/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern query(g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0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, 12 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 (21) PiDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) PiDP (24) PiDPToQDPProof [SOUND, 0 ms] (25) QDP (26) MRRProof [EQUIVALENT, 0 ms] (27) QDP (28) PisEmptyProof [EQUIVALENT, 0 ms] (29) YES ---------------------------------------- (0) Obligation: Clauses: append(nil, XS, XS). append(cons(X, XS), YS, cons(X, ZS)) :- append(XS, YS, ZS). reverse(nil, nil). reverse(cons(X, nil), cons(X, nil)). reverse(cons(X, XS), YS) :- ','(reverse(XS, ZS), append(ZS, cons(X, nil), YS)). shuffle(nil, nil). shuffle(cons(X, XS), cons(X, YS)) :- ','(reverse(XS, ZS), shuffle(ZS, YS)). query(XS) :- shuffle(cons(X, XS), YS). Query: query(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: query_in_1: (b) shuffle_in_2: (b,f) reverse_in_2: (b,f) append_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: query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) The argument filtering Pi contains the following mapping: query_in_g(x1) = query_in_g(x1) U6_g(x1, x2) = U6_g(x2) shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) cons(x1, x2) = cons(x2) nil = nil shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) U4_ga(x1, x2, x3, x4) = U4_ga(x4) reverse_in_ga(x1, x2) = reverse_in_ga(x1) reverse_out_ga(x1, x2) = reverse_out_ga(x2) U2_ga(x1, x2, x3, x4) = U2_ga(x4) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) append_out_gga(x1, x2, x3) = append_out_gga(x3) U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) query_out_g(x1) = query_out_g Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) The argument filtering Pi contains the following mapping: query_in_g(x1) = query_in_g(x1) U6_g(x1, x2) = U6_g(x2) shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) cons(x1, x2) = cons(x2) nil = nil shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) U4_ga(x1, x2, x3, x4) = U4_ga(x4) reverse_in_ga(x1, x2) = reverse_in_ga(x1) reverse_out_ga(x1, x2) = reverse_out_ga(x2) U2_ga(x1, x2, x3, x4) = U2_ga(x4) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) append_out_gga(x1, x2, x3) = append_out_gga(x3) U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) query_out_g(x1) = query_out_g ---------------------------------------- (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: QUERY_IN_G(XS) -> U6_G(XS, shuffle_in_ga(cons(X, XS), YS)) QUERY_IN_G(XS) -> SHUFFLE_IN_GA(cons(X, XS), YS) SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> U4_GA(X, XS, YS, reverse_in_ga(XS, ZS)) SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> REVERSE_IN_GA(XS, ZS) REVERSE_IN_GA(cons(X, XS), YS) -> U2_GA(X, XS, YS, reverse_in_ga(XS, ZS)) REVERSE_IN_GA(cons(X, XS), YS) -> REVERSE_IN_GA(XS, ZS) U2_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_GA(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) U2_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> APPEND_IN_GGA(ZS, cons(X, nil), YS) APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> U1_GGA(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> APPEND_IN_GGA(XS, YS, ZS) U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_GA(X, XS, YS, shuffle_in_ga(ZS, YS)) U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> SHUFFLE_IN_GA(ZS, YS) The TRS R consists of the following rules: query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) The argument filtering Pi contains the following mapping: query_in_g(x1) = query_in_g(x1) U6_g(x1, x2) = U6_g(x2) shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) cons(x1, x2) = cons(x2) nil = nil shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) U4_ga(x1, x2, x3, x4) = U4_ga(x4) reverse_in_ga(x1, x2) = reverse_in_ga(x1) reverse_out_ga(x1, x2) = reverse_out_ga(x2) U2_ga(x1, x2, x3, x4) = U2_ga(x4) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) append_out_gga(x1, x2, x3) = append_out_gga(x3) U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) query_out_g(x1) = query_out_g QUERY_IN_G(x1) = QUERY_IN_G(x1) U6_G(x1, x2) = U6_G(x2) SHUFFLE_IN_GA(x1, x2) = SHUFFLE_IN_GA(x1) U4_GA(x1, x2, x3, x4) = U4_GA(x4) REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1) U2_GA(x1, x2, x3, x4) = U2_GA(x4) U3_GA(x1, x2, x3, x4) = U3_GA(x4) APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2) U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x5) U5_GA(x1, x2, x3, x4) = U5_GA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: