/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern rotate(g,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND,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, 3 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: rotate(X, Y) :- ','(append2(A, B, X), append1(B, A, Y)). append1(.(X, Xs), Ys, .(X, Zs)) :- append1(Xs, Ys, Zs). append1([], Ys, Ys). append2(.(X, Xs), Ys, .(X, Zs)) :- append2(Xs, Ys, Zs). append2([], Ys, Ys). Query: rotate(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: rotate_in_2: (b,f) append2_in_3: (f,f,b) append1_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: rotate_in_ga(X, Y) -> U1_ga(X, Y, append2_in_aag(A, B, X)) append2_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U4_aag(X, Xs, Ys, Zs, append2_in_aag(Xs, Ys, Zs)) append2_in_aag([], Ys, Ys) -> append2_out_aag([], Ys, Ys) U4_aag(X, Xs, Ys, Zs, append2_out_aag(Xs, Ys, Zs)) -> append2_out_aag(.(X, Xs), Ys, .(X, Zs)) U1_ga(X, Y, append2_out_aag(A, B, X)) -> U2_ga(X, Y, append1_in_gga(B, A, Y)) append1_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, append1_in_gga(Xs, Ys, Zs)) append1_in_gga([], Ys, Ys) -> append1_out_gga([], Ys, Ys) U3_gga(X, Xs, Ys, Zs, append1_out_gga(Xs, Ys, Zs)) -> append1_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Y, append1_out_gga(B, A, Y)) -> rotate_out_ga(X, Y) The argument filtering Pi contains the following mapping: rotate_in_ga(x1, x2) = rotate_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) append2_in_aag(x1, x2, x3) = append2_in_aag(x3) .(x1, x2) = .(x1, x2) U4_aag(x1, x2, x3, x4, x5) = U4_aag(x1, x5) append2_out_aag(x1, x2, x3) = append2_out_aag(x1, x2) U2_ga(x1, x2, x3) = U2_ga(x3) append1_in_gga(x1, x2, x3) = append1_in_gga(x1, x2) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x5) [] = [] append1_out_gga(x1, x2, x3) = append1_out_gga(x3) rotate_out_ga(x1, x2) = rotate_out_ga(x2) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: rotate_in_ga(X, Y) -> U1_ga(X, Y, append2_in_aag(A, B, X)) append2_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U4_aag(X, Xs, Ys, Zs, append2_in_aag(Xs, Ys, Zs)) append2_in_aag([], Ys, Ys) -> append2_out_aag([], Ys, Ys) U4_aag(X, Xs, Ys, Zs, append2_out_aag(Xs, Ys, Zs)) -> append2_out_aag(.(X, Xs), Ys, .(X, Zs)) U1_ga(X, Y, append2_out_aag(A, B, X)) -> U2_ga(X, Y, append1_in_gga(B, A, Y)) append1_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, append1_in_gga(Xs, Ys, Zs)) append1_in_gga([], Ys, Ys) -> append1_out_gga([], Ys, Ys) U3_gga(X, Xs, Ys, Zs, append1_out_gga(Xs, Ys, Zs)) -> append1_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Y, append1_out_gga(B, A, Y)) -> rotate_out_ga(X, Y) The argument filtering Pi contains the following mapping: rotate_in_ga(x1, x2) = rotate_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) append2_in_aag(x1, x2, x3) = append2_in_aag(x3) .(x1, x2) = .(x1, x2) U4_aag(x1, x2, x3, x4, x5) = U4_aag(x1, x5) append2_out_aag(x1, x2, x3) = append2_out_aag(x1, x2) U2_ga(x1, x2, x3) = U2_ga(x3) append1_in_gga(x1, x2, x3) = append1_in_gga(x1, x2) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x5) [] = [] append1_out_gga(x1, x2, x3) = append1_out_gga(x3) rotate_out_ga(x1, x2) = rotate_out_ga(x2) ---------------------------------------- (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: ROTATE_IN_GA(X, Y) -> U1_GA(X, Y, append2_in_aag(A, B, X)) ROTATE_IN_GA(X, Y) -> APPEND2_IN_AAG(A, B, X) APPEND2_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U4_AAG(X, Xs, Ys, Zs, append2_in_aag(Xs, Ys, Zs)) APPEND2_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND2_IN_AAG(Xs, Ys, Zs) U1_GA(X, Y, append2_out_aag(A, B, X)) -> U2_GA(X, Y, append1_in_gga(B, A, Y)) U1_GA(X, Y, append2_out_aag(A, B, X)) -> APPEND1_IN_GGA(B, A, Y) APPEND1_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> U3_GGA(X, Xs, Ys, Zs, append1_in_gga(Xs, Ys, Zs)) APPEND1_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APPEND1_IN_GGA(Xs, Ys, Zs) The TRS R consists of the following rules: rotate_in_ga(X, Y) -> U1_ga(X, Y, append2_in_aag(A, B, X)) append2_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U4_aag(X, Xs, Ys, Zs, append2_in_aag(Xs, Ys, Zs)) append2_in_aag([], Ys, Ys) -> append2_out_aag([], Ys, Ys) U4_aag(X, Xs, Ys, Zs, append2_out_aag(Xs, Ys, Zs)) -> append2_out_aag(.