/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern p(g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) CutEliminatorProof [SOUND, 0 ms] (2) Prolog (3) PrologToPiTRSProof [SOUND, 0 ms] (4) PiTRS (5) DependencyPairsProof [EQUIVALENT, 7 ms] (6) PiDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) PiDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) PiDP (11) PiDPToQDPProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Clauses: p(0). p(s(X)) :- ','(q(X), ','(!, r)). p(X) :- r. q(0). q(s(X)) :- ','(p(X), ','(!, r)). q(X) :- r. r. Query: p(g) ---------------------------------------- (1) CutEliminatorProof (SOUND) Eliminated all cuts by simply ignoring them[PROLOG]. ---------------------------------------- (2) Obligation: Clauses: p(0). p(s(X)) :- ','(q(X), r). p(X) :- r. q(0). q(s(X)) :- ','(p(X), r). q(X) :- r. r. Query: p(g) ---------------------------------------- (3) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: p_in_1: (b) q_in_1: (b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_g(0) -> p_out_g(0) p_in_g(s(X)) -> U1_g(X, q_in_g(X)) q_in_g(0) -> q_out_g(0) q_in_g(s(X)) -> U4_g(X, p_in_g(X)) p_in_g(X) -> U3_g(X, r_in_) r_in_ -> r_out_ U3_g(X, r_out_) -> p_out_g(X) U4_g(X, p_out_g(X)) -> U5_g(X, r_in_) U5_g(X, r_out_) -> q_out_g(s(X)) q_in_g(X) -> U6_g(X, r_in_) U6_g(X, r_out_) -> q_out_g(X) U1_g(X, q_out_g(X)) -> U2_g(X, r_in_) U2_g(X, r_out_) -> p_out_g(s(X)) The argument filtering Pi contains the following mapping: p_in_g(x1) = p_in_g(x1) 0 = 0 p_out_g(x1) = p_out_g s(x1) = s(x1) U1_g(x1, x2) = U1_g(x2) q_in_g(x1) = q_in_g(x1) q_out_g(x1) = q_out_g U4_g(x1, x2) = U4_g(x2) U3_g(x1, x2) = U3_g(x2) r_in_ = r_in_ r_out_ = r_out_ U5_g(x1, x2) = U5_g(x2) U6_g(x1, x2) = U6_g(x2) U2_g(x1, x2) = U2_g(x2) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (4) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_g(0) -> p_out_g(0) p_in_g(s(X)) -> U1_g(X, q_in_g(X)) q_in_g(0) -> q_out_g(0) q_in_g(s(X)) -> U4_g(X, p_in_g(X)) p_in_g(X) -> U3_g(X, r_in_) r_in_ -> r_out_ U3_g(X, r_out_) -> p_out_g(X) U4_g(X, p_out_g(X)) -> U5_g(X, r_in_) U5_g(X, r_out_) -> q_out_g(s(X)) q_in_g(X) -> U6_g(X, r_in_) U6_g(X, r_out_) -> q_out_g(X) U1_g(X, q_out_g(X)) -> U2_g(X, r_in_) U2_g(X, r_out_) -> p_out_g(s(X)) The argument filtering Pi contains the following mapping: p_in_g(x1) = p_in_g(x1) 0 = 0 p_out_g(x1) = p_out_g s(x1) = s(x1) U1_g(x1, x2) = U1_g(x2) q_in_g(x1) = q_in_g(x1) q_out_g(x1) = q_out_g U4_g(x1, x2) = U4_g(x2) U3_g(x1, x2) = U3_g(x2) r_in_ = r_in_ r_out_ = r_out_ U5_g(x1, x2) = U5_g(x2) U6_g(x1, x2) = U6_g(x2) U2_g(x1, x2) = U2_g(x2) ---------------------------------------- (5) 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: P_IN_G(s(X)) -> U1_G(X, q_in_g(X)) P_IN_G(s(X)) -> Q_IN_G(X) Q_IN_G(s(X)) -> U4_G(X, p_in_g(X)) Q_IN_G(s(X)) -> P_IN_G(X) P_IN_G(X) -> U3_G(X, r_in_) P_IN_G(X) -> R_IN_ U4_G(X, p_out_g(X)) -> U5_G(X, r_in_) U4_G(X, p_out_g(X)) -> R_IN_ Q_IN_G(X) -> U6_G(X, r_in_) Q_IN_G(X) -> R_IN_ U1_G(X, q_out_g(X)) -> U2_G(X, r_in_) U1_G(X, q_out_g(X)) -> R_IN_ The TRS R consists of the following rules: p_in_g(0) -> p_out_g(0) p_in_g(s(X)) -> U1_g(X, q_in_g(X)) q_in_g(0) -> q_out_g(0) q_in_g(s(X)) -> U4_g(X, p_in_g(X)) p_in_g(X) -> U3_g(X, r_in_) r_in_ -> r_out_ U3_g(X, r_out_) -> p_out_g(X) U4_g(X, p_out_g(X)) -> U5_g(X, r_in_) U5_g(X, r_out_) -> q_out_g(s(X)) q_in_g(X) -> U6_g(X, r_in_) U6_g(X, r_out_) -> q_out_g(X) U1_g(X, q_out_g(X)) -> U2_g(X, r_in_) U2_g(X, r_out_) -> p_out_g(s(X)) The argument filtering Pi contains the following mapping: p_in_g(x1) = p_in_g(x1) 0 = 0 p_out_g(x1) = p_out_g