/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern reverse(g,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof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, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Clauses: reverse(L, LR) :- revacc(L, LR, []). revacc([], L, L). revacc(.(EL, T), R, A) :- revacc(T, R, .(EL, A)). Query: reverse(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: reverse_in_2: (b,f) revacc_in_3: (b,f,b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: reverse_in_ga(L, LR) -> U1_ga(L, LR, revacc_in_gag(L, LR, [])) revacc_in_gag([], L, L) -> revacc_out_gag([], L, L) revacc_in_gag(.(EL, T), R, A) -> U2_gag(EL, T, R, A, revacc_in_gag(T, R, .(EL, A))) U2_gag(EL, T, R, A, revacc_out_gag(T, R, .(EL, A))) -> revacc_out_gag(.(EL, T), R, A) U1_ga(L, LR, revacc_out_gag(L, LR, [])) -> reverse_out_ga(L, LR) The argument filtering Pi contains the following mapping: reverse_in_ga(x1, x2) = reverse_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) revacc_in_gag(x1, x2, x3) = revacc_in_gag(x1, x3) [] = [] revacc_out_gag(x1, x2, x3) = revacc_out_gag(x2) .(x1, x2) = .(x1, x2) U2_gag(x1, x2, x3, x4, x5) = U2_gag(x5) reverse_out_ga(x1, x2) = reverse_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: reverse_in_ga(L, LR) -> U1_ga(L, LR, revacc_in_gag(L, LR, [])) revacc_in_gag([], L, L) -> revacc_out_gag([], L, L) revacc_in_gag(.(EL, T), R, A) -> U2_gag(EL, T, R, A, revacc_in_gag(T, R, .(EL, A))) U2_gag(EL, T, R, A, revacc_out_gag(T, R, .(EL, A))) -> revacc_out_gag(.(EL, T), R, A) U1_ga(L, LR, revacc_out_gag(L, LR, [])) -> reverse_out_ga(L, LR) The argument filtering Pi contains the following mapping: reverse_in_ga(x1, x2) = reverse_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) revacc_in_gag(x1, x2, x3) = revacc_in_gag(x1, x3) [] = [] revacc_out_gag(x1, x2, x3) = revacc_out_gag(x2) .(x1, x2) = .(x1, x2) U2_gag(x1, x2, x3, x4, x5) = U2_gag(x5) reverse_out_ga(x1, x2) = reverse_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: REVERSE_IN_GA(L, LR) -> U1_GA(L, LR, revacc_in_gag(L, LR, [])) REVERSE_IN_GA(L, LR) -> REVACC_IN_GAG(L, LR, []) REVACC_IN_GAG(.(EL, T), R, A) -> U2_GAG(EL, T, R, A, revacc_in_gag(T, R, .(EL, A))) REVACC_IN_GAG(.(EL, T), R, A) -> REVACC_IN_GAG(T, R, .(EL, A)) The TRS R consists of the following rules: reverse_in_ga(L, LR) -> U1_ga(L, LR, revacc_in_gag(L, LR, [])) revacc_in_gag([], L, L) -> revacc_out_gag([], L, L) revacc_in_gag(.(EL, T), R, A) -> U2_gag(EL, T, R, A, revacc_in_gag(T, R, .(EL, A))) U2_gag(EL, T, R, A, revacc_out_gag(T, R, .(EL, A))) -> revacc_out_gag(.(EL, T), R, A) U1_ga(L, LR, revacc_out_gag(L, LR, [])) -> reverse_out_ga(L, LR) The argument filtering Pi contains the following mapping: reverse_in_ga(x1, x2) = reverse_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) revacc_in_gag(x1, x2, x3) = revacc_in_gag(x1, x3) [] = [] revacc_out_gag(x1, x2, x3) = revacc_out_gag(x2) .(x1, x2) = .