/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern plus(g,a,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 2 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, 2 ms] (18) QDP (19) MRRProof [EQUIVALENT, 3 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Clauses: p(s(0), 0). p(s(s(X)), s(s(Y))) :- p(s(X), s(Y)). plus(0, Y, Y). plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z)). Query: plus(g,a,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: plus_in_3: (b,f,f) p_in_2: (b,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: plus_in_gaa(0, Y, Y) -> plus_out_gaa(0, Y, Y) plus_in_gaa(s(X), Y, s(Z)) -> U2_gaa(X, Y, Z, p_in_ga(s(X), U)) p_in_ga(s(0), 0) -> p_out_ga(s(0), 0) p_in_ga(s(s(X)), s(s(Y))) -> U1_ga(X, Y, p_in_ga(s(X), s(Y))) U1_ga(X, Y, p_out_ga(s(X), s(Y))) -> p_out_ga(s(s(X)), s(s(Y))) U2_gaa(X, Y, Z, p_out_ga(s(X), U)) -> U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z)) U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) -> plus_out_gaa(s(X), Y, s(Z)) The argument filtering Pi contains the following mapping: plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1) 0 = 0 plus_out_gaa(x1, x2, x3) = plus_out_gaa s(x1) = s(x1) U2_gaa(x1, x2, x3, x4) = U2_gaa(x4) p_in_ga(x1, x2) = p_in_ga(x1) p_out_ga(x1, x2) = p_out_ga(x2) U1_ga(x1, x2, x3) = U1_ga(x3) U3_gaa(x1, x2, x3, x4) = U3_gaa(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: plus_in_gaa(0, Y, Y) -> plus_out_gaa(0, Y, Y) plus_in_gaa(s(X), Y, s(Z)) -> U2_gaa(X, Y, Z, p_in_ga(s(X), U)) p_in_ga(s(0), 0) -> p_out_ga(s(0), 0) p_in_ga(s(s(X)), s(s(Y))) -> U1_ga(X, Y, p_in_ga(s(X), s(Y))) U1_ga(X, Y, p_out_ga(s(X), s(Y))) -> p_out_ga(s(s(X)), s(s(Y))) U2_gaa(X, Y, Z, p_out_ga(s(X), U)) -> U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z)) U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) -> plus_out_gaa(s(X), Y, s(Z)) The argument filtering Pi contains the following mapping: plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1) 0 = 0 plus_out_gaa(x1, x2, x3) = plus_out_gaa s(x1) = s(x1) U2_gaa(x1, x2, x3, x4) = U2_gaa(x4) p_in_ga(x1, x2) = p_in_ga(x1) p_out_ga(x1, x2) = p_out_ga(x2) U1_ga(x1, x2, x3) = U1_ga(x3) U3_gaa(x1, x2, x3, x4) = U3_gaa(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: PLUS_IN_GAA(s(X), Y, s(Z)) -> U2_GAA(X, Y, Z, p_in_ga(s(X), U)) PLUS_IN_GAA(s(X), Y, s(Z)) -> P_IN_GA(s(X), U) P_IN_GA(s(s(X)), s(s(Y))) -> U1_GA(X, Y, p_in_ga(s(X), s(Y))) P_IN_GA(s(s(X)), s(s(Y))) -> P_IN_GA(s(X), s(Y)) U2_GAA(X, Y, Z, p_out_ga(s(X), U)) -> U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z)) U2_GAA(X, Y, Z, p_out_ga(s(X), U)) -> PLUS_IN_GAA(U, Y, Z) The TRS R consists of the following rules: plus_in_gaa(0, Y, Y) -> plus_out_gaa(0, Y, Y) plus_in_gaa(s(X), Y, s(Z)) -> U2_gaa(X, Y, Z, p_in_ga(s(X), U)) p_in_ga(s(0), 0) -> p_out_ga(s(0), 0) p_in_ga(s(s(X)), s(s(Y))) -> U1_ga(X, Y, p_in_ga(s(X), s(Y))) U1_ga(X, Y, p_out_ga(s(X), s(Y))) -> p_out_ga(s(s(X)), s(s(Y))) U2_gaa(X, Y, Z, p_out_ga(s(X), U)) -> U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z)) U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) -> plus_out_gaa(s(X), Y, s(Z)) The argument filtering Pi contains the following