s(x1) = s(x1) U1_g(x1, x2) = U1_g(x2) q_in_g(x1) = q_in_g(x1) q_out_g(x1) = q_out_g U4_g(x1, x2) = U4_g(x2) U3_g(x1, x2) = U3_g(x2) r_in_ = r_in_ r_out_ = r_out_ U5_g(x1, x2) = U5_g(x2) U6_g(x1, x2) = U6_g(x2) U2_g(x1, x2) = U2_g(x2) P_IN_G(x1) = P_IN_G(x1) U1_G(x1, x2) = U1_G(x2) Q_IN_G(x1) = Q_IN_G(x1) U4_G(x1, x2) = U4_G(x2) U3_G(x1, x2) = U3_G(x2) R_IN_ = R_IN_ U5_G(x1, x2) = U5_G(x2) U6_G(x1, x2) = U6_G(x2) U2_G(x1, x2) = U2_G(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_G(s(X)) -> U1_G(X, q_in_g(X)) P_IN_G(s(X)) -> Q_IN_G(X) Q_IN_G(s(X)) -> U4_G(X, p_in_g(X)) Q_IN_G(s(X)) -> P_IN_G(X) P_IN_G(X) -> U3_G(X, r_in_) P_IN_G(X) -> R_IN_ U4_G(X, p_out_g(X)) -> U5_G(X, r_in_) U4_G(X, p_out_g(X)) -> R_IN_ Q_IN_G(X) -> U6_G(X, r_in_) Q_IN_G(X) -> R_IN_ U1_G(X, q_out_g(X)) -> U2_G(X, r_in_) U1_G(X, q_out_g(X)) -> R_IN_ The TRS R consists of the following rules: p_in_g(0) -> p_out_g(0) p_in_g(s(X)) -> U1_g(X, q_in_g(X)) q_in_g(0) -> q_out_g(0) q_in_g(s(X)) -> U4_g(X, p_in_g(X)) p_in_g(X) -> U3_g(X, r_in_) r_in_ -> r_out_ U3_g(X, r_out_) -> p_out_g(X) U4_g(X, p_out_g(X)) -> U5_g(X, r_in_) U5_g(X, r_out_) -> q_out_g(s(X)) q_in_g(X) -> U6_g(X, r_in_) U6_g(X, r_out_) -> q_out_g(X) U1_g(X, q_out_g(X)) -> U2_g(X, r_in_) U2_g(X, r_out_) -> p_out_g(s(X)) The argument filtering Pi contains the following mapping: p_in_g(x1) = p_in_g(x1) 0 = 0 p_out_g(x1) = p_out_g s(x1) = s(x1) U1_g(x1, x2) = U1_g(x2) q_in_g(x1) = q_in_g(x1) q_out_g(x1) = q_out_g U4_g(x1, x2) = U4_g(x2) U3_g(x1, x2) = U3_g(x2) r_in_ = r_in_ r_out_ = r_out_ U5_g(x1, x2) = U5_g(x2) U6_g(x1, x2) = U6_g(x2) U2_g(x1, x2) = U2_g(x2) P_IN_G(x1) = P_IN_G(x1) U1_G(x1, x2) = U1_G(x2) Q_IN_G(x1) = Q_IN_G(x1) U4_G(x1, x2) = U4_G(x2) U3_G(x1, x2) = U3_G(x2) R_IN_ = R_IN_ U5_G(x1, x2) = U5_G(x2) U6_G(x1, x2) = U6_G(x2) U2_G(x1, x2) = U2_G(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 10 less nodes. ---------------------------------------- (8) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_G(s(X)) -> Q_IN_G(X) Q_IN_G(s(X)) -> P_IN_G(X) The TRS R consists of the following rules: p_in_g(0) -> p_out_g(0) p_in_g(s(X)) -> U1_g(X, q_in_g(X)) q_in_g(0) -> q_out_g(0) q_in_g(s(X)) -> U4_g(X, p_in_g(X)) p_in_g(X) -> U3_g(X, r_in_) r_in_ -> r_out_ U3_g(X, r_out_) -> p_out_g(X) U4_g(X, p_out_g(X)) -> U5_g(X, r_in_) U5_g(X, r_out_) -> q_out_g(s(X)) q_in_g(X) -> U6_g(X, r_in_) U6_g(X, r_out_) -> q_out_g(X) U1_g(X, q_out_g(X)) -> U2_g(X, r_in_) U2_g(X, r_out_) -> p_out_g(s(X)) The argument filtering Pi contains the following mapping: p_in_g(x1) = p_in_g(x1) 0 = 0 p_out_g(x1) = p_out_g s(x1) = s(x1) U1_g(x1, x2) = U1_g(x2) q_in_g(x1) = q_in_g(x1) q_out_g(x1) = q_out_g U4_g(x1, x2) = U4_g(x2) U3_g(x1, x2) = U3_g(x2) r_in_ = r_in_ r_out_ = r_out_ U5_g(x1, x2) = U5_g(x2) U6_g(x1, x2) = U6_g(x2) U2_g(x1, x2) = U2_g(x2) P_IN_G(x1) = P_IN_G(x1) Q_IN_G(x1) = Q_IN_G(x1) We have to consider all (P,R,Pi)-chains ---------------------------------------- (9) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (10) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_G(s(X)) -> Q_IN_G(X) Q_IN_G(s(X)) -> P_IN_G(X) R is empty. Pi is empty. We have to consider all (P,R,Pi)-chains ---------------------------------------- (11) PiDPToQDPProof (EQUIVALENT) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: P_IN_G(s(X)) -> Q_IN_G(X) Q_IN_G(s(X)) -> P_IN_G(X) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (13) 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: *Q_IN_G(s(X)) -> P_IN_G(X) The graph contains the following edges 1 > 1 *P_IN_G(s(X)) -> Q_IN_G(X) The graph contains the following edges 1 > 1 ---------------------------------------- (14) YES