/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern dis(g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0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) PiDP (7) PiDPToQDPProof [SOUND, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Clauses: dis(or(B1, B2)) :- ','(con(B1), dis(B2)). dis(B) :- con(B). con(and(B1, B2)) :- ','(dis(B1), con(B2)). con(B) :- bool(B). bool(0). bool(1). Query: dis(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: dis_in_1: (b) con_in_1: (b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1)) con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1)) dis_in_g(B) -> U3_g(B, con_in_g(B)) con_in_g(B) -> U6_g(B, bool_in_g(B)) bool_in_g(0) -> bool_out_g(0) bool_in_g(1) -> bool_out_g(1) U6_g(B, bool_out_g(B)) -> con_out_g(B) U3_g(B, con_out_g(B)) -> dis_out_g(B) U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2)) U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2)) U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2)) U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2)) The argument filtering Pi contains the following mapping: dis_in_g(x1) = dis_in_g(x1) or(x1, x2) = or(x1, x2) U1_g(x1, x2, x3) = U1_g(x2, x3) con_in_g(x1) = con_in_g(x1) and(x1, x2) = and(x1, x2) U4_g(x1, x2, x3) = U4_g(x2, x3) U3_g(x1, x2) = U3_g(x2) U6_g(x1, x2) = U6_g(x2) bool_in_g(x1) = bool_in_g(x1) 0 = 0 bool_out_g(x1) = bool_out_g 1 = 1 con_out_g(x1) = con_out_g dis_out_g(x1) = dis_out_g U5_g(x1, x2, x3) = U5_g(x3) U2_g(x1, x2, x3) = U2_g(x3) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1)) con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1)) dis_in_g(B) -> U3_g(B, con_in_g(B)) con_in_g(B) -> U6_g(B, bool_in_g(B)) bool_in_g(0) -> bool_out_g(0) bool_in_g(1) -> bool_out_g(1) U6_g(B, bool_out_g(B)) -> con_out_g(B) U3_g(B, con_out_g(B)) -> dis_out_g(B) U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2)) U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2)) U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2)) U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2)) The argument filtering Pi contains the following mapping: dis_in_g(x1) = dis_in_g(x1) or(x1, x2) = or(x1, x2) U1_g(x1, x2, x3) = U1_g(x2, x3) con_in_g(x1) = con_in_g(x1) and(x1, x2) = and(x1, x2) U4_g(x1, x2, x3) = U4_g(x2, x3) U3_g(x1, x2) = U3_g(x2) U6_g(x1, x2) = U6_g(x2) bool_in_g(x1) = bool_in_g(x1) 0 = 0 bool_out_g(x1) = bool_out_g 1 = 1 con_out_g(x1) = con_out_g dis_out_g(x1) = dis_out_g U5_g(x1, x2, x3) = U5_g(x3) U2_g(x1, x2, x3) = U2_g(x3) ---------------------------------------- (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: DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1)) DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1) CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1)) CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1) DIS_IN_G(B) -> U3_G(B, con_in_g(B)) DIS_IN_G(B) -> CON_IN_G(B) CON_IN_G(B) -> U6_G(B, bool_in_g(B)) CON_IN_G(B) -> BOOL_IN_G(B) U4_G(B1, B2, dis_out_g(B1)) -> U5_G(B1, B2, con_in_g(B2)) U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2) U1_G(B1, B2, con_out_g(B1)) -> U2_G(B1, B2, dis_in_g(B2)) U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2) The TRS R consists of the following rules: dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1)) con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1)) dis_in_g(B) -> U3_g(B, con_in_g(B)) con_in_g(B) -> U6_g(B, bool_in_g(B)) bool_in_g(0) -> bool_out_g(0) bool_in_g(1) -> bool_out_g(1) U6_g(B, bool_out_g(B)) -> con_out_g(B) U3_g(B, con_out_g(B)) -> dis_out_g(B) U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2)) U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2)) U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2)) U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2)) The argument filtering Pi contains the following mapping: dis_in_g(x1) = dis_in_g(x1) or(x1, x2) = or(x1, x2) U1_g(x1, x2, x3) = U1_g(x2, x3) con_in_g(x1) = con_in_g(x1) and(x1, x2) = and(x1, x2) U4_g(x1, x2, x3) = U4_g(x2, x3) U3_g(x1, x2) = U3_g(x2) U6_g(x1, x2) = U6_g(x2) bool_in_g(x1) = bool_in_g(x1) 0 = 0 bool_out_g(x1) = bool_out_g 1 = 1 con_out_g(x1) = con_out_g dis_out_g(x1) = dis_out_g U5_g(x1, x2, x3) = U5_g(x3) U2_g(x1, x2, x3) = U2_g(x3) DIS_IN_G(x1) = DIS_IN_G(x1) U1_G(x1, x2, x3) = U1_G(x2, x3) CON_IN_G(x1) = CON_IN_G(x1) U4_G(x1, x2, x3) = U4_G(x2, x3) U3_G(x1, x2) = U3_G(x2) U6_G(x1, x2) = U6_G(x2) BOOL_IN_G(x1) = BOOL_IN_G(x1) U5_G(x1, x2, x3) = U5_G(x3) U2_G(x1, x2, x3) = U2_G(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1)) DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1) CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1)) CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1) DIS_IN_G(B) -> U3_G(B, con_in_g(B)) DIS_IN_G(B) -> CON_IN_G(B) CON_IN_G(B) -> U6_G(B, bool_in_g(B)) CON_IN_G(B) -> BOOL_IN_G(B) U4_G(B1, B2, dis_out_g(B1)) -> U5_G(B1, B2, con_in_g(B2)) U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2) U1_G(B1, B2, con_out_g(B1)) -> U2_G(B1, B2, dis_in_g(B2)) U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2) The