/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern div(g,g,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND,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, 0 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) PiDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) PiDP (9) PiDPToQDPProof [SOUND, 1 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Clauses: div(X, Y, Z) :- quot(X, Y, Y, Z). quot(0, s(Y), s(Z), 0). quot(s(X), s(Y), Z, U) :- quot(X, Y, Z, U). quot(X, 0, s(Z), s(U)) :- quot(X, s(Z), s(Z), U). Query: div(g,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: div_in_3: (b,b,f) quot_in_4: (b,b,b,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: div_in_gga(X, Y, Z) -> U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z)) quot_in_ggga(0, s(Y), s(Z), 0) -> quot_out_ggga(0, s(Y), s(Z), 0) quot_in_ggga(s(X), s(Y), Z, U) -> U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U)) quot_in_ggga(X, 0, s(Z), s(U)) -> U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U)) U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) -> quot_out_ggga(X, 0, s(Z), s(U)) U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) -> quot_out_ggga(s(X), s(Y), Z, U) U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) -> div_out_gga(X, Y, Z) The argument filtering Pi contains the following mapping: div_in_gga(x1, x2, x3) = div_in_gga(x1, x2) U1_gga(x1, x2, x3, x4) = U1_gga(x4) quot_in_ggga(x1, x2, x3, x4) = quot_in_ggga(x1, x2, x3) 0 = 0 s(x1) = s(x1) quot_out_ggga(x1, x2, x3, x4) = quot_out_ggga(x4) U2_ggga(x1, x2, x3, x4, x5) = U2_ggga(x5) U3_ggga(x1, x2, x3, x4) = U3_ggga(x4) div_out_gga(x1, x2, x3) = div_out_gga(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: div_in_gga(X, Y, Z) -> U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z)) quot_in_ggga(0, s(Y), s(Z), 0) -> quot_out_ggga(0, s(Y), s(Z), 0) quot_in_ggga(s(X), s(Y), Z, U) -> U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U)) quot_in_ggga(X, 0, s(Z), s(U)) -> U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U)) U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) -> quot_out_ggga(X, 0, s(Z), s(U)) U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) -> quot_out_ggga(s(X), s(Y), Z, U) U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) -> div_out_gga(X, Y, Z) The argument filtering Pi contains the following mapping: div_in_gga(x1, x2, x3) = div_in_gga(x1, x2) U1_gga(x1, x2, x3, x4) = U1_gga(x4) quot_in_ggga(x1, x2, x3, x4) = quot_in_ggga(x1, x2, x3) 0 = 0 s(x1) = s(x1) quot_out_ggga(x1, x2, x3, x4) = quot_out_ggga(x4) U2_ggga(x1, x2, x3, x4, x5) = U2_ggga(x5) U3_ggga(x1, x2, x3, x4) = U3_ggga(x4) div_out_gga(x1, x2, x3) = div_out_gga(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: DIV_IN_GGA(X, Y, Z) -> U1_GGA(X, Y, Z, quot_in_ggga(X, Y, Y, Z)) DIV_IN_GGA(X, Y, Z) -> QUOT_IN_GGGA(X, Y, Y, Z) QUOT_IN_GGGA(s(X), s(Y), Z, U) -> U2_GGGA(X, Y, Z, U, quot_in_ggga(X, Y, Z, U)) QUOT_IN_GGGA(s(X), s(Y), Z, U) -> QUOT_IN_GGGA(X, Y, Z, U) QUOT_IN_GGGA(X, 0, s(Z), s(U)) -> U3_GGGA(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U)) QUOT_IN_GGGA(X, 0, s(Z), s(U)) -> QUOT_IN_GGGA(X, s(Z), s(Z), U) The TRS R consists of the following rules: div_in_gga(X, Y, Z) -> U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z)) quot_in_ggga(0, s(Y), s(Z), 0) -> quot_out_ggga(0, s(Y), s(Z), 0) quot_in_ggga(s(X), s(Y), Z, U) -> U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U)) quot_in_ggga(X, 0, s(Z), s(U)) -> U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U)) U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) -> quot_out_ggga(X, 0, s(Z), s(U)) U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) -> quot_out_ggga(s(X), s(Y), Z, U) U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) -> div_out_gga(X, Y, Z) The argument filtering Pi contains the following mapping: div_in_gga(x1, x2, x3) = div_in_gga(x1, x2) U1_gga(x1, x2, x3, x4) = U1_gga(x4) quot_in_ggga(x1, x2, x3, x4) = quot_in_ggga(x1, x2, x3) 0 = 0 