QUERY_IN_G(XS) -> U6_G(XS, shuffle_in_ga(cons(X, XS), YS)) QUERY_IN_G(XS) -> SHUFFLE_IN_GA(cons(X, XS), YS) SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> U4_GA(X, XS, YS, reverse_in_ga(XS, ZS)) SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> REVERSE_IN_GA(XS, ZS) REVERSE_IN_GA(cons(X, XS), YS) -> U2_GA(X, XS, YS, reverse_in_ga(XS, ZS)) REVERSE_IN_GA(cons(X, XS), YS) -> REVERSE_IN_GA(XS, ZS) U2_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_GA(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) U2_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> APPEND_IN_GGA(ZS, cons(X, nil), YS) APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> U1_GGA(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> APPEND_IN_GGA(XS, YS, ZS) U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_GA(X, XS, YS, shuffle_in_ga(ZS, YS)) U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> SHUFFLE_IN_GA(ZS, YS) The TRS R consists of the following rules: query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) The argument filtering Pi contains the following mapping: query_in_g(x1) = query_in_g(x1) U6_g(x1, x2) = U6_g(x2) shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) cons(x1, x2) = cons(x2) nil = nil shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) U4_ga(x1, x2, x3, x4) = U4_ga(x4) reverse_in_ga(x1, x2) = reverse_in_ga(x1) reverse_out_ga(x1, x2) = reverse_out_ga(x2) U2_ga(x1, x2, x3, x4) = U2_ga(x4) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) append_out_gga(x1, x2, x3) = append_out_gga(x3) U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) query_out_g(x1) = query_out_g QUERY_IN_G(x1) = QUERY_IN_G(x1) U6_G(x1, x2) = U6_G(x2) SHUFFLE_IN_GA(x1, x2) = SHUFFLE_IN_GA(x1) U4_GA(x1, x2, x3, x4) = U4_GA(x4) REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1) U2_GA(x1, x2, x3, x4) = U2_GA(x4) U3_GA(x1, x2, x3, x4) = U3_GA(x4) APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2) U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x5) U5_GA(x1, x2, x3, x4) = U5_GA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 8 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> APPEND_IN_GGA(XS, YS, ZS) The TRS R consists of the following rules: query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) The argument filtering Pi contains the following mapping: query_in_g(x1) = query_in_g(x1) U6_g(x1, x2) = U6_g(x2) shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) cons(x1, x2) = cons(x2) nil = nil shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) U4_ga(x1, x2, x3, x4) = U4_ga(x4) reverse_in_ga(x1, x2) = reverse_in_ga(x1) reverse_out_ga(x1, x2) = reverse_out_ga(x2) U2_ga(x1, x2, x3, x4) = U2_ga(x4) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) append_out_gga(x1, x2, x3) = append_out_gga(x3) U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) query_out_g(x1) = query_out_g APPEND_IN_GGA(x1, x2, x3) = APPEND_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: APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> APPEND_IN_GGA(XS, YS, ZS) R is empty. The argument filtering Pi contains the following mapping: cons(x1, x2) = cons(x2) APPEND_IN_GGA(x1, x2, x3) = APPEND_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: APPEND_IN_GGA(cons(XS), YS) -> APPEND_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: *APPEND_IN_GGA(cons(XS), YS) -> APPEND_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: REVERSE_IN_GA(cons(X, XS), YS) -> REVERSE_IN_GA(XS, ZS) The TRS R consists of the following rules: query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) The argument filtering Pi contains the following mapping: query_in_g(x1) = query_in_g(x1) U6_g(x1, x2) = U6_g(x2) shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) cons(x1, x2) = cons(x2) nil = nil shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) U4_ga(x1, x2, x3, x4) = U4_ga(x4) reverse_in_ga(x1, x2) = reverse_in_ga(x1) reverse_out_ga(x1, x2) = reverse_out_ga(x2) U2_ga(x1, x2, x3, x4) = U2_ga(x4) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) append_out_gga(x1, x2, x3) = append_out_gga(x3) U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) query_out_g(x1) = query_out_g REVERSE_IN_GA(x1, x2) = REVERSE_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: REVERSE_IN_GA(cons(X, XS), YS) -> REVERSE_IN_GA(XS, ZS) R is empty. The argument filtering Pi contains the following mapping: