/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern log2(g,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0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) MRRProof [EQUIVALENT, 34 ms] (20) QDP (21) DependencyGraphProof [EQUIVALENT, 0 ms] (22) TRUE ---------------------------------------- (0) Obligation: Clauses: log2(X, Y) :- log2(X, 0, Y). log2(0, I, I). log2(s(0), I, I). log2(s(s(X)), I, Y) :- ','(half(s(s(X)), X1), log2(X1, s(I), Y)). half(0, 0). half(s(0), 0). half(s(s(X)), s(Y)) :- half(X, Y). Query: log2(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: log2_in_2: (b,f) log2_in_3: (b,b,f) half_in_2: (b,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: log2_in_ga(X, Y) -> U1_ga(X, Y, log2_in_gga(X, 0, Y)) log2_in_gga(0, I, I) -> log2_out_gga(0, I, I) log2_in_gga(s(0), I, I) -> log2_out_gga(s(0), I, I) log2_in_gga(s(s(X)), I, Y) -> U2_gga(X, I, Y, half_in_ga(s(s(X)), X1)) half_in_ga(0, 0) -> half_out_ga(0, 0) half_in_ga(s(0), 0) -> half_out_ga(s(0), 0) half_in_ga(s(s(X)), s(Y)) -> U4_ga(X, Y, half_in_ga(X, Y)) U4_ga(X, Y, half_out_ga(X, Y)) -> half_out_ga(s(s(X)), s(Y)) U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) -> U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y)) U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) -> log2_out_gga(s(s(X)), I, Y) U1_ga(X, Y, log2_out_gga(X, 0, Y)) -> log2_out_ga(X, Y) The argument filtering Pi contains the following mapping: log2_in_ga(x1, x2) = log2_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) log2_in_gga(x1, x2, x3) = log2_in_gga(x1, x2) 0 = 0 log2_out_gga(x1, x2, x3) = log2_out_gga(x3) s(x1) = s(x1) U2_gga(x1, x2, x3, x4) = U2_gga(x2, x4) half_in_ga(x1, x2) = half_in_ga(x1) half_out_ga(x1, x2) = half_out_ga(x2) U4_ga(x1, x2, x3) = U4_ga(x3) U3_gga(x1, x2, x3, x4) = U3_gga(x4) log2_out_ga(x1, x2) = log2_out_ga(x2) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: log2_in_ga(X, Y) -> U1_ga(X, Y, log2_in_gga(X, 0, Y)) log2_in_gga(0, I, I) -> log2_out_gga(0, I, I) log2_in_gga(s(0), I, I) -> log2_out_gga(s(0), I, I) log2_in_gga(s(s(X)), I, Y) -> U2_gga(X, I, Y, half_in_ga(s(s(X)), X1)) half_in_ga(0, 0) -> half_out_ga(0, 0) half_in_ga(s(0), 0) -> half_out_ga(s(0), 0) half_in_ga(s(s(X)), s(Y)) -> U4_ga(X, Y, half_in_ga(X, Y)) U4_ga(X, Y, half_out_ga(X, Y)) -> half_out_ga(s(s(X)), s(Y)) U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) -> U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y)) U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) -> log2_out_gga(s(s(X)), I, Y) U1_ga(X, Y, log2_out_gga(X, 0, Y)) -> log2_out_ga(X, Y) The argument filtering Pi contains the following mapping: log2_in_ga(x1, x2) = log2_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) log2_in_gga(x1, x2, x3) = log2_in_gga(x1, x2) 0 = 0 log2_out_gga(x1, x2, x3) = log2_out_gga(x3) s(x1) = s(x1) U2_gga(x1, x2, x3, x4) = U2_gga(x2, x4) half_in_ga(x1, x2) = half_in_ga(x1) half_out_ga(x1, x2) = half_out_ga(x2) U4_ga(x1, x2, x3) = U4_ga(x3) U3_gga(x1, x2, x3, x4) = U3_gga(x4) log2_out_ga(x1, x2) = log2_out_ga(x2) ---------------------------------------- (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: LOG2_IN_GA(X, Y) -> U1_GA(X, Y, log2_in_gga(X, 0, Y)) LOG2_IN_GA(X, Y) -> LOG2_IN_GGA(X, 0, Y) LOG2_IN_GGA(s(s(X)), I, Y) -> U2_GGA(X, I, Y, half_in_ga(s(s(X)), X1)) LOG2_IN_GGA(s(s(X)), I, Y) -> HALF_IN_GA(s(s(X)), X1) HALF_IN_GA(s(s(X)), s(Y)) -> U4_GA(X, Y, half_in_ga(X, Y)) HALF_IN_GA(s(s(X)), s(Y)) -> HALF_IN_GA(X, Y) U2_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) -> U3_GGA(X, I, Y, log2_in_gga(X1, s(I), Y)) U2_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) -> LOG2_IN_GGA(X1, s(I), Y) The TRS R consists of the following rules: log2_in_ga(X, Y) -> U1_ga(X, Y, log2_in_gga(X, 0, Y)) log2_in_gga(0, I, I) -> log2_out_gga(0, I, I) log2_in_gga(s(0), I, I) -> log2_out_gga(s(0), I, I) log2_in_gga(s(s(X)), I, Y) -> U2_gga(X, I, Y, half_in_ga(s(s(X)), X1)) half_in_ga(0, 0) -> half_out_ga(0, 0) half_in_ga(s(0), 0) -> half_out_ga(s(0), 0) half_in_ga(s(s(X)), s(Y)) -> U4_ga(X, Y, half_in_ga(X, Y)) U4_ga(X, Y, half_out_ga(X, Y)) -> half_out_ga(s(s(X)), s(Y)) U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) -> U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y)) U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) -> log2_out_gga(s(s(X)), I, Y) U1_ga(X, Y, log2_out_gga(X, 0, Y)) -> log2_out_ga(X, Y) The argument filtering Pi contains the following mapping: log2_in_ga(x1, x2) = log2_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) log2_in_gga(x1, x2, x3) = log2_in_gga(x1, x2) 0 = 0 log2_out_gga(x1, x2, x3) = log2_out_gga(x3) s(x1) = s(x1) U2_gga(x1, x2, x3, x4) = U2_gga(x2, x4) half_in_ga(x1, x2) = half_in_ga(x1) half_out_ga(x1, x2) = half_out_ga(x2) U4_ga(x1, x2, x3) = U4_ga(x3) U3_gga(x1, x2, x3, x4) = U3_gga(x4) log2_out_ga(x1, x2) = log2_out_ga(x2) LOG2_IN_GA(x1, x2) = LOG2_IN_GA(x1) U1_GA(x1, x2, x3) = U1_GA(x3) LOG2_IN_GGA(x1, x2, x3) = LOG2_IN_GGA(x1, x2) U2_GGA(x1, x2, x3, x4) = U2_GGA(x2, x4) HALF_IN_GA(x1, x2) = HALF_IN_GA(x1) U4_GA(x1, x2, x3) = U4_GA(x3) U3_GGA(x1, x2, x3, x4) = U3_GGA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: LOG2_IN_GA(X, Y) -> U1_GA(X, Y, log2_in_gga(X, 0, Y)) LOG2_IN_GA(X, Y) -> LOG2_IN_GGA(X, 0, Y) LOG2_IN_GGA(s(s(X)), I, Y) -> U2_GGA(X, I, Y, half_in_ga(s(s(X)), X1)) LOG2_IN_GGA(s(s(X)), I, Y) -> HALF_IN_GA(s(s(X)), X1) HALF_IN_GA(s(s(X)), s(Y)) -> U4_GA(X, Y, half_in_ga(X, Y)) HALF_IN_GA(s(s(X)), s(Y)) -> HALF_IN_GA(X, Y) U2_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) -> U3_GGA(X, I, Y, log2_in_gga(X1, s(I), Y)) U2_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) -> LOG2_IN_GGA(X1, s(I), Y) The TRS R consists of the following rules: log2_in_ga(X, Y) -> U1_ga(X, Y, log2_in_gga(X, 0, Y)) log2_in_gga(0, I, I) -> log2_out_gga(0, I, I) log2_in_gga(s(0), I, I) -> log2_out_gga(s(0), I, I) log2_in_gga(s(s(X)), I, Y) -> U2_gga(X, I, Y, half_in_ga(s(s(X)), X1)) half_in_ga(0, 0) -> half_out_ga(0, 0) half_in_ga(s(0), 0) -> half_out_ga(s(0), 