(x1, x2) U2_gag(x1, x2, x3, x4, x5) = U2_gag(x5) reverse_out_ga(x1, x2) = reverse_out_ga(x2) REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1) U1_GA(x1, x2, x3) = U1_GA(x3) REVACC_IN_GAG(x1, x2, x3) = REVACC_IN_GAG(x1, x3) U2_GAG(x1, x2, x3, x4, x5) = U2_GAG(x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: REVERSE_IN_GA(L, LR) -> U1_GA(L, LR, revacc_in_gag(L, LR, [])) REVERSE_IN_GA(L, LR) -> REVACC_IN_GAG(L, LR, []) REVACC_IN_GAG(.(EL, T), R, A) -> U2_GAG(EL, T, R, A, revacc_in_gag(T, R, .(EL, A))) REVACC_IN_GAG(.(EL, T), R, A) -> REVACC_IN_GAG(T, R, .(EL, A)) The TRS R consists of the following rules: reverse_in_ga(L, LR) -> U1_ga(L, LR, revacc_in_gag(L, LR, [])) revacc_in_gag([], L, L) -> revacc_out_gag([], L, L) revacc_in_gag(.(EL, T), R, A) -> U2_gag(EL, T, R, A, revacc_in_gag(T, R, .(EL, A))) U2_gag(EL, T, R, A, revacc_out_gag(T, R, .(EL, A))) -> revacc_out_gag(.(EL, T), R, A) U1_ga(L, LR, revacc_out_gag(L, LR, [])) -> reverse_out_ga(L, LR) The argument filtering Pi contains the following mapping: reverse_in_ga(x1, x2) = reverse_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) revacc_in_gag(x1, x2, x3) = revacc_in_gag(x1, x3) [] = [] revacc_out_gag(x1, x2, x3) = revacc_out_gag(x2) .(x1, x2) = .(x1, x2) U2_gag(x1, x2, x3, x4, x5) = U2_gag(x5) reverse_out_ga(x1, x2) = reverse_out_ga(x2) REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1) U1_GA(x1, x2, x3) = U1_GA(x3) REVACC_IN_GAG(x1, x2, x3) = REVACC_IN_GAG(x1, x3) U2_GAG(x1, x2, x3, x4, x5) = U2_GAG(x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes. ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: REVACC_IN_GAG(.(EL, T), R, A) -> REVACC_IN_GAG(T, R, .(EL, A)) The TRS R consists of the following rules: reverse_in_ga(L, LR) -> U1_ga(L, LR, revacc_in_gag(L, LR, [])) revacc_in_gag([], L, L) -> revacc_out_gag([], L, L) revacc_in_gag(.(EL, T), R, A) -> U2_gag(EL, T, R, A, revacc_in_gag(T, R, .(EL, A))) U2_gag(EL, T, R, A, revacc_out_gag(T, R, .(EL, A))) -> revacc_out_gag(.(EL, T), R, A) U1_ga(L, LR, revacc_out_gag(L, LR, [])) -> reverse_out_ga(L, LR) The argument filtering Pi contains the following mapping: reverse_in_ga(x1, x2) = reverse_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x3) revacc_in_gag(x1, x2, x3) = revacc_in_gag(x1, x3) [] = [] revacc_out_gag(x1, x2, x3) = revacc_out_gag(x2) .(x1, x2) = .(x1, x2) U2_gag(x1, x2, x3, x4, x5) = U2_gag(x5) reverse_out_ga(x1, x2) = reverse_out_ga(x2) REVACC_IN_GAG(x1, x2, x3) = REVACC_IN_GAG(x1, 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: REVACC_IN_GAG(.(EL, T), R, A) -> REVACC_IN_GAG(T, R, .(EL, A)) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) REVACC_IN_GAG(x1, x2, x3) = REVACC_IN_GAG(x1, 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: REVACC_IN_GAG(.(EL, T), A) -> REVACC_IN_GAG(T, .(EL, A)) 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: *REVACC_IN_GAG(.(EL, T), A) -> REVACC_IN_GAG(T, .(EL, A)) The graph contains the following edges 1 > 1 ---------------------------------------- (12) YES