cons(x1, x2) = cons(x2) REVERSE_IN_GA(x1, x2) = REVERSE_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: REVERSE_IN_GA(cons(XS)) -> REVERSE_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: *REVERSE_IN_GA(cons(XS)) -> REVERSE_IN_GA(XS) The graph contains the following edges 1 > 1 ---------------------------------------- (20) YES ---------------------------------------- (21) Obligation: Pi DP problem: The TRS P consists of the following rules: U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> SHUFFLE_IN_GA(ZS, YS) SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> U4_GA(X, XS, YS, reverse_in_ga(XS, ZS)) The TRS R consists of the following rules: query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) The argument filtering Pi contains the following mapping: query_in_g(x1) = query_in_g(x1) U6_g(x1, x2) = U6_g(x2) shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) cons(x1, x2) = cons(x2) nil = nil shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) U4_ga(x1, x2, x3, x4) = U4_ga(x4) reverse_in_ga(x1, x2) = reverse_in_ga(x1) reverse_out_ga(x1, x2) = reverse_out_ga(x2) U2_ga(x1, x2, x3, x4) = U2_ga(x4) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) append_out_gga(x1, x2, x3) = append_out_gga(x3) U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) query_out_g(x1) = query_out_g SHUFFLE_IN_GA(x1, x2) = SHUFFLE_IN_GA(x1) U4_GA(x1, x2, x3, x4) = U4_GA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (22) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (23) Obligation: Pi DP problem: The TRS P consists of the following rules: U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> SHUFFLE_IN_GA(ZS, YS) SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> U4_GA(X, XS, YS, reverse_in_ga(XS, ZS)) The TRS R consists of the following rules: reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) The argument filtering Pi contains the following mapping: cons(x1, x2) = cons(x2) nil = nil reverse_in_ga(x1, x2) = reverse_in_ga(x1) reverse_out_ga(x1, x2) = reverse_out_ga(x2) U2_ga(x1, x2, x3, x4) = U2_ga(x4) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) append_out_gga(x1, x2, x3) = append_out_gga(x3) U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) SHUFFLE_IN_GA(x1, x2) = SHUFFLE_IN_GA(x1) U4_GA(x1, x2, x3, x4) = U4_GA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (24) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: U4_GA(reverse_out_ga(ZS)) -> SHUFFLE_IN_GA(ZS) SHUFFLE_IN_GA(cons(XS)) -> U4_GA(reverse_in_ga(XS)) The TRS R consists of the following rules: reverse_in_ga(nil) -> reverse_out_ga(nil) reverse_in_ga(cons(nil)) -> reverse_out_ga(cons(nil)) reverse_in_ga(cons(XS)) -> U2_ga(reverse_in_ga(XS)) U2_ga(reverse_out_ga(ZS)) -> U3_ga(append_in_gga(ZS, cons(nil))) U3_ga(append_out_gga(YS)) -> reverse_out_ga(YS) append_in_gga(nil, XS) -> append_out_gga(XS) append_in_gga(cons(XS), YS) -> U1_gga(append_in_gga(XS, YS)) U1_gga(append_out_gga(ZS)) -> append_out_gga(cons(ZS)) The set Q consists of the following terms: reverse_in_ga(x0) U2_ga(x0) U3_ga(x0) append_in_gga(x0, x1) U1_gga(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (26) 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: U4_GA(reverse_out_ga(ZS)) -> SHUFFLE_IN_GA(ZS) SHUFFLE_IN_GA(cons(XS)) -> U4_GA(reverse_in_ga(XS)) Strictly oriented rules of the TRS R: reverse_in_ga(nil) -> reverse_out_ga(nil) reverse_in_ga(cons(nil)) -> reverse_out_ga(cons(nil)) reverse_in_ga(cons(XS)) -> U2_ga(reverse_in_ga(XS)) U2_ga(reverse_out_ga(ZS)) -> U3_ga(append_in_gga(ZS, cons(nil))) U3_ga(append_out_gga(YS)) -> reverse_out_ga(YS) append_in_gga(nil, XS) -> append_out_gga(XS) append_in_gga(cons(XS), YS) -> U1_gga(append_in_gga(XS, YS)) U1_gga(append_out_gga(ZS)) -> append_out_gga(cons(ZS)) Used ordering: Knuth-Bendix order [KBO] with precedence:SHUFFLE_IN_GA_1 > append_in_gga_2 > U1_gga_1 > nil > U4_GA_1 > append_out_gga_1 > reverse_in_ga_1 > U2_ga_1 > U3_ga_1 > cons_1 > reverse_out_ga_1 and weight map: nil=1 reverse_in_ga_1=4 reverse_out_ga_1=4 cons_1=3 U2_ga_1=3 U3_ga_1=2 append_out_gga_1=2 U1_gga_1=3 U4_GA_1=1 SHUFFLE_IN_GA_1=2 append_in_gga_2=1 The variable weight is 1 ---------------------------------------- (27) Obligation: Q DP problem: P is empty. R is empty. The set Q consists of the following terms: reverse_in_ga(x0) U2_ga(x0) U3_ga(x0) append_in_gga(x0, x1) U1_gga(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (28) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (29) YES