TRS R consists of the following rules: dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1)) con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1)) dis_in_g(B) -> U3_g(B, con_in_g(B)) con_in_g(B) -> U6_g(B, bool_in_g(B)) bool_in_g(0) -> bool_out_g(0) bool_in_g(1) -> bool_out_g(1) U6_g(B, bool_out_g(B)) -> con_out_g(B) U3_g(B, con_out_g(B)) -> dis_out_g(B) U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2)) U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2)) U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2)) U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2)) The argument filtering Pi contains the following mapping: dis_in_g(x1) = dis_in_g(x1) or(x1, x2) = or(x1, x2) U1_g(x1, x2, x3) = U1_g(x2, x3) con_in_g(x1) = con_in_g(x1) and(x1, x2) = and(x1, x2) U4_g(x1, x2, x3) = U4_g(x2, x3) U3_g(x1, x2) = U3_g(x2) U6_g(x1, x2) = U6_g(x2) bool_in_g(x1) = bool_in_g(x1) 0 = 0 bool_out_g(x1) = bool_out_g 1 = 1 con_out_g(x1) = con_out_g dis_out_g(x1) = dis_out_g U5_g(x1, x2, x3) = U5_g(x3) U2_g(x1, x2, x3) = U2_g(x3) DIS_IN_G(x1) = DIS_IN_G(x1) U1_G(x1, x2, x3) = U1_G(x2, x3) CON_IN_G(x1) = CON_IN_G(x1) U4_G(x1, x2, x3) = U4_G(x2, x3) U3_G(x1, x2) = U3_G(x2) U6_G(x1, x2) = U6_G(x2) BOOL_IN_G(x1) = BOOL_IN_G(x1) U5_G(x1, x2, x3) = U5_G(x3) U2_G(x1, x2, x3) = U2_G(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes. ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2) DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1)) DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1) CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1)) U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2) CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1) DIS_IN_G(B) -> CON_IN_G(B) The TRS R consists of the following rules: dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1)) con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1)) dis_in_g(B) -> U3_g(B, con_in_g(B)) con_in_g(B) -> U6_g(B, bool_in_g(B)) bool_in_g(0) -> bool_out_g(0) bool_in_g(1) -> bool_out_g(1) U6_g(B, bool_out_g(B)) -> con_out_g(B) U3_g(B, con_out_g(B)) -> dis_out_g(B) U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2)) U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2)) U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2)) U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2)) The argument filtering Pi contains the following mapping: dis_in_g(x1) = dis_in_g(x1) or(x1, x2) = or(x1, x2) U1_g(x1, x2, x3) = U1_g(x2, x3) con_in_g(x1) = con_in_g(x1) and(x1, x2) = and(x1, x2) U4_g(x1, x2, x3) = U4_g(x2, x3) U3_g(x1, x2) = U3_g(x2) U6_g(x1, x2) = U6_g(x2) bool_in_g(x1) = bool_in_g(x1) 0 = 0 bool_out_g(x1) = bool_out_g 1 = 1 con_out_g(x1) = con_out_g dis_out_g(x1) = dis_out_g U5_g(x1, x2, x3) = U5_g(x3) U2_g(x1, x2, x3) = U2_g(x3) DIS_IN_G(x1) = DIS_IN_G(x1) U1_G(x1, x2, x3) = U1_G(x2, x3) CON_IN_G(x1) = CON_IN_G(x1) U4_G(x1, x2, x3) = U4_G(x2, x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (7) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: U1_G(B2, con_out_g) -> DIS_IN_G(B2) DIS_IN_G(or(B1, B2)) -> U1_G(B2, con_in_g(B1)) DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1) CON_IN_G(and(B1, B2)) -> U4_G(B2, dis_in_g(B1)) U4_G(B2, dis_out_g) -> CON_IN_G(B2) CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1) DIS_IN_G(B) -> CON_IN_G(B) The TRS R consists of the following rules: dis_in_g(or(B1, B2)) -> U1_g(B2, con_in_g(B1)) con_in_g(and(B1, B2)) -> U4_g(B2, dis_in_g(B1)) dis_in_g(B) -> U3_g(con_in_g(B)) con_in_g(B) -> U6_g(bool_in_g(B)) bool_in_g(0) -> bool_out_g bool_in_g(1) -> bool_out_g U6_g(bool_out_g) -> con_out_g U3_g(con_out_g) -> dis_out_g U4_g(B2, dis_out_g) -> U5_g(con_in_g(B2)) U5_g(con_out_g) -> con_out_g U1_g(B2, con_out_g) -> U2_g(dis_in_g(B2)) U2_g(dis_out_g) -> dis_out_g The set Q consists of the following terms: dis_in_g(x0) con_in_g(x0) bool_in_g(x0) U6_g(x0) U3_g(x0) U4_g(x0, x1) U5_g(x0) U1_g(x0, x1) U2_g(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (9) 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: *DIS_IN_G(or(B1, B2)) -> U1_G(B2, con_in_g(B1)) The graph contains the following edges 1 > 1 *U1_G(B2, con_out_g) -> DIS_IN_G(B2) The graph contains the following edges 1 >= 1 *CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1) The graph contains the following edges 1 > 1 *CON_IN_G(and(B1, B2)) -> U4_G(B2, dis_in_g(B1)) The graph contains the following edges 1 > 1 *U4_G(B2, dis_out_g) -> CON_IN_G(B2) The graph contains the following edges 1 >= 1 *DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1) The graph contains the following edges 1 > 1 *DIS_IN_G(B) -> CON_IN_G(B) The graph contains the following edges 1 >= 1 ---------------------------------------- (10) YES