/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.pl /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.pl # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern ordered(g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 3 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) PiDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) PiDP (10) PiDPToQDPProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) PiDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) PiDP (17) PiDPToQDPProof [SOUND, 0 ms] (18) QDP (19) UsableRulesReductionPairsProof [EQUIVALENT, 15 ms] (20) QDP (21) MRRProof [EQUIVALENT, 10 ms] (22) QDP (23) PisEmptyProof [EQUIVALENT, 0 ms] (24) YES ---------------------------------------- (0) Obligation: Clauses: ordered([]). ordered(.(X1, [])). ordered(.(X, .(Y, Xs))) :- ','(less(X, s(Y)), ordered(.(Y, Xs))). less(0, s(X2)). less(s(X), s(Y)) :- less(X, Y). Query: ordered(g) ---------------------------------------- (1) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: ordered_in_1: (b) less_in_2: (b,b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: ordered_in_g([]) -> ordered_out_g([]) ordered_in_g(.(X1, [])) -> ordered_out_g(.(X1, [])) ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, less_in_gg(X, s(Y))) less_in_gg(0, s(X2)) -> less_out_gg(0, s(X2)) less_in_gg(s(X), s(Y)) -> U3_gg(X, Y, less_in_gg(X, Y)) U3_gg(X, Y, less_out_gg(X, Y)) -> less_out_gg(s(X), s(Y)) U1_g(X, Y, Xs, less_out_gg(X, s(Y))) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs))) U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs))) The argument filtering Pi contains the following mapping: ordered_in_g(x1) = ordered_in_g(x1) [] = [] ordered_out_g(x1) = ordered_out_g .(x1, x2) = .(x1, x2) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) less_in_gg(x1, x2) = less_in_gg(x1, x2) 0 = 0 s(x1) = s(x1) less_out_gg(x1, x2) = less_out_gg U3_gg(x1, x2, x3) = U3_gg(x3) U2_g(x1, x2, x3, x4) = U2_g(x4) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: ordered_in_g([]) -> ordered_out_g([]) ordered_in_g(.(X1, [])) -> ordered_out_g(.(X1, [])) ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, less_in_gg(X, s(Y))) less_in_gg(0, s(X2)) -> less_out_gg(0, s(X2)) less_in_gg(s(X), s(Y)) -> U3_gg(X, Y, less_in_gg(X, Y)) U3_gg(X, Y, less_out_gg(X, Y)) -> less_out_gg(s(X), s(Y)) U1_g(X, Y, Xs, less_out_gg(X, s(Y))) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs))) U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs))) The argument filtering Pi contains the following mapping: ordered_in_g(x1) = ordered_in_g(x1) [] = [] ordered_out_g(x1) = ordered_out_g .(x1, x2) = .(x1, x2) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) less_in_gg(x1, x2) = less_in_gg(x1, x2) 0 = 0 s(x1) = s(x1) less_out_gg(x1, x2) = less_out_gg U3_gg(x1, x2, x3) = U3_gg(x3) U2_g(x1, x2, x3, x4) = U2_g(x4) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: Pi DP problem: The TRS P consists of the following rules: ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(X, Y, Xs, less_in_gg(X, s(Y))) ORDERED_IN_G(.(X, .(Y, Xs))) -> LESS_IN_GG(X, s(Y)) LESS_IN_GG(s(X), s(Y)) -> U3_GG(X, Y, less_in_gg(X, Y)) LESS_IN_GG(s(X), s(Y)) -> LESS_IN_GG(X, Y) U1_G(X, Y, Xs, less_out_gg(X, s(Y))) -> U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs))) U1_G(X, Y, Xs, less_out_gg(X, s(Y))) -> ORDERED_IN_G(.(Y, Xs)) The TRS R consists of the following rules: ordered_in_g([]) -> ordered_out_g([]) ordered_in_g(.(X1, [])) -> ordered_out_g(.(X1, [])) ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, less_in_gg(X, s(Y))) less_in_gg(0, s(X2)) -> less_out_gg(0, s(X2)) less_in_gg(s(X), s(Y)) -> U3_gg(X, Y, less_in_gg(X, Y)) U3_gg(X, Y, less_out_gg(X, Y)) -> less_out_gg(s(X), s(Y)) U1_g(X, Y, Xs, less_out_gg(X, s(Y))) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs))) U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs))) The argument filtering Pi contains the following mapping: ordered_in_g(x1) = ordered_in_g(x1) [] = [] ordered_out_g(x1) = ordered_out_g .