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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 62 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 15 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) MRRProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 62 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 0 ms] (12) QDP (13) TransformationProof [EQUIVALENT, 0 ms] (14) QDP (15) NonTerminationLoopProof [COMPLETE, 0 ms] (16) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: nats -> adx(zeros) zeros -> cons(n__0, n__zeros) incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) hd(cons(X, Y)) -> activate(X) tl(cons(X, Y)) -> activate(Y) 0 -> n__0 zeros -> n__zeros s(X) -> n__s(X) incr(X) -> n__incr(X) adx(X) -> n__adx(X) activate(n__0) -> 0 activate(n__zeros) -> zeros activate(n__s(X)) -> s(X) activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) 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)) = 2*x_1 + 2*x_2 POL(hd(x_1)) = 1 + 2*x_1 POL(incr(x_1)) = x_1 POL(n__0) = 0 POL(n__adx(x_1)) = x_1 POL(n__incr(x_1)) = x_1 POL(n__s(x_1)) = x_1 POL(n__zeros) = 0 POL(nats) = 2 POL(s(x_1)) = x_1 POL(tl(x_1)) = 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: nats -> adx(zeros) hd(cons(X, Y)) -> activate(X) tl(cons(X, Y)) -> activate(Y) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: zeros -> cons(n__0, n__zeros) incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) 0 -> n__0 zeros -> n__zeros s(X) -> n__s(X) incr(X) -> n__incr(X) adx(X) -> n__adx(X) activate(n__0) -> 0 activate(n__zeros) -> zeros activate(n__s(X)) -> s(X) activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(X) -> X Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(cons(X, Y)) -> ACTIVATE(X) INCR(cons(X, Y)) -> ACTIVATE(Y) ADX(cons(X, Y)) -> INCR(cons(activate(X), n__adx(activate(Y)))) ADX(cons(X, Y)) -> ACTIVATE(X) ADX(cons(X, Y)) -> ACTIVATE(Y) ACTIVATE(n__0) -> 0^1 ACTIVATE(n__zeros) -> ZEROS ACTIVATE(n__s(X)) -> S(X) ACTIVATE(n__incr(X)) -> INCR(X) ACTIVATE(n__adx(X)) -> ADX(X) The TRS R consists of the following rules: zeros -> cons(n__0, n__zeros) incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) 0 -> n__0 zeros -> n__zeros s(X) -> n__s(X) incr(X) -> n__incr(X) adx(X) -> n__adx(X) activate(n__0) -> 0 activate(n__zeros) -> zeros activate(n__s(X)) -> s(X) activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__incr(X)) -> INCR(X) INCR(cons(X, Y)) -> ACTIVATE(X) ACTIVATE(n__adx(X)) -> ADX(X) ADX(cons(X, Y)) -> INCR(cons(activate(X), n__adx(activate(Y)))) INCR(cons(X, Y)) -> ACTIVATE(Y) ADX(cons(X, Y)) -> ACTIVATE(X) ADX(cons(X, Y)) -> ACTIVATE(Y) The TRS R consists of the following rules: zeros -> cons(n__0, n__zeros) incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) 0 -> n__0 zeros -> n__zeros s(X) -> n__s(X) incr(X) -> n__incr(X) adx(X) -> n__adx(X) activate(n__0) -> 0 activate(n__zeros) -> zeros activate(n__s(X)) -> s(X) activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) 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, Y)) -> ACTIVATE(X) ADX(cons(X, Y)) -> ACTIVATE(Y) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVATE(x_1)) = x_1 POL(ADX(x_1)) = 2 + x_1 POL(INCR(x_1)) = x_1 POL(activate(x_1)) = x_1 POL(adx(x_1)) = 2 + x_1 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(incr(x_1)) = x_1 POL(n__0) = 0 POL(n__adx(x_1)) = 2 + x_1 POL(n__incr(x_1)) = x_1 POL(n__s(x_1)) = x_1 POL(n__zeros) = 0 POL(s(x_1)) = x_1 POL(zeros) = 0 ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__incr(X)) -> INCR(X) INCR(cons(X, Y)) -> ACTIVATE(X) ACTIVATE(n__adx(X)) -> ADX(X) ADX(cons(X, Y)) -> INCR(cons(activate(X), n__adx(activate(Y)))) INCR(cons(X, Y)) -> ACTIVATE(Y) The TRS R consists of the following rules: zeros -> cons(n__0, n__zeros) incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) 0 -> n__0 zeros -> n__zeros s(X) -> n__s(X) incr(X) -> n__incr(X) adx(X) -> n__adx(X) activate(n__0) -> 0 activate(n__zeros) -> zeros activate(n__s(X)) -> s(X) activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. INCR(cons(X, Y)) -> ACTIVATE(X) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(ACTIVATE(x_1)) = [[-I]] + [[0A]] * x_1 >>> <<< POL(n__incr(x_1)) = [[-I]] + [[0A]] * x_1 >>> <<< POL(INCR(x_1)) = [[-I]] + [[0A]] * x_1 >>> <<< POL(cons(x_1, x_2)) = [[-I]] + [[1A]] * x_1 + [[0A]] * x_2 >>> <<< POL(n__adx(x_1)) = [[-I]] + [[0A]] * x_1 >>> <<< POL(ADX(x_1)) = [[-I]] + [[0A]] * x_1 >>> <<< POL(activate(x_1)) = [[-I]] + [[0A]] * x_1 >>> <<< POL(n__0) = [[0A]] >>> <<< POL(0) = [[0A]] >>> <<< POL(n__zeros) = [[1A]] >>> <<< POL(zeros) = [[1A]] >>> <<< POL(n__s(x_1)) = [[-I]] + [[0A]] * x_1 >>> <<< POL(s(x_1)) = [[-I]] + [[0A]] * x_1 >>> <<< POL(incr(x_1)) = [[-I]] + [[0A]] * x_1 >>> <<< POL(adx(x_1)) = [[-I]] + [[0A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: activate(n__0) -> 0 activate(n__zeros) -> zeros activate(n__s(X)) -> s(X) activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(X) -> X incr(X) -> n__incr(X) incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) adx(X) -> n__adx(X) adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) zeros -> cons(n__0, n__zeros) zeros -> n__zeros 0 -> n__0 s(X) -> n__s(X) ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__incr(X)) -> INCR(X) ACTIVATE(n__adx(X)) -> ADX(X) ADX(cons(X, Y)) -> INCR(cons(activate(X), n__adx(activate(Y)))) INCR(cons(X, Y)) -> ACTIVATE(Y) The TRS R consists of the following rules: zeros -> cons(n__0, n__zeros) incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) 0 -> n__0 zeros -> n__zeros s(X) -> n__s(X) incr(X) -> n__incr(X) adx(X) -> n__adx(X) activate(n__0) -> 0 activate(n__zeros) -> zeros activate(n__s(X)) -> s(X) activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) 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: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( INCR_1(x_1) ) = x_1 + 2 POL( cons_2(x_1, x_2) ) = max{0, 2x_2 - 2} POL( n__adx_1(x_1) ) = 1 POL( activate_1(x_1) ) = 0 POL( n__0 ) = 0 POL( 0 ) = 2 POL( n__zeros ) = 0 POL( zeros ) = 0 POL( n__s_1(x_1) ) = max{0, -2} POL( s_1(x_1) ) = 2x_1 + 1 POL( n__incr_1(x_1) ) = 2x_1 + 2 POL( incr_1(x_1) ) = 1 POL( adx_1(x_1) ) = x_1 + 1 POL( ACTIVATE_1(x_1) ) = 2x_1 POL( ADX_1(x_1) ) = 2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: none ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__adx(X)) -> ADX(X) ADX(cons(X, Y)) -> INCR(cons(activate(X), n__adx(activate(Y)))) INCR(cons(X, Y)) -> ACTIVATE(Y) The TRS R consists of the following rules: zeros -> cons(n__0, n__zeros) incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) 0 -> n__0 zeros -> n__zeros s(X) -> n__s(X) incr(X) -> n__incr(X) adx(X) -> n__adx(X) activate(n__0) -> 0 activate(n__zeros) -> zeros activate(n__s(X)) -> s(X) activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule INCR(cons(X, Y)) -> ACTIVATE(Y) we obtained the following new rules [LPAR04]: (INCR(cons(y_1, n__adx(y_3))) -> ACTIVATE(n__adx(y_3)),INCR(cons(y_1, n__adx(y_3))) -> ACTIVATE(n__adx(y_3))) ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__adx(X)) -> ADX(X) ADX(cons(X, Y)) -> INCR(cons(activate(X), n__adx(activate(Y)))) INCR(cons(y_1, n__adx(y_3))) -> ACTIVATE(n__adx(y_3)) The TRS R consists of the following rules: zeros -> cons(n__0, n__zeros) incr(cons(X, Y)) -> cons(n__s(activate(X)), n__incr(activate(Y))) adx(cons(X, Y)) -> incr(cons(activate(X), n__adx(activate(Y)))) 0 -> n__0 zeros -> n__zeros s(X) -> n__s(X) incr(X) -> n__incr(X) adx(X) -> n__adx(X) activate(n__0) -> 0 activate(n__zeros) -> zeros activate(n__s(X)) -> s(X) activate(n__incr(X)) -> incr(X) activate(n__adx(X)) -> adx(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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(n__0, n__zeros)) with rule zeros -> cons(n__0, n__zeros) at position [0] and matcher [ ] ADX(cons(n__0, n__zeros)) -> INCR(cons(activate(n__0), n__adx(activate(n__zeros)))) with rule ADX(cons(X, Y)) -> INCR(cons(activate(X), n__adx(activate(Y)))) at position [] and matcher [X / n__0, Y / n__zeros] INCR(cons(activate(n__0), n__adx(activate(n__zeros)))) -> ACTIVATE(n__adx(activate(n__zeros))) with rule INCR(cons(y_1, n__adx(y_3))) -> ACTIVATE(n__adx(y_3)) at position [] and matcher [y_1 / activate(n__0), y_3 / 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. ---------------------------------------- (16) NO