0) half_in_ga(s(s(X)), s(Y)) -> U4_ga(X, Y, half_in_ga(X, Y)) U4_ga(X, Y, half_out_ga(X, Y)) -> half_out_ga(s(s(X)), s(Y)) U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) -> U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y)) U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) -> log2_out_gga(s(s(X)), I, Y) U1_ga(X, Y, log2_out_gga(X, 0, Y)) -> log2_out_ga(X, Y) The argument filtering Pi contains the following mapping: log2_in_ga(x1, x2) = log2_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) log2_in_gga(x1, x2, x3) = log2_in_gga(x1, x2) 0 = 0 log2_out_gga(x1, x2, x3) = log2_out_gga(x3) s(x1) = s(x1) U2_gga(x1, x2, x3, x4) = U2_gga(x2, x4) half_in_ga(x1, x2) = half_in_ga(x1) half_out_ga(x1, x2) = half_out_ga(x2) U4_ga(x1, x2, x3) = U4_ga(x3) U3_gga(x1, x2, x3, x4) = U3_gga(x4) log2_out_ga(x1, x2) = log2_out_ga(x2) LOG2_IN_GA(x1, x2) = LOG2_IN_GA(x1) U1_GA(x1, x2, x3) = U1_GA(x3) LOG2_IN_GGA(x1, x2, x3) = LOG2_IN_GGA(x1, x2) U2_GGA(x1, x2, x3, x4) = U2_GGA(x2, x4) HALF_IN_GA(x1, x2) = HALF_IN_GA(x1) U4_GA(x1, x2, x3) = U4_GA(x3) U3_GGA(x1, x2, x3, x4) = U3_GGA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 5 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: HALF_IN_GA(s(s(X)), s(Y)) -> HALF_IN_GA(X, Y) The TRS R consists of the following rules: log2_in_ga(X, Y) -> U1_ga(X, Y, log2_in_gga(X, 0, Y)) log2_in_gga(0, I, I) -> log2_out_gga(0, I, I) log2_in_gga(s(0), I, I) -> log2_out_gga(s(0), I, I) log2_in_gga(s(s(X)), I, Y) -> U2_gga(X, I, Y, half_in_ga(s(s(X)), X1)) half_in_ga(0, 0) -> half_out_ga(0, 0) half_in_ga(s(0), 0) -> half_out_ga(s(0), 0) half_in_ga(s(s(X)), s(Y)) -> U4_ga(X, Y, half_in_ga(X, Y)) U4_ga(X, Y, half_out_ga(X, Y)) -> half_out_ga(s(s(X)), s(Y)) U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) -> U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y)) U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) -> log2_out_gga(s(s(X)), I, Y) U1_ga(X, Y, log2_out_gga(X, 0, Y)) -> log2_out_ga(X, Y) The argument filtering Pi contains the following mapping: log2_in_ga(x1, x2) = log2_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) log2_in_gga(x1, x2, x3) = log2_in_gga(x1, x2) 0 = 0 log2_out_gga(x1, x2, x3) = log2_out_gga(x3) s(x1) = s(x1) U2_gga(x1, x2, x3, x4) = U2_gga(x2, x4) half_in_ga(x1, x2) = half_in_ga(x1) half_out_ga(x1, x2) = half_out_ga(x2) U4_ga(x1, x2, x3) = U4_ga(x3) U3_gga(x1, x2, x3, x4) = U3_gga(x4) log2_out_ga(x1, x2) = log2_out_ga(x2) HALF_IN_GA(x1, x2) = HALF_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: HALF_IN_GA(s(s(X)), s(Y)) -> HALF_IN_GA(X, Y) R is empty. The argument filtering Pi contains the following mapping: s(x1) = s(x1) HALF_IN_GA(x1, x2) = HALF_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: HALF_IN_GA(s(s(X))) -> HALF_IN_GA(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: *HALF_IN_GA(s(s(X))) -> HALF_IN_GA(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_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) -> LOG2_IN_GGA(X1, s(I), Y) LOG2_IN_GGA(s(s(X)), I, Y) -> U2_GGA(X, I, Y, half_in_ga(s(s(X)), X1)) The TRS R consists of the following rules: log2_in_ga(X, Y) -> U1_ga(X, Y, log2_in_gga(X, 0, Y)) log2_in_gga(0, I, I) -> log2_out_gga(0, I, I) log2_in_gga(s(0), I, I) -> log2_out_gga(s(0), I, I) log2_in_gga(s(s(X)), I, Y) -> U2_gga(X, I, Y, half_in_ga(s(s(X)), X1)) half_in_ga(0, 