/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern sublist(g,g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 15 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) PiDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) PiDP (10) PiDPToQDPProof [SOUND, 8 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: append1([], Ys, Ys). append1(.(X, Xs), Ys, .(X, Zs)) :- append1(Xs, Ys, Zs). append2([], Ys, Ys). append2(.(X, Xs), Ys, .(X, Zs)) :- append2(Xs, Ys, Zs). sublist(X, Y) :- ','(append1(P, X1, Y), append2(X2, X, P)). Query: sublist(g,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: sublist_in_2: (b,b) append1_in_3: (f,f,b) append2_in_3: (f,b,b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: sublist_in_gg(X, Y) -> U3_gg(X, Y, append1_in_aag(P, X1, Y)) append1_in_aag([], Ys, Ys) -> append1_out_aag([], Ys, Ys) append1_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs)) U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) -> append1_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_gg(X, Y, append1_out_aag(P, X1, Y)) -> U4_gg(X, Y, append2_in_agg(X2, X, P)) append2_in_agg([], Ys, Ys) -> append2_out_agg([], Ys, Ys) append2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs)) U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) -> append2_out_agg(.(X, Xs), Ys, .(X, Zs)) U4_gg(X, Y, append2_out_agg(X2, X, P)) -> sublist_out_gg(X, Y) The argument filtering Pi contains the following mapping: sublist_in_gg(x1, x2) = sublist_in_gg(x1, x2) U3_gg(x1, x2, x3) = U3_gg(x1, x3) append1_in_aag(x1, x2, x3) = append1_in_aag(x3) append1_out_aag(x1, x2, x3) = append1_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x1, x5) U4_gg(x1, x2, x3) = U4_gg(x3) append2_in_agg(x1, x2, x3) = append2_in_agg(x2, x3) append2_out_agg(x1, x2, x3) = append2_out_agg(x1) U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) sublist_out_gg(x1, x2) = sublist_out_gg Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: sublist_in_gg(X, Y) -> U3_gg(X, Y, append1_in_aag(P, X1, Y)) append1_in_aag([], Ys, Ys) -> append1_out_aag([], Ys, Ys) append1_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs)) U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) -> append1_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_gg(X, Y, append1_out_aag(P, X1, Y)) -> U4_gg(X, Y, append2_in_agg(X2, X, P)) append2_in_agg([], Ys, Ys) -> append2_out_agg([], Ys, Ys) append2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs)) U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) -> append2_out_agg(.(X, Xs), Ys, .(X, Zs)) U4_gg(X, Y, append2_out_agg(X2, X, P)) -> sublist_out_gg(X, Y) The argument filtering Pi contains the following mapping: sublist_in_gg(x1, x2) = sublist_in_gg(x1, x2) U3_gg(x1, x2, x3) = U3_gg(x1, x3) append1_in_aag(x1, x2, x3) = append1_in_aag(x3) append1_out_aag(x1, x2, x3) = append1_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x1, x5) U4_gg(x1, x2, x3) = U4_gg(x3) append2_in_agg(x1, x2, x3) = append2_in_agg(x2, x3) append2_out_agg(x1, x2, x3) = append2_out_agg(x1) U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) sublist_out_gg(x1, x2) = sublist_out_gg ---------------------------------------- (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: SUBLIST_IN_GG(X, Y) -> U3_GG(X, Y, append1_in_aag(P, X1, Y)) SUBLIST_IN_GG(X, Y) -> APPEND1_IN_AAG(P, X1, Y) APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U1_AAG(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs)) APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND1_IN_AAG(Xs, Ys, Zs) U3_GG(X, Y, append1_out_aag(P, X1, Y)) -> U4_GG(X, Y, append2_in_agg(X2, X, P)) U3_GG(X, Y, append1_out_aag(P, X1, Y)) -> APPEND2_IN_AGG(X2, X, P) APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> U2_AGG(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs)) APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APPEND2_IN_AGG(Xs, Ys, Zs) The