(x1, x2) = .(x1, x2) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) less_in_gg(x1, x2) = less_in_gg(x1, x2) 0 = 0 s(x1) = s(x1) less_out_gg(x1, x2) = less_out_gg U3_gg(x1, x2, x3) = U3_gg(x3) U2_g(x1, x2, x3, x4) = U2_g(x4) ORDERED_IN_G(x1) = ORDERED_IN_G(x1) U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4) LESS_IN_GG(x1, x2) = LESS_IN_GG(x1, x2) U3_GG(x1, x2, x3) = U3_GG(x3) U2_G(x1, x2, x3, x4) = U2_G(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(X, Y, Xs, less_in_gg(X, s(Y))) ORDERED_IN_G(.(X, .(Y, Xs))) -> LESS_IN_GG(X, s(Y)) LESS_IN_GG(s(X), s(Y)) -> U3_GG(X, Y, less_in_gg(X, Y)) LESS_IN_GG(s(X), s(Y)) -> LESS_IN_GG(X, Y) U1_G(X, Y, Xs, less_out_gg(X, s(Y))) -> U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs))) U1_G(X, Y, Xs, less_out_gg(X, s(Y))) -> ORDERED_IN_G(.(Y, Xs)) The TRS R consists of the following rules: ordered_in_g([]) -> ordered_out_g([]) ordered_in_g(.(X1, [])) -> ordered_out_g(.(X1, [])) ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, less_in_gg(X, s(Y))) less_in_gg(0, s(X2)) -> less_out_gg(0, s(X2)) less_in_gg(s(X), s(Y)) -> U3_gg(X, Y, less_in_gg(X, Y)) U3_gg(X, Y, less_out_gg(X, Y)) -> less_out_gg(s(X), s(Y)) U1_g(X, Y, Xs, less_out_gg(X, s(Y))) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs))) U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs))) The argument filtering Pi contains the following mapping: ordered_in_g(x1) = ordered_in_g(x1) [] = [] ordered_out_g(x1) = ordered_out_g .(x1, x2) = .(x1, x2) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) less_in_gg(x1, x2) = less_in_gg(x1, x2) 0 = 0 s(x1) = s(x1) less_out_gg(x1, x2) = less_out_gg U3_gg(x1, x2, x3) = U3_gg(x3) U2_g(x1, x2, x3, x4) = U2_g(x4) ORDERED_IN_G(x1) = ORDERED_IN_G(x1) U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4) LESS_IN_GG(x1, x2) = LESS_IN_GG(x1, x2) U3_GG(x1, x2, x3) = U3_GG(x3) U2_G(x1, x2, x3, x4) = U2_G(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: LESS_IN_GG(s(X), s(Y)) -> LESS_IN_GG(X, Y) The TRS R consists of the following rules: ordered_in_g([]) -> ordered_out_g([]) ordered_in_g(.(X1, [])) -> ordered_out_g(.(X1, [])) ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, less_in_gg(X, s(Y))) less_in_gg(0, s(X2)) -> less_out_gg(0, s(X2)) less_in_gg(s(X), s(Y)) -> U3_gg(X, Y, less_in_gg(X, Y)) U3_gg(X, Y, less_out_gg(X, Y)) -> less_out_gg(s(X), s(Y)) U1_g(X, Y, Xs, less_out_gg(X, s(Y))) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs))) U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs))) The argument filtering Pi contains the following mapping: ordered_in_g(x1) = ordered_in_g(x1) [] = [] ordered_out_g(x1) = ordered_out_g .(x1, x2) = .(x1, x2) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) less_in_gg(x1, x2) = less_in_gg(x1, x2) 0 = 0 s(x1) = s(x1) less_out_gg(x1, x2) = less_out_gg U3_gg(x1, x2, x3) = U3_gg(x3) U2_g(x1, x2, x3, x4) = U2_g(x4) LESS_IN_GG(x1, x2) = LESS_IN_GG(x1, x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (8) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (9) Obligation: Pi DP problem: The TRS P consists of the following rules: LESS_IN_GG(s(X), s(Y)) -> LESS_IN_GG(X, Y) R is empty. Pi is empty. We have to consider all (P,R,Pi)-chains ---------------------------------------- (10) PiDPToQDPProof (EQUIVALENT) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: LESS_IN_GG(s(X), s(Y)) -> LESS_IN_GG(X, Y) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (12) 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: *LESS_IN_GG(s(X), s(Y)) -> LESS_IN_GG(X, Y) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Pi DP problem: The TRS P consists of the following rules: U1_G(X, Y, Xs, less_out_gg(X, s(Y))) -> ORDERED_IN_G(.