/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern perm(g,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0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, 10 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 (21) PiDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) PiDP (24) PiDPToQDPProof [SOUND, 0 ms] (25) QDP (26) MRRProof [EQUIVALENT, 2 ms] (27) QDP (28) PisEmptyProof [EQUIVALENT, 0 ms] (29) YES ---------------------------------------- (0) Obligation: Clauses: append(nil, XS, XS). append(cons(X, XS1), XS2, cons(X, YS)) :- append(XS1, XS2, YS). split(XS, nil, XS). split(cons(X, XS), cons(X, YS1), YS2) :- split(XS, YS1, YS2). perm(nil, nil). perm(XS, cons(Y, YS)) :- ','(split(XS, YS1, cons(Y, YS2)), ','(append(YS1, YS2, ZS), perm(ZS, YS))). Query: perm(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: perm_in_2: (b,f) split_in_3: (b,f,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: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) split_in_gaa(x1, x2, x3) = split_in_gaa(x1) cons(x1, x2) = cons(x2) split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 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) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) split_in_gaa(x1, x2, x3) = split_in_gaa(x1) cons(x1, x2) = cons(x2) split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 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) ---------------------------------------- (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: PERM_IN_GA(XS, cons(Y, YS)) -> U3_GA(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) PERM_IN_GA(XS, cons(Y, YS)) -> SPLIT_IN_GAA(XS, YS1, cons(Y, YS2)) SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> U2_GAA(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> SPLIT_IN_GAA(XS, YS1, YS2) U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_GA(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> APPEND_IN_GGA(YS1, YS2, ZS) APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> U1_GGA(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> APPEND_IN_GGA(XS1, XS2, YS) U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_GA(XS, Y, YS, perm_in_ga(ZS, YS)) U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> PERM_IN_GA(ZS, YS) The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) split_in_gaa(x1, x2, x3) = split_in_gaa(x1) cons(x1, x2) = cons(x2) split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 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) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) SPLIT_IN_GAA(x1, x2, x3) = SPLIT_IN_GAA(x1) U2_GAA(x1, x2, x3, x4, x5) = U2_GAA(x5) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) 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: PERM_IN_GA(XS, cons(Y, YS)) -> U3_GA(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) PERM_IN_GA(XS, cons(Y, YS)) -> SPLIT_IN_GAA(XS, YS1, cons(Y, YS2)) SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> U2_GAA(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> SPLIT_IN_GAA(XS, YS1, YS2) U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_GA(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> APPEND_IN_GGA(YS1, YS2, ZS) APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> U1_GGA(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> APPEND_IN_GGA(XS1, XS2, YS) U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_GA(XS, Y, YS, perm_in_ga(ZS, YS)) U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> PERM_IN_GA(ZS, YS) The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) split_in_gaa(x1, x2, x3) = split_in_gaa(x1) cons(x1, x2) = cons(x2) split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 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) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) SPLIT_IN_GAA(x1, x2, x3) = SPLIT_IN_GAA(x1) U2_GAA(x1, x2, x3, x4, x5) = U2_GAA(x5) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) 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 5 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> APPEND_IN_GGA(XS1, XS2, YS) The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) split_in_gaa(x1, x2, x3) = split_in_gaa(x1) cons(x1, x2) = cons(x2) split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 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) 