TRS R consists of the following rules: sublist_in_gg(X, Y) -> U3_gg(X, Y, append1_in_aag(P, X1, Y)) append1_in_aag([], Ys, Ys) -> append1_out_aag([], Ys, Ys) append1_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs)) U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) -> append1_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_gg(X, Y, append1_out_aag(P, X1, Y)) -> U4_gg(X, Y, append2_in_agg(X2, X, P)) append2_in_agg([], Ys, Ys) -> append2_out_agg([], Ys, Ys) append2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs)) U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) -> append2_out_agg(.(X, Xs), Ys, .(X, Zs)) U4_gg(X, Y, append2_out_agg(X2, X, P)) -> sublist_out_gg(X, Y) The argument filtering Pi contains the following mapping: sublist_in_gg(x1, x2) = sublist_in_gg(x1, x2) U3_gg(x1, x2, x3) = U3_gg(x1, x3) append1_in_aag(x1, x2, x3) = append1_in_aag(x3) append1_out_aag(x1, x2, x3) = append1_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x1, x5) U4_gg(x1, x2, x3) = U4_gg(x3) append2_in_agg(x1, x2, x3) = append2_in_agg(x2, x3) append2_out_agg(x1, x2, x3) = append2_out_agg(x1) U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) sublist_out_gg(x1, x2) = sublist_out_gg SUBLIST_IN_GG(x1, x2) = SUBLIST_IN_GG(x1, x2) U3_GG(x1, x2, x3) = U3_GG(x1, x3) APPEND1_IN_AAG(x1, x2, x3) = APPEND1_IN_AAG(x3) U1_AAG(x1, x2, x3, x4, x5) = U1_AAG(x1, x5) U4_GG(x1, x2, x3) = U4_GG(x3) APPEND2_IN_AGG(x1, x2, x3) = APPEND2_IN_AGG(x2, x3) U2_AGG(x1, x2, x3, x4, x5) = U2_AGG(x1, x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: SUBLIST_IN_GG(X, Y) -> U3_GG(X, Y, append1_in_aag(P, X1, Y)) SUBLIST_IN_GG(X, Y) -> APPEND1_IN_AAG(P, X1, Y) APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U1_AAG(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs)) APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND1_IN_AAG(Xs, Ys, Zs) U3_GG(X, Y, append1_out_aag(P, X1, Y)) -> U4_GG(X, Y, append2_in_agg(X2, X, P)) U3_GG(X, Y, append1_out_aag(P, X1, Y)) -> APPEND2_IN_AGG(X2, X, P) APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> U2_AGG(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs)) APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APPEND2_IN_AGG(Xs, Ys, Zs) The TRS R consists of the following rules: sublist_in_gg(X, Y) -> U3_gg(X, Y, append1_in_aag(P, X1, Y)) append1_in_aag([], Ys, Ys) -> append1_out_aag([], Ys, Ys) append1_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs)) U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) -> append1_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_gg(X, Y, append1_out_aag(P, X1, Y)) -> U4_gg(X, Y, append2_in_agg(X2, X, P)) append2_in_agg([], Ys, Ys) -> append2_out_agg([], Ys, Ys) append2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs)) U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) -> append2_out_agg(.(X, Xs), Ys, .(X, Zs)) U4_gg(X, Y, append2_out_agg(X2, X, P)) -> sublist_out_gg(X, Y) The argument filtering Pi contains the following mapping: sublist_in_gg(x1, x2) = sublist_in_gg(x1, x2) U3_gg(x1, x2, x3) = U3_gg(x1, x3) append1_in_aag(x1, x2, x3) = append1_in_aag(x3) append1_out_aag(x1, x2, x3) = append1_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x1, x5) U4_gg(x1, x2, x3) = U4_gg(x3) append2_in_agg(x1, x2, x3) = append2_in_agg(x2, x3) append2_out_agg(x1, x2, x3) = append2_out_agg(x1) U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) sublist_out_gg(x1, x2) = sublist_out_gg SUBLIST_IN_GG(x1, x2) = SUBLIST_IN_GG(x1, x2) U3_GG(x1, x2, x3) = U3_GG(x1, x3) APPEND1_IN_AAG(x1, x2, x3) = APPEND1_IN_AAG'(x3) U1_AAG(x1, x2, x3, x4, x5) = U1_AAG(x1, x5) U4_GG(x1, x2, x3) = U4_GG(x3) APPEND2_IN_AGG(x1, x2, x3) = APPEND2_IN_AGG(x2, x3) U2_AGG(x1, x2, x3, x4, x5) = U2_AGG(x1, x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APPEND2_IN_AGG(Xs, Ys, Zs) The