(Y, Xs)) ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(X, Y, Xs, less_in_gg(X, s(Y))) The TRS R consists of the following rules: ordered_in_g([]) -> ordered_out_g([]) ordered_in_g(.(X1, [])) -> ordered_out_g(.(X1, [])) ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, less_in_gg(X, s(Y))) less_in_gg(0, s(X2)) -> less_out_gg(0, s(X2)) less_in_gg(s(X), s(Y)) -> U3_gg(X, Y, less_in_gg(X, Y)) U3_gg(X, Y, less_out_gg(X, Y)) -> less_out_gg(s(X), s(Y)) U1_g(X, Y, Xs, less_out_gg(X, s(Y))) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs))) U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs))) The argument filtering Pi contains the following mapping: ordered_in_g(x1) = ordered_in_g(x1) [] = [] ordered_out_g(x1) = ordered_out_g .(x1, x2) = .(x1, x2) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) less_in_gg(x1, x2) = less_in_gg(x1, x2) 0 = 0 s(x1) = s(x1) less_out_gg(x1, x2) = less_out_gg U3_gg(x1, x2, x3) = U3_gg(x3) U2_g(x1, x2, x3, x4) = U2_g(x4) ORDERED_IN_G(x1) = ORDERED_IN_G(x1) U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (15) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (16) Obligation: Pi DP problem: The TRS P consists of the following rules: U1_G(X, Y, Xs, less_out_gg(X, s(Y))) -> ORDERED_IN_G(.(Y, Xs)) ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(X, Y, Xs, less_in_gg(X, s(Y))) The TRS R consists of the following rules: less_in_gg(0, s(X2)) -> less_out_gg(0, s(X2)) less_in_gg(s(X), s(Y)) -> U3_gg(X, Y, less_in_gg(X, Y)) U3_gg(X, Y, less_out_gg(X, Y)) -> less_out_gg(s(X), s(Y)) The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) less_in_gg(x1, x2) = less_in_gg(x1, x2) 0 = 0 s(x1) = s(x1) less_out_gg(x1, x2) = less_out_gg U3_gg(x1, x2, x3) = U3_gg(x3) ORDERED_IN_G(x1) = ORDERED_IN_G(x1) U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (17) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: U1_G(Y, Xs, less_out_gg) -> ORDERED_IN_G(.(Y, Xs)) ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(Y, Xs, less_in_gg(X, s(Y))) The TRS R consists of the following rules: less_in_gg(0, s(X2)) -> less_out_gg less_in_gg(s(X), s(Y)) -> U3_gg(less_in_gg(X, Y)) U3_gg(less_out_gg) -> less_out_gg The set Q consists of the following terms: less_in_gg(x0, x1) U3_gg(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (19) UsableRulesReductionPairsProof (EQUIVALENT) 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: less_in_gg(0, s(X2)) -> less_out_gg Used ordering: POLO with Polynomial interpretation [POLO]: POL(.(x_1, x_2)) = 2*x_1 + 2*x_2 POL(0) = 0 POL(ORDERED_IN_G(x_1)) = x_1 POL(U1_G(x_1, x_2, x_3)) = 2*x_1 + 2*x_2 + 2*x_3 POL(U3_gg(x_1)) = x_1 POL(less_in_gg(x_1, x_2)) = x_1 + x_2 POL(less_out_gg) = 0 POL(s(x_1)) = x_1 ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: U1_G(Y, Xs, less_out_gg) -> ORDERED_IN_G(.(Y, Xs)) ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(Y, Xs, less_in_gg(X, s(Y))) The TRS R consists of the following rules: less_in_gg(s(X), s(Y)) -> U3_gg(less_in_gg(X, Y)) U3_gg(less_out_gg) -> less_out_gg The set Q consists of the following terms: less_in_gg(x0, x1) U3_gg(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (21) 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: U1_G(Y, Xs, less_out_gg) -> ORDERED_IN_G(.(Y, Xs)) ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(Y, Xs, less_in_gg(X, s(Y))) Used ordering: Polynomial interpretation [POLO]: POL(.(x_1, x_2)) = 2*x_1 + 2*x_2 POL(ORDERED_IN_G(x_1)) = 2 + x_1 POL(U1_G(x_1, x_2, x_3)) = 2*x_1 + 2*x_2 + 2*x_3 POL(U3_gg(x_1)) = x_1 POL(less_in_gg(x_1, x_2)) = x_1 + x_2 POL(less_out_gg) = 2 POL(s(x_1)) = x_1 ---------------------------------------- (22) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: less_in_gg(s(X), s(Y)) -> U3_gg(less_in_gg(X, Y)) U3_gg(less_out_gg) -> less_out_gg The set Q consists of the following terms: less_in_gg(x0, x1) U3_gg(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (23) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (24) YES