mapping: plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1) 0 = 0 plus_out_gaa(x1, x2, x3) = plus_out_gaa s(x1) = s(x1) U2_gaa(x1, x2, x3, x4) = U2_gaa(x4) p_in_ga(x1, x2) = p_in_ga(x1) p_out_ga(x1, x2) = p_out_ga(x2) U1_ga(x1, x2, x3) = U1_ga(x3) U3_gaa(x1, x2, x3, x4) = U3_gaa(x4) PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1) U2_GAA(x1, x2, x3, x4) = U2_GAA(x4) P_IN_GA(x1, x2) = P_IN_GA(x1) U1_GA(x1, x2, x3) = U1_GA(x3) U3_GAA(x1, x2, x3, x4) = U3_GAA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: PLUS_IN_GAA(s(X), Y, s(Z)) -> U2_GAA(X, Y, Z, p_in_ga(s(X), U)) PLUS_IN_GAA(s(X), Y, s(Z)) -> P_IN_GA(s(X), U) P_IN_GA(s(s(X)), s(s(Y))) -> U1_GA(X, Y, p_in_ga(s(X), s(Y))) P_IN_GA(s(s(X)), s(s(Y))) -> P_IN_GA(s(X), s(Y)) U2_GAA(X, Y, Z, p_out_ga(s(X), U)) -> U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z)) U2_GAA(X, Y, Z, p_out_ga(s(X), U)) -> PLUS_IN_GAA(U, Y, Z) The TRS R consists of the following rules: plus_in_gaa(0, Y, Y) -> plus_out_gaa(0, Y, Y) plus_in_gaa(s(X), Y, s(Z)) -> U2_gaa(X, Y, Z, p_in_ga(s(X), U)) p_in_ga(s(0), 0) -> p_out_ga(s(0), 0) p_in_ga(s(s(X)), s(s(Y))) -> U1_ga(X, Y, p_in_ga(s(X), s(Y))) U1_ga(X, Y, p_out_ga(s(X), s(Y))) -> p_out_ga(s(s(X)), s(s(Y))) U2_gaa(X, Y, Z, p_out_ga(s(X), U)) -> U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z)) U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) -> plus_out_gaa(s(X), Y, s(Z)) The argument filtering Pi contains the following mapping: plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1) 0 = 0 plus_out_gaa(x1, x2, x3) = plus_out_gaa s(x1) = s(x1) U2_gaa(x1, x2, x3, x4) = U2_gaa(x4) p_in_ga(x1, x2) = p_in_ga(x1) p_out_ga(x1, x2) = p_out_ga(x2) U1_ga(x1, x2, x3) = U1_ga(x3) U3_gaa(x1, x2, x3, x4) = U3_gaa(x4) PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1) U2_GAA(x1, x2, x3, x4) = U2_GAA(x4) P_IN_GA(x1, x2) = P_IN_GA(x1) U1_GA(x1, x2, x3) = U1_GA(x3) U3_GAA(x1, x2, x3, x4) = U3_GAA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_GA(s(s(X)), s(s(Y))) -> P_IN_GA(s(X), s(Y)) The TRS R consists of the following rules: plus_in_gaa(0, Y, Y) -> plus_out_gaa(0, Y, Y) plus_in_gaa(s(X), Y, s(Z)) -> U2_gaa(X, Y, Z, p_in_ga(s(X), U)) p_in_ga(s(0), 0) -> p_out_ga(s(0), 0) p_in_ga(s(s(X)), s(s(Y))) -> U1_ga(X, Y, p_in_ga(s(X), s(Y))) U1_ga(X, Y, p_out_ga(s(X), s(Y))) -> p_out_ga(s(s(X)), s(s(Y))) U2_gaa(X, Y, Z, p_out_ga(s(X), U)) -> U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z)) U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) -> plus_out_gaa(s(X), Y, s(Z)) The argument filtering Pi contains the following mapping: plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1) 0 = 0 plus_out_gaa(x1, x2, x3) = plus_out_gaa s(x1) = s(x1) U2_gaa(x1, x2, x3, x4) = U2_gaa(x4) p_in_ga(x1, x2) = p_in_ga(x1) p_out_ga(x1, x2) = p_out_ga(x2) U1_ga(x1, x2, x3) = U1_ga(x3) U3_gaa(x1, x2, x3, x4) = U3_gaa(x4) P_IN_GA(x1, x2) = P_IN_GA(x1) 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: P_IN_GA(s(s(X)), s(s(Y))) -> P_IN_GA(s(X), s(Y)) R is empty. The argument filtering Pi contains the following mapping: s(x1) = s(x1) P_IN_GA(x1, x2) = P_IN_GA(x1) 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: P_IN_GA(s(s(X))) -> P_IN_GA(s(X)) 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: *P_IN_GA(s(s(X))) -> P_IN_GA(s(X)) The graph contains the following edges 1 > 1 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Pi DP problem: The TRS P consists of the following rules: U2_GAA(X, Y, Z, p_out_ga(s(X), U)) -> PLUS_IN_GAA(U, Y, Z) PLUS_IN_GAA(s(X), Y, s(Z)) -> U2_GAA(X, Y, Z, p_in_ga(s(X), U)) The TRS R consists of the following rules: plus_in_gaa(0, Y, Y) -> plus_out_gaa(0, Y, Y) plus_in_gaa(s(X), Y, s(Z)) -> U2_gaa(X, Y, Z, p_in_ga(s(X), U)) p_in_ga(s(0), 0) -> p_out_ga(s(0), 0) p_in_ga(s(s(X)), s(s(Y))) -> U1_ga(X, Y, p_in_ga(s(X), s(Y))) U1_ga(X, Y, p_out_ga(s(X), s(Y))) -> p_out_ga(s(s(X)), s(s(Y))) U2_gaa(X, Y, Z, p_out_ga(s(X), U)) -> U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z)) U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) -> plus_out_gaa(s(X), Y, s(Z)) The argument filtering Pi contains the following mapping: plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1) 0 = 0 plus_out_gaa(x1, x2, x3) = plus_out_gaa s(x1) = s(x1) U2_gaa(x1, x2, x3, x4) = U2_gaa(x4) p_in_ga(x1, x2) = p_in_ga(x1) p_out_ga(x1, x2) = p_out_ga(x2) U1_ga(x1, x2, x3) = U1_ga(x3) U3_gaa(x1, x2, x3, x4) = U3_gaa(x4) PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1) U2_GAA(x1, x2, x3, x4) = U2_GAA(x4) 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: U2_GAA(X, Y, Z, p_out_ga(s(X), U)) -> PLUS_IN_GAA(U, Y, Z) PLUS_IN_GAA(s(X), Y, s(Z)) -> U2_GAA(X, Y, Z, p_in_ga(s(X), U)) The TRS R consists of the following rules: p_in_ga(s(0), 0) -> p_out_ga(s(0), 0) p_in_ga(s(s(X)), s(s(Y))) -> U1_ga(X, Y, p_in_ga(s(X), s(Y))) U1_ga(X, Y, p_out_ga(s(X), s(Y))) -> p_out_ga(s(s(X)), s(s(Y))) The argument filtering Pi contains the following mapping: 0 = 0 s(x1) = s(x1) p_in_ga(x1, x2) = p_in_ga(x1) p_out_ga(x1, x2) = p_out_ga(x2) U1_ga(x1, x2, x3) = U1_ga(x3) PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1) U2_GAA(x1, x2, x3, x4) = U2_GAA(x4) 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: U2_GAA(p_out_ga(U)) -> PLUS_IN_GAA(U) PLUS_IN_GAA(s(X)) -> U2_GAA(p_in_ga(s(X))) The TRS R consists of the following rules: p_in_ga(s(0)) -> p_out_ga(0) p_in_ga(s(s(X))) -> U1_ga(p_in_ga(s(X))) U1_ga(p_out_ga(s(Y))) -> p_out_ga(s(s(Y))) The set Q consists of the following terms: p_in_ga(x0) U1_ga(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (19) 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: U2_GAA(p_out_ga(U)) -> PLUS_IN_GAA(U) PLUS_IN_GAA(s(X)) -> U2_GAA(p_in_ga(s(X))) Strictly oriented rules of the TRS R: p_in_ga(s(0)) -> p_out_ga(0) p_in_ga(s(s(X))) -> U1_ga(p_in_ga(s(X))) U1_ga(p_out_ga(s(Y))) -> p_out_ga(s(s(Y))) Used ordering: Knuth-Bendix order [KBO] with precedence:0 > s_1 > p_in_ga_1 > U1_ga_1 > U2_GAA_1 > PLUS_IN_GAA_1 > p_out_ga_1 and weight map: 0=1 p_in_ga_1=1 s_1=4 p_out_ga_1=3 U1_ga_1=4 U2_GAA_1=1 PLUS_IN_GAA_1=3 The variable weight is 1 ---------------------------------------- (20) Obligation: Q DP problem: P is empty. R is empty. The set Q consists of the following terms: p_in_ga(x0) U1_ga(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (21) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (22) YES