(X, Xs), Ys, .(X, Zs)) U1_ga(X, Y, append2_out_aag(A, B, X)) -> U2_ga(X, Y, append1_in_gga(B, A, Y)) append1_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, append1_in_gga(Xs, Ys, Zs)) append1_in_gga([], Ys, Ys) -> append1_out_gga([], Ys, Ys) U3_gga(X, Xs, Ys, Zs, append1_out_gga(Xs, Ys, Zs)) -> append1_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Y, append1_out_gga(B, A, Y)) -> rotate_out_ga(X, Y) The argument filtering Pi contains the following mapping: rotate_in_ga(x1, x2) = rotate_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) append2_in_aag(x1, x2, x3) = append2_in_aag(x3) .(x1, x2) = .(x1, x2) U4_aag(x1, x2, x3, x4, x5) = U4_aag(x1, x5) append2_out_aag(x1, x2, x3) = append2_out_aag(x1, x2) U2_ga(x1, x2, x3) = U2_ga(x3) append1_in_gga(x1, x2, x3) = append1_in_gga(x1, x2) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x5) [] = [] append1_out_gga(x1, x2, x3) = append1_out_gga(x3) rotate_out_ga(x1, x2) = rotate_out_ga(x2) ROTATE_IN_GA(x1, x2) = ROTATE_IN_GA(x1) U1_GA(x1, x2, x3) = U1_GA(x3) APPEND2_IN_AAG(x1, x2, x3) = APPEND2_IN_AAG(x3) U4_AAG(x1, x2, x3, x4, x5) = U4_AAG(x1, x5) U2_GA(x1, x2, x3) = U2_GA(x3) APPEND1_IN_GGA(x1, x2, x3) = APPEND1_IN_GGA(x1, x2) U3_GGA(x1, x2, x3, x4, x5) = U3_GGA(x1, x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: ROTATE_IN_GA(X, Y) -> U1_GA(X, Y, append2_in_aag(A, B, X)) ROTATE_IN_GA(X, Y) -> APPEND2_IN_AAG(A, B, X) APPEND2_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U4_AAG(X, Xs, Ys, Zs, append2_in_aag(Xs, Ys, Zs)) APPEND2_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND2_IN_AAG(Xs, Ys, Zs) U1_GA(X, Y, append2_out_aag(A, B, X)) -> U2_GA(X, Y, append1_in_gga(B, A, Y)) U1_GA(X, Y, append2_out_aag(A, B, X)) -> APPEND1_IN_GGA(B, A, Y) APPEND1_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> U3_GGA(X, Xs, Ys, Zs, append1_in_gga(Xs, Ys, Zs)) APPEND1_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APPEND1_IN_GGA(Xs, Ys, Zs) The TRS R consists of the following rules: rotate_in_ga(X, Y) -> U1_ga(X, Y, append2_in_aag(A, B, X)) append2_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U4_aag(X, Xs, Ys, Zs, append2_in_aag(Xs, Ys, Zs)) append2_in_aag([], Ys, Ys) -> append2_out_aag([], Ys, Ys) U4_aag(X, Xs, Ys, Zs, append2_out_aag(Xs, Ys, Zs)) -> append2_out_aag(.(X, Xs), Ys, .(X, Zs)) U1_ga(X, Y, append2_out_aag(A, B, X)) -> U2_ga(X, Y, append1_in_gga(B, A, Y)) append1_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, append1_in_gga(Xs, Ys, Zs)) append1_in_gga([], Ys, Ys) -> append1_out_gga([], Ys, Ys) U3_gga(X, Xs, Ys, Zs, append1_out_gga(Xs, Ys, Zs)) -> append1_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Y, append1_out_gga(B, A, Y)) -> rotate_out_ga(X, Y) The argument filtering Pi contains the following mapping: rotate_in_ga(x1, x2) = rotate_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) append2_in_aag(x1, x2, x3) = append2_in_aag(x3) .(x1, x2) = .(x1, x2) U4_aag(x1, x2, x3, x4, x5) = U4_aag(x1, x5) append2_out_aag(x1, x2, x3) = append2_out_aag(x1, x2) U2_ga(x1, x2, x3) = U2_ga(x3) append1_in_gga(x1, x2, x3) = append1_in_gga(x1, x2) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x5) [] = [] append1_out_gga(x1, x2, x3) = append1_out_gga(x3) rotate_out_ga(x1, x2) = rotate_out_ga(x2) ROTATE_IN_GA(x1, x2) = ROTATE_IN_GA(x1) U1_GA(x1, x2, x3) = U1_GA(x3) APPEND2_IN_AAG(x1, x2, x3) = APPEND2_IN_AAG(x3) U4_AAG(x1, x2, x3, x4, x5) = U4_AAG(x1, x5) U2_GA(x1, x2, x3) = U2_GA(x3) APPEND1_IN_GGA(x1, x2, x3) = APPEND1_IN_GGA(x1, x2) U3_GGA(x1, x2, x3, x4, x5) = U3_GGA(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: APPEND1_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APPEND1_IN_GGA(Xs, Ys, Zs) The TRS R consists of the following rules: rotate_in_ga(X, Y) -> U1_ga(X, Y, append2_in_aag(A, B, X)) append2_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U4_aag(X, Xs, Ys, Zs, append2_in_aag(Xs, Ys, Zs)) append2_in_aag([], Ys, Ys) -> append2_out_aag([], Ys, Ys) U4_aag(X, Xs, Ys, Zs, append2_out_aag(Xs, Ys, Zs)) -> append2_out_aag(.(X, Xs), Ys, .(X, Zs)) U1_ga(X, Y, append2_out_aag(A, B, X)) -> U2_ga(X, Y, append1_in_gga(B, A, Y)) append1_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, append1_in_gga(Xs, Ys, Zs)) append1_in_gga([], Ys, Ys) -> append1_out_gga([], Ys, Ys) U3_gga(X, Xs, Ys, Zs, append1_out_gga(Xs, Ys, Zs)) -> append1_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Y, append1_out_gga(B, A, Y)) -> rotate_out_ga(X, Y) The argument filtering Pi contains the following mapping: rotate_in_ga(x1, x2) = rotate_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) append2_in_aag(x1, x2, x3) = append2_in_aag(x3) .(x1, x2) = .(x1, x2) U4_aag(x1, x2, x3, x4, x5) = U4_aag(x1, x5) append2_out_aag(x1, x2, x3) = append2_out_aag(x1, x2) U2_ga(x1, x2, x3) = U2_ga(x3) append1_in_gga(x1, x2, x3) = append1_in_gga(x1, x2) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x5) [] = [] append1_out_gga(x1, x2, x3) = append1_out_gga(x3) rotate_out_ga(x1, x2) = rotate_out_ga(x2) APPEND1_IN_GGA(x1, x2, x3) = APPEND1_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: APPEND1_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APPEND1_IN_GGA(Xs, Ys, Zs) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) APPEND1_IN_GGA(x1, x2, x3) = APPEND1_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: APPEND1_IN_GGA(.(X, Xs), Ys) -> APPEND1_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: *APPEND1_IN_GGA(.(X, Xs), Ys) -> APPEND1_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: APPEND2_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND2_IN_AAG(Xs, Ys, Zs) The TRS R consists of the following rules: rotate_in_ga(X, Y) -> U1_ga(X, Y, append2_in_aag(A, B, X)) append2_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U4_aag(X, Xs, Ys, Zs, append2_in_aag(Xs, Ys, Zs)) append2_in_aag([], Ys, Ys) -> append2_out_aag([], Ys, Ys) U4_aag(X, Xs, Ys, Zs, append2_out_aag(Xs, Ys, Zs)) -> append2_out_aag(.(X, Xs), Ys, .(X, Zs)) U1_ga(X, Y, append2_out_aag(A, B, X)) -> U2_ga(X, Y, append1_in_gga(B, A, Y)) append1_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U3_gga(X, Xs, Ys, Zs, append1_in_gga(Xs, Ys, Zs)) append1_in_gga([], Ys, Ys) -> append1_out_gga([], Ys, Ys) U3_gga(X, Xs, Ys, Zs, append1_out_gga(Xs, Ys, Zs)) -> append1_out_gga(.(X, Xs), Ys, .(X, Zs)) U2_ga(X, Y, append1_out_gga(B, A, Y)) -> rotate_out_ga(X, Y) The argument filtering Pi contains the following mapping: rotate_in_ga(x1, x2) = rotate_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) append2_in_aag(x1, x2, x3) = append2_in_aag(x3) .(x1, x2) = .(x1, x2) U4_aag(x1, x2, x3, x4, x5) = U4_aag(x1, x5) append2_out_aag(x1, x2, x3) = append2_out_aag(x1, x2) U2_ga(x1, x2, x3) = U2_ga(x3) append1_in_gga(x1, x2, x3) = append1_in_gga(x1, x2) U3_gga(x1, x2, x3, x4, x5) = U3_gga(x1, x5) [] = [] append1_out_gga(x1, x2, x3) = append1_out_gga(x3) rotate_out_ga(x1, x2) = rotate_out_ga(x2) APPEND2_IN_AAG(x1, x2, x3) = APPEND2_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: APPEND2_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND2_IN_AAG(Xs, Ys, Zs) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) APPEND2_IN_AAG(x1, x2, x3) = APPEND2_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: APPEND2_IN_AAG(.(X, Zs)) -> APPEND2_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: *APPEND2_IN_AAG(.(X, Zs)) -> APPEND2_IN_AAG(Zs) The graph contains the following edges 1 > 1 ---------------------------------------- (20) YES