0) -> half_out_ga(0, 0) half_in_ga(s(0), 0) -> half_out_ga(s(0), 0) half_in_ga(s(s(X)), s(Y)) -> U4_ga(X, Y, half_in_ga(X, Y)) U4_ga(X, Y, half_out_ga(X, Y)) -> half_out_ga(s(s(X)), s(Y)) U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) -> U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y)) U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) -> log2_out_gga(s(s(X)), I, Y) U1_ga(X, Y, log2_out_gga(X, 0, Y)) -> log2_out_ga(X, Y) The argument filtering Pi contains the following mapping: log2_in_ga(x1, x2) = log2_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) log2_in_gga(x1, x2, x3) = log2_in_gga(x1, x2) 0 = 0 log2_out_gga(x1, x2, x3) = log2_out_gga(x3) s(x1) = s(x1) U2_gga(x1, x2, x3, x4) = U2_gga(x2, x4) half_in_ga(x1, x2) = half_in_ga(x1) half_out_ga(x1, x2) = half_out_ga(x2) U4_ga(x1, x2, x3) = U4_ga(x3) U3_gga(x1, x2, x3, x4) = U3_gga(x4) log2_out_ga(x1, x2) = log2_out_ga(x2) LOG2_IN_GGA(x1, x2, x3) = LOG2_IN_GGA(x1, x2) U2_GGA(x1, x2, x3, x4) = U2_GGA(x2, 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_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) -> LOG2_IN_GGA(X1, s(I), Y) LOG2_IN_GGA(s(s(X)), I, Y) -> U2_GGA(X, I, Y, half_in_ga(s(s(X)), X1)) The TRS R consists of the following rules: half_in_ga(s(s(X)), s(Y)) -> U4_ga(X, Y, half_in_ga(X, Y)) U4_ga(X, Y, half_out_ga(X, Y)) -> half_out_ga(s(s(X)), s(Y)) half_in_ga(0, 0) -> half_out_ga(0, 0) half_in_ga(s(0), 0) -> half_out_ga(s(0), 0) The argument filtering Pi contains the following mapping: 0 = 0 s(x1) = s(x1) half_in_ga(x1, x2) = half_in_ga(x1) half_out_ga(x1, x2) = half_out_ga(x2) U4_ga(x1, x2, x3) = U4_ga(x3) LOG2_IN_GGA(x1, x2, x3) = LOG2_IN_GGA(x1, x2) U2_GGA(x1, x2, x3, x4) = U2_GGA(x2, 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_GGA(I, half_out_ga(X1)) -> LOG2_IN_GGA(X1, s(I)) LOG2_IN_GGA(s(s(X)), I) -> U2_GGA(I, half_in_ga(s(s(X)))) The TRS R consists of the following rules: half_in_ga(s(s(X))) -> U4_ga(half_in_ga(X)) U4_ga(half_out_ga(Y)) -> half_out_ga(s(Y)) half_in_ga(0) -> half_out_ga(0) half_in_ga(s(0)) -> half_out_ga(0) The set Q consists of the following terms: half_in_ga(x0) U4_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: LOG2_IN_GGA(s(s(X)), I) -> U2_GGA(I, half_in_ga(s(s(X)))) Strictly oriented rules of the TRS R: half_in_ga(s(0)) -> half_out_ga(0) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(LOG2_IN_GGA(x_1, x_2)) = 2*x_1 + x_2 POL(U2_GGA(x_1, x_2)) = 1 + x_1 + x_2 POL(U4_ga(x_1)) = 2 + x_1 POL(half_in_ga(x_1)) = x_1 POL(half_out_ga(x_1)) = 2*x_1 POL(s(x_1)) = 1 + x_1 ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: U2_GGA(I, half_out_ga(X1)) -> LOG2_IN_GGA(X1, s(I)) The TRS R consists of the following rules: half_in_ga(s(s(X))) -> U4_ga(half_in_ga(X)) U4_ga(half_out_ga(Y)) -> half_out_ga(s(Y)) half_in_ga(0) -> half_out_ga(0) The set Q consists of the following terms: half_in_ga(x0) U4_ga(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (21) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (22) TRUE