeros -> cons(0, n__zeros) incr(X) -> n__incr(X) adx(X) -> n__adx(X) zeros -> n__zeros activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(n__zeros) -> zeros activate(X) -> X Q is empty. ---------------------------------------- (7) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(cons(X, L)) -> ACTIVATE(L) ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L)))) ADX(cons(X, L)) -> ACTIVATE(L) ACTIVATE(n__incr(X)) -> INCR(X) ACTIVATE(n__adx(X)) -> ADX(X) ACTIVATE(n__zeros) -> ZEROS The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) zeros -> cons(0, n__zeros) incr(X) -> n__incr(X) adx(X) -> n__adx(X) zeros -> n__zeros activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(n__zeros) -> zeros activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__incr(X)) -> INCR(X) INCR(cons(X, L)) -> ACTIVATE(L) ACTIVATE(n__adx(X)) -> ADX(X) ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L)))) ADX(cons(X, L)) -> ACTIVATE(L) The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) zeros -> cons(0, n__zeros) incr(X) -> n__incr(X) adx(X) -> n__adx(X) zeros -> n__zeros activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(n__zeros) -> zeros activate(X) -> X 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: ADX(cons(X, L)) -> ACTIVATE(L) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVATE(x_1)) = x_1 POL(ADX(x_1)) = 1 + x_1 POL(INCR(x_1)) = x_1 POL(activate(x_1)) = x_1 POL(adx(x_1)) = 1 + x_1 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(incr(x_1)) = x_1 POL(n__adx(x_1)) = 1 + x_1 POL(n__incr(x_1)) = x_1 POL(n__zeros) = 0 POL(nil) = 0 POL(s(x_1)) = x_1 POL(zeros) = 0 ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__incr(X)) -> INCR(X) INCR(cons(X, L)) -> ACTIVATE(L) ACTIVATE(n__adx(X)) -> ADX(X) ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L)))) The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) zeros -> cons(0, n__zeros) incr(X) -> n__incr(X) adx(X) -> n__adx(X) zeros -> n__zeros activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(n__zeros) -> zeros activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. ACTIVATE(n__incr(X)) -> INCR(X) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. ACTIVATE(x1) = x1 n__incr(x1) = n__incr(x1) INCR(x1) = x1 cons(x1, x2) = x2 n__adx(x1) = n__adx ADX(x1) = ADX activate(x1) = activate incr(x1) = x1 adx(x1) = adx n__zeros = n__zeros zeros = zeros s(x1) = s(x1) nil = nil 0 = 0 Recursive path order with status [RPO]. Quasi-Precedence: [n__adx, ADX] > [activate, adx] zeros > n__zeros zeros > 0 Status: n__incr_1: [1] n__adx: [] ADX: [] activate: multiset status adx: multiset status n__zeros: multiset status zeros: multiset status s_1: multiset status nil: multiset status 0: multiset status The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: none ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(cons(X, L)) -> ACTIVATE(L) ACTIVATE(n__adx(X)) -> ADX(X) ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L)))) The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) zeros -> cons(0, n__zeros) incr(X) -> n__incr(X) adx(X) -> n__adx(X) zeros -> n__zeros activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(n__zeros) -> zeros activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule INCR(cons(X, L)) -> ACTIVATE(L) we obtained the following new rules [LPAR04]: (INCR(cons(y_0, n__adx(y_2))) -> ACTIVATE(n__adx(y_2)),INCR(cons(y_0, n__adx(y_2))) -> ACTIVATE(n__adx(y_2))) ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__adx(X)) -> ADX(X) ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L)))) INCR(cons(y_0, n__adx(y_2))) -> ACTIVATE(n__adx(y_2)) The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), n__incr(activate(L))) adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L)))) zeros -> cons(0, n__zeros) incr(X) -> n__incr(X) adx(X) -> n__adx(X) zeros -> n__zeros activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(n__zeros) -> zeros activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = ADX(activate(n__zeros)) evaluates to t =ADX(activate(n__zeros)) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence ADX(activate(n__zeros)) -> ADX(zeros) with rule activate(n__zeros) -> zeros at position [0] and matcher [ ] ADX(zeros) -> ADX(cons(0, n__zeros)) with rule zeros -> cons(0, n__zeros) at position [0] and matcher [ ] ADX(cons(0, n__zeros)) -> INCR(cons(0, n__adx(activate(n__zeros)))) with rule ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L)))) at position [] and matcher [X / 0, L / n__zeros] INCR(cons(0, n__adx(activate(n__zeros)))) -> ACTIVATE(n__adx(activate(n__zeros))) with rule INCR(cons(y_0, n__adx(y_2))) -> ACTIVATE(n__adx(y_2)) at position [] and matcher [y_0 / 0, y_2 / activate(n__zeros)] ACTIVATE(n__adx(activate(n__zeros))) -> ADX(activate(n__zeros)) with rule ACTIVATE(n__adx(X)) -> ADX(X) Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (18) NO