s(x1) = s(x1) quot_out_ggga(x1, x2, x3, x4) = quot_out_ggga(x4) U2_ggga(x1, x2, x3, x4, x5) = U2_ggga(x5) U3_ggga(x1, x2, x3, x4) = U3_ggga(x4) div_out_gga(x1, x2, x3) = div_out_gga(x3) DIV_IN_GGA(x1, x2, x3) = DIV_IN_GGA(x1, x2) U1_GGA(x1, x2, x3, x4) = U1_GGA(x4) QUOT_IN_GGGA(x1, x2, x3, x4) = QUOT_IN_GGGA(x1, x2, x3) U2_GGGA(x1, x2, x3, x4, x5) = U2_GGGA(x5) U3_GGGA(x1, x2, x3, x4) = U3_GGGA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: DIV_IN_GGA(X, Y, Z) -> U1_GGA(X, Y, Z, quot_in_ggga(X, Y, Y, Z)) DIV_IN_GGA(X, Y, Z) -> QUOT_IN_GGGA(X, Y, Y, Z) QUOT_IN_GGGA(s(X), s(Y), Z, U) -> U2_GGGA(X, Y, Z, U, quot_in_ggga(X, Y, Z, U)) QUOT_IN_GGGA(s(X), s(Y), Z, U) -> QUOT_IN_GGGA(X, Y, Z, U) QUOT_IN_GGGA(X, 0, s(Z), s(U)) -> U3_GGGA(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U)) QUOT_IN_GGGA(X, 0, s(Z), s(U)) -> QUOT_IN_GGGA(X, s(Z), s(Z), U) The TRS R consists of the following rules: div_in_gga(X, Y, Z) -> U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z)) quot_in_ggga(0, s(Y), s(Z), 0) -> quot_out_ggga(0, s(Y), s(Z), 0) quot_in_ggga(s(X), s(Y), Z, U) -> U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U)) quot_in_ggga(X, 0, s(Z), s(U)) -> U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U)) U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) -> quot_out_ggga(X, 0, s(Z), s(U)) U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) -> quot_out_ggga(s(X), s(Y), Z, U) U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) -> div_out_gga(X, Y, Z) The argument filtering Pi contains the following mapping: div_in_gga(x1, x2, x3) = div_in_gga(x1, x2) U1_gga(x1, x2, x3, x4) = U1_gga(x4) quot_in_ggga(x1, x2, x3, x4) = quot_in_ggga(x1, x2, x3) 0 = 0 s(x1) = s(x1) quot_out_ggga(x1, x2, x3, x4) = quot_out_ggga(x4) U2_ggga(x1, x2, x3, x4, x5) = U2_ggga(x5) U3_ggga(x1, x2, x3, x4) = U3_ggga(x4) div_out_gga(x1, x2, x3) = div_out_gga(x3) DIV_IN_GGA(x1, x2, x3) = DIV_IN_GGA(x1, x2) U1_GGA(x1, x2, x3, x4) = U1_GGA(x4) QUOT_IN_GGGA(x1, x2, x3, x4) = QUOT_IN_GGGA(x1, x2, x3) U2_GGGA(x1, x2, x3, x4, x5) = U2_GGGA(x5) U3_GGGA(x1, x2, x3, x4) = U3_GGGA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes. ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: QUOT_IN_GGGA(X, 0, s(Z), s(U)) -> QUOT_IN_GGGA(X, s(Z), s(Z), U) QUOT_IN_GGGA(s(X), s(Y), Z, U) -> QUOT_IN_GGGA(X, Y, Z, U) The TRS R consists of the following rules: div_in_gga(X, Y, Z) -> U1_gga(X, Y, Z, quot_in_ggga(X, Y, Y, Z)) quot_in_ggga(0, s(Y), s(Z), 0) -> quot_out_ggga(0, s(Y), s(Z), 0) quot_in_ggga(s(X), s(Y), Z, U) -> U2_ggga(X, Y, Z, U, quot_in_ggga(X, Y, Z, U)) quot_in_ggga(X, 0, s(Z), s(U)) -> U3_ggga(X, Z, U, quot_in_ggga(X, s(Z), s(Z), U)) U3_ggga(X, Z, U, quot_out_ggga(X, s(Z), s(Z), U)) -> quot_out_ggga(X, 0, s(Z), s(U)) U2_ggga(X, Y, Z, U, quot_out_ggga(X, Y, Z, U)) -> quot_out_ggga(s(X), s(Y), Z, U) U1_gga(X, Y, Z, quot_out_ggga(X, Y, Y, Z)) -> div_out_gga(X, Y, Z) The argument filtering Pi contains the following mapping: div_in_gga(x1, x2, x3) = div_in_gga(x1, x2) U1_gga(x1, x2, x3, x4) = U1_gga(x4) quot_in_ggga(x1, x2, x3, x4) = quot_in_ggga(x1, x2, x3) 0 = 0 s(x1) = s(x1) quot_out_ggga(x1, x2, x3, x4) = quot_out_ggga(x4) U2_ggga(x1, x2, x3, x4, x5) = U2_ggga(x5) U3_ggga(x1, x2, x3, x4) = U3_ggga(x4) div_out_gga(x1, x2, x3) = div_out_gga(x3) QUOT_IN_GGGA(x1, x2, x3, x4) = QUOT_IN_GGGA(x1, x2, x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (7) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (8) Obligation: Pi DP problem: The TRS P consists of the following rules: QUOT_IN_GGGA(X, 0, s(Z), s(U)) -> QUOT_IN_GGGA(X, s(Z), s(Z), U) QUOT_IN_GGGA(s(X), s(Y), Z, U) -> QUOT_IN_GGGA(X, Y, Z, U) R is empty. The argument filtering Pi contains the following mapping: 0 = 0 s(x1) = s(x1) QUOT_IN_GGGA(x1, x2, x3, x4) = QUOT_IN_GGGA(x1, x2, x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (9) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: QUOT_IN_GGGA(X, 0, s(Z)) -> QUOT_IN_GGGA(X, s(Z), s(Z)) QUOT_IN_GGGA(s(X), s(Y), Z) -> QUOT_IN_GGGA(X, Y, Z) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (11) 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: *QUOT_IN_GGGA(s(X), s(Y), Z) -> QUOT_IN_GGGA(X, Y, Z) The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3 *QUOT_IN_GGGA(X, 0, s(Z)) -> QUOT_IN_GGGA(X, s(Z), s(Z)) The graph contains the following edges 1 >= 1, 3 >= 2, 3 >= 3 ---------------------------------------- (12) YES