/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern append3(a,a,g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof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: append1([], X, X). append1(.(X, Y), U, .(X, Z)) :- append1(Y, U, Z). append2([], X, X). append2(.(X, Y), U, .(X, Z)) :- append2(Y, U, Z). append3([], X, X). append3(.(X, Y), U, .(X, Z)) :- append3(Y, U, Z). Query: append3(a,a,g) ---------------------------------------- (1) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: append3_in_3: (f,f,b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: append3_in_aag([], X, X) -> append3_out_aag([], X, X) append3_in_aag(.(X, Y), U, .(X, Z)) -> U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z)) U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) -> append3_out_aag(.(X, Y), U, .(X, Z)) The argument filtering Pi contains the following mapping: append3_in_aag(x1, x2, x3) = append3_in_aag(x3) append3_out_aag(x1, x2, x3) = append3_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U3_aag(x1, x2, x3, x4, x5) = U3_aag(x1, x5) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: append3_in_aag([], X, X) -> append3_out_aag([], X, X) append3_in_aag(.(X, Y), U, .(X, Z)) -> U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z)) U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) -> append3_out_aag(.(X, Y), U, .(X, Z)) The argument filtering Pi contains the following mapping: append3_in_aag(x1, x2, x3) = append3_in_aag(x3) append3_out_aag(x1, x2, x3) = append3_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U3_aag(x1, x2, x3, x4, x5) = U3_aag(x1, x5) ---------------------------------------- (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: APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> U3_AAG(X, Y, U, Z, append3_in_aag(Y, U, Z)) APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> APPEND3_IN_AAG(Y, U, Z) The TRS R consists of the following rules: append3_in_aag([], X, X) -> append3_out_aag([], X, X) append3_in_aag(.(X, Y), U, .(X, Z)) -> U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z)) U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) -> append3_out_aag(.(X, Y), U, .(X, Z)) The argument filtering Pi contains the following mapping: append3_in_aag(x1, x2, x3) = append3_in_aag(x3) append3_out_aag(x1, x2, x3) = append3_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U3_aag(x1, x2, x3, x4, x5) = U3_aag(x1, x5) APPEND3_IN_AAG(x1, x2, x3) = APPEND3_IN_AAG(x3) U3_AAG(x1, x2, x3, x4, x5) = U3_AAG(x1, x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> U3_AAG(X, Y, U, Z, append3_in_aag(Y, U, Z)) APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> APPEND3_IN_AAG(Y, U, Z) The TRS R consists of the following rules: append3_in_aag([], X, X) -> append3_out_aag([], X, X) append3_in_aag(.(X, Y), U, .(X, Z)) -> U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z)) U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) -> append3_out_aag(.(X, Y), U, .(X, Z)) The argument filtering Pi contains the following mapping: append3_in_aag(x1, x2, x3) = append3_in_aag(x3) append3_out_aag(x1, x2, x3) = append3_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U3_aag(x1, x2, x3, x4, x5) = U3_aag(x1, x5) APPEND3_IN_AAG(x1, x2, x3) = APPEND3_IN_AAG(x3) U3_AAG(x1, x2, x3, x4, x5) = U3_AAG(x1, x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node. ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> APPEND3_IN_AAG(Y, U, Z) The TRS R consists of the following rules: append3_in_aag([], X, X) -> append3_out_aag([], X, X) append3_in_aag(.(X, Y), U, .(X, Z)) -> U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z)) U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) -> append3_out_aag(.(X, Y), U, .(X, Z)) The argument filtering Pi contains the following mapping: append3_in_aag(x1, x2, x3) = append3_in_aag(x3) append3_out_aag(x1, x2, x3) = append3_out_aag(x1, x2) .(x1, x2) = .(x1, x2) U3_aag(x1, x2, x3, x4, x5) = U3_aag(x1, x5) APPEND3_IN_AAG(x1, x2, x3) = APPEND3_IN_AAG(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: APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> APPEND3_IN_AAG(Y, U, Z) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) APPEND3_IN_AAG(x1, x2, x3) = APPEND3_IN_AAG(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: APPEND3_IN_AAG(