/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.pl /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Left Termination of the query pattern rev(g,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) PiDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) PiDP (10) PiDPToQDPProof [SOUND, 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) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Clauses: rev([], []). rev(.(X, Xs), Ys) :- ','(rev(Xs, Zs), app(Zs, .(X, []), Ys)). app([], X, X). app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs). Query: rev(g,a) ---------------------------------------- (1) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: rev_in_2: (b,f) app_in_3: (b,b,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: rev_in_ga([], []) -> rev_out_ga([], []) rev_in_ga(.(X, Xs), Ys) -> U1_ga(X, Xs, Ys, rev_in_ga(Xs, Zs)) U1_ga(X, Xs, Ys, rev_out_ga(Xs, Zs)) -> U2_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys)) app_in_gga([], X, X) -> app_out_gga([], X, X) app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs)) U3_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> rev_out_ga(.(X, Xs), Ys) The argument filtering Pi contains the following mapping: rev_in_ga(x1, x2) = rev_in_ga(x1) [] = [] rev_out_ga(x1, x2) = rev_out_ga(x1, x2) .(x1, x2) = .(x1, x2) U1_ga(x1, x2, x3, x4) = U1_ga(x1, x2, x4) U2_ga(x1, x2, x3, x4) = U2_ga(x1, x2, x4) app_in_gga(x1, x2, x3) = app_in_gga(x1, x2) app_out_gga(x1, x2, x3) = app_out_gga(x1, x2, x3) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x2, x3, x5) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: rev_in_ga([], []) -> rev_out_ga([], []) rev_in_ga(.(X, Xs), Ys) -> U1_ga(X, Xs, Ys, rev_in_ga(Xs, Zs)) U1_ga(X, Xs, Ys, rev_out_ga(Xs, Zs)) -> U2_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys)) app_in_gga([], X, X) -> app_out_gga([], X, X) app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs)) U3_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> rev_out_ga(.(X, Xs), Ys) The argument filtering Pi contains the following mapping: rev_in_ga(x1, x2) = rev_in_ga(x1) [] = [] rev_out_ga(x1, x2) = rev_out_ga(x1, x2) .(x1, x2) = .(x1, x2) U1_ga(x1, x2, x3, x4) = U1_ga(x1, x2, x4) U2_ga(x1, x2, x3, x4) = U2_ga(x1, x2, x4) app_in_gga(x1, x2, x3) = app_in_gga(x1, x2) app_out_gga(x1, x2, x3) = app_out_gga(x1, x2, x3) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x2, x3, x5) ---------------------------------------- (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: REV_IN_GA(.(X, Xs), Ys) -> U1_GA(X, Xs, Ys, rev_in_ga(Xs, Zs)) REV_IN_GA(.(X, Xs), Ys) -> REV_IN_GA(Xs, Zs) U1_GA(X, Xs, Ys, rev_out_ga(Xs, Zs)) -> U2_GA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys)) U1_GA(X, Xs, Ys, rev_out_ga(Xs, Zs)) -> APP_IN_GGA(Zs, .(X, []), Ys) APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> U3_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs)) APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs) The TRS R consists of the following rules: rev_in_ga([], []) -> rev_out_ga([], []) rev_in_ga(.(X, Xs), Ys) -> U1_ga(X, Xs, Ys, rev_in_ga(Xs, Zs)) U1_ga(X, Xs, Ys, rev_out_ga(Xs, Zs)) -> U2_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys)) app_in_gga([], X, X) -> app_out_gga([], X, X) app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs)) U3_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> rev_out_ga(.(X, Xs), Ys) The argument filtering Pi contains the following mapping: rev_in_ga(x1, x2) = rev_in_ga(x1) [] = [] rev_out_ga(x1, x2) = rev_out_ga(x1, x2) .(x1, x2) = .(x1, x2) U1_ga(x1, x2, x3, x4) = U1_ga(x1, x2, x4) U2_ga(x1, x2, x3, x4) = U2_ga(x1, x2, x4) app_in_gga(x1, x2, x3) = app_in_gga(x1, x2) app_out_gga(x1, x2, x3) = app_out_gga(x1, x2, x3) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x2, x3, x5) REV_IN_GA(x1, x2) = REV_IN_GA(x1) U1_GA(x1, x2, x3, x4) = U1_GA(x1, x2, x4) U2_GA(x1, x2, x3, x4) = U2_GA(x1, x2, x4) APP_IN_GGA(x1, x2, x3) = APP_IN_GGA(x1, x2) U3_GGA(x1, x2, x3, x4, x5) = U3_GGA(x1, x2, x3, x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: REV_IN_GA(.(X, Xs), Ys) -> U1_GA(X, Xs, Ys, rev_in_ga(Xs, Zs)) REV_IN_GA(.