TRS R consists of the following rules: sublist_in_gg(X, Y) -> U3_gg(X, Y, append1_in_aag(P, X1, Y)) append1_in_aag([], Ys, Ys) -> append1_out_aag([], Ys, Ys) append1_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs)) U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) -> append1_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_gg(X, Y, append1_out_aag(P, X1, Y)) -> U4_gg(X, Y, append2_in_agg(X2, X, P)) append2_in_agg([], Ys, Ys) -> append2_out_agg([], Ys, Ys) append2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs)) U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) -> append2_out_agg(.(X, Xs), Ys, .(X, Zs)) U4_gg(X, Y, append2_out_agg(X2, X, P)) -> sublist_out_gg(X, Y) The argument filtering Pi contains the following mapping: sublist_in_gg(x1, x2) = sublist_in_gg(x1, x2) U3_gg(x1, x2, x3) = U3_gg(x1, x3) append1_in_aag(x1, x2, x3) = append1_in_aag(x3) append1_out_aag(x1, x2, x3) = append1_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x1, x5) U4_gg(x1, x2, x3) = U4_gg(x3) append2_in_agg(x1, x2, x3) = append2_in_agg(x2, x3) append2_out_agg(x1, x2, x3) = append2_out_agg(x1) U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) sublist_out_gg(x1, x2) = sublist_out_gg APPEND2_IN_AGG(x1, x2, x3) = APPEND2_IN_AGG(x2, x3) 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: APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APPEND2_IN_AGG(Xs, Ys, Zs) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) APPEND2_IN_AGG(x1, x2, x3) = APPEND2_IN_AGG(x2, x3) 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: APPEND2_IN_AGG(Ys, .(X, Zs)) -> APPEND2_IN_AGG(Ys, Zs) 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: *APPEND2_IN_AGG(Ys, .(X, Zs)) -> APPEND2_IN_AGG(Ys, Zs) 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: APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND1_IN_AAG(Xs, Ys, Zs) The TRS R consists of the following rules: sublist_in_gg(X, Y) -> U3_gg(X, Y, append1_in_aag(P, X1, Y)) append1_in_aag([], Ys, Ys) -> append1_out_aag([], Ys, Ys) append1_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs)) U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) -> append1_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_gg(X, Y, append1_out_aag(P, X1, Y)) -> U4_gg(X, Y, append2_in_agg(X2, X, P)) append2_in_agg([], Ys, Ys) -> append2_out_agg([], Ys, Ys) append2_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs)) U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) -> append2_out_agg(.(X, Xs), Ys, .(X, Zs)) U4_gg(X, Y, append2_out_agg(X2, X, P)) -> sublist_out_gg(X, Y) The argument filtering Pi contains the following mapping: sublist_in_gg(x1, x2) = sublist_in_gg(x1, x2) U3_gg(x1, x2, x3) = U3_gg(x1, x3) append1_in_aag(x1, x2, x3) = append1_in_aag(x3) append1_out_aag(x1, x2, x3) = append1_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x1, x5) U4_gg(x1, x2, x3) = U4_gg(x3) append2_in_agg(x1, x2, x3) = append2_in_agg(x2, x3) append2_out_agg(x1, x2, x3) = append2_out_agg(x1) U2_agg(x1, x2, x3, x4, x5) = U2_agg(x1, x5) sublist_out_gg(x1, x2) = sublist_out_gg APPEND1_IN_AAG(x1, x2, x3) = APPEND1_IN_AAG(x3) 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: APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND1_IN_AAG(Xs, Ys, Zs) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) APPEND1_IN_AAG(x1, x2, x3) = APPEND1_IN_AAG(x3) 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: APPEND1_IN_AAG(.(X, Zs)) -> APPEND1_IN_AAG(Zs) 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: *APPEND1_IN_AAG(.(X, Zs)) -> APPEND1_IN_AAG(Zs) The graph contains the following edges 1 > 1 ---------------------------------------- (20) YES