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, XS1), XS2, cons(X, YS)) -> APPEND_IN_GGA(XS1, XS2, YS) 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(XS1), XS2) -> APPEND_IN_GGA(XS1, XS2) 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(XS1), XS2) -> APPEND_IN_GGA(XS1, XS2) 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: SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> SPLIT_IN_GAA(XS, YS1, YS2) The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) split_in_gaa(x1, x2, x3) = split_in_gaa(x1) cons(x1, x2) = cons(x2) split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 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) SPLIT_IN_GAA(x1, x2, x3) = SPLIT_IN_GAA(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: SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> SPLIT_IN_GAA(XS, YS1, YS2) R is empty. The argument filtering Pi contains the following mapping: cons(x1, x2) = cons(x2) SPLIT_IN_GAA(x1, x2, x3) = SPLIT_IN_GAA(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: SPLIT_IN_GAA(cons(XS)) -> SPLIT_IN_GAA(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: *SPLIT_IN_GAA(cons(XS)) -> SPLIT_IN_GAA(XS) The graph contains the following edges 1 > 1 ---------------------------------------- (20) YES ---------------------------------------- (21) Obligation: Pi DP problem: The TRS P consists of the following rules: U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_GA(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> PERM_IN_GA(ZS, YS) PERM_IN_GA(XS, cons(Y, YS)) -> U3_GA(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) split_in_gaa(x1, x2, x3) = split_in_gaa(x1) cons(x1, x2) = cons(x2) split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 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) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) 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: U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_GA(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> PERM_IN_GA(ZS, YS) PERM_IN_GA(XS, cons(Y, YS)) -> U3_GA(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) The TRS R consists of the following rules: append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) The argument filtering Pi contains the following mapping: nil = nil split_in_gaa(x1, x2, x3) = split_in_gaa(x1) cons(x1, x2) = cons(x2) split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) 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) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) 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: U3_GA(split_out_gaa(YS1, cons(YS2))) -> U4_GA(append_in_gga(YS1, YS2)) U4_GA(append_out_gga(ZS)) -> PERM_IN_GA(ZS) PERM_IN_GA(XS) -> U3_GA(split_in_gaa(XS)) The TRS R consists of the following rules: append_in_gga(nil, XS) -> append_out_gga(XS) append_in_gga(cons(XS1), XS2) -> U1_gga(append_in_gga(XS1, XS2)) split_in_gaa(XS) -> split_out_gaa(nil, XS) split_in_gaa(cons(XS)) -> U2_gaa(split_in_gaa(XS)) U1_gga(append_out_gga(YS)) -> append_out_gga(cons(YS)) U2_gaa(split_out_gaa(YS1, YS2)) -> split_out_gaa(cons(YS1), YS2) The set Q consists of the following terms: append_in_gga(x0, x1) split_in_gaa(x0) U1_gga(x0) U2_gaa(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: U3_GA(split_out_gaa(YS1, cons(YS2))) -> U4_GA(append_in_gga(YS1, YS2)) U4_GA(append_out_gga(ZS)) -> PERM_IN_GA(ZS) PERM_IN_GA(XS) -> U3_GA(split_in_gaa(XS)) Strictly oriented rules of the TRS R: append_in_gga(nil, XS) -> append_out_gga(XS) append_in_gga(cons(XS1), XS2) -> U1_gga(append_in_gga(XS1, XS2)) split_in_gaa(XS) -> split_out_gaa(nil, XS) split_in_gaa(cons(XS)) -> U2_gaa(split_in_gaa(XS)) U1_gga(append_out_gga(YS)) -> append_out_gga(cons(YS)) U2_gaa(split_out_gaa(YS1, YS2)) -> split_out_gaa(cons(YS1), YS2) Used ordering: Knuth-Bendix order [KBO] with precedence:append_in_gga_2 > nil > PERM_IN_GA_1 > cons_1 > U4_GA_1 > split_in_gaa_1 > U3_GA_1 > U2_gaa_1 > split_out_gaa_2 > U1_gga_1 > append_out_gga_1 and weight map: nil=2 append_out_gga_1=6 cons_1=7 U1_gga_1=7 split_in_gaa_1=2 U2_gaa_1=7 U3_GA_1=3 U4_GA_1=1 PERM_IN_GA_1=6 append_in_gga_2=8 split_out_gaa_2=0 The variable weight is 1 ---------------------------------------- (27) Obligation: Q DP problem: P is empty. R is empty. The set Q consists of the following terms: append_in_gga(x0, x1) split_in_gaa(x0) U1_gga(x0) U2_gaa(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