(X, Xs), Ys) -> REV_IN_GA(Xs, Zs) U1_GA(X, Xs, Ys, rev_out_ga(Xs, Zs)) -> U2_GA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys)) U1_GA(X, Xs, Ys, rev_out_ga(Xs, Zs)) -> APP_IN_GGA(Zs, .(X, []), Ys) APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> U3_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs)) APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs) The TRS R consists of the following rules: rev_in_ga([], []) -> rev_out_ga([], []) rev_in_ga(.(X, Xs), Ys) -> U1_ga(X, Xs, Ys, rev_in_ga(Xs, Zs)) U1_ga(X, Xs, Ys, rev_out_ga(Xs, Zs)) -> U2_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys)) app_in_gga([], X, X) -> app_out_gga([], X, X) app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs)) U3_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> rev_out_ga(.(X, Xs), Ys) The argument filtering Pi contains the following mapping: rev_in_ga(x1, x2) = rev_in_ga(x1) [] = [] rev_out_ga(x1, x2) = rev_out_ga(x1, x2) .(x1, x2) = .(x1, x2) U1_ga(x1, x2, x3, x4) = U1_ga(x1, x2, x4) U2_ga(x1, x2, x3, x4) = U2_ga(x1, x2, x4) app_in_gga(x1, x2, x3) = app_in_gga(x1, x2) app_out_gga(x1, x2, x3) = app_out_gga(x1, x2, x3) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x2, x3, x5) REV_IN_GA(x1, x2) = REV_IN_GA(x1) U1_GA(x1, x2, x3, x4) = U1_GA(x1, x2, x4) U2_GA(x1, x2, x3, x4) = U2_GA(x1, x2, x4) APP_IN_GGA(x1, x2, x3) = APP_IN_GGA(x1, x2) U3_GGA(x1, x2, x3, x4, x5) = U3_GGA(x1, x2, x3, x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs) The TRS R consists of the following rules: rev_in_ga([], []) -> rev_out_ga([], []) rev_in_ga(.(X, Xs), Ys) -> U1_ga(X, Xs, Ys, rev_in_ga(Xs, Zs)) U1_ga(X, Xs, Ys, rev_out_ga(Xs, Zs)) -> U2_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys)) app_in_gga([], X, X) -> app_out_gga([], X, X) app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs)) U3_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> rev_out_ga(.(X, Xs), Ys) The argument filtering Pi contains the following mapping: rev_in_ga(x1, x2) = rev_in_ga(x1) [] = [] rev_out_ga(x1, x2) = rev_out_ga(x1, x2) .(x1, x2) = .(x1, x2) U1_ga(x1, x2, x3, x4) = U1_ga(x1, x2, x4) U2_ga(x1, x2, x3, x4) = U2_ga(x1, x2, x4) app_in_gga(x1, x2, x3) = app_in_gga(x1, x2) app_out_gga(x1, x2, x3) = app_out_gga(x1, x2, x3) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x2, x3, x5) APP_IN_GGA(x1, x2, x3) = APP_IN_GGA(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: APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) APP_IN_GGA(x1, x2, x3) = APP_IN_GGA(x1, x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (10) PiDPToQDPProof (SOUND) 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: APP_IN_GGA(.(X, Xs), Ys) -> APP_IN_GGA(Xs, Ys) 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: *APP_IN_GGA(.(X, Xs), Ys) -> APP_IN_GGA(Xs, Ys) 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: REV_IN_GA(.(X, Xs), Ys) -> REV_IN_GA(Xs, Zs) The TRS R consists of the following rules: rev_in_ga([], []) -> rev_out_ga([], []) rev_in_ga(.(X, Xs), Ys) -> U1_ga(X, Xs, Ys, rev_in_ga(Xs, Zs)) U1_ga(X, Xs, Ys, rev_out_ga(Xs, Zs)) -> U2_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys)) app_in_gga([], X, X) -> app_out_gga([], X, X) app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs)) U3_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> rev_out_ga(.(X, Xs), Ys) The argument filtering Pi contains the following mapping: rev_in_ga(x1, x2) = rev_in_ga(x1) [] = [] rev_out_ga(x1, x2) = rev_out_ga(x1, x2) .(x1, x2) = .(x1, x2) U1_ga(x1, x2, x3, x4) = U1_ga(x1, x2, x4) U2_ga(x1, x2, x3, x4) = U2_ga(x1, x2, x4) app_in_gga(x1, x2, x3) = app_in_gga(x1, x2) app_out_gga(x1, x2, x3) = app_out_gga(x1, x2, x3) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x2, x3, x5) REV_IN_GA(x1, x2) = REV_IN_GA(x1) 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: REV_IN_GA(.(X, Xs), Ys) -> REV_IN_GA(Xs, Zs) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) REV_IN_GA(x1, x2) = REV_IN_GA(x1) 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: REV_IN_GA(.(X, Xs)) -> REV_IN_GA(Xs) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (19) 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: *REV_IN_GA(.(X, Xs)) -> REV_IN_GA(Xs) The graph contains the following edges 1 > 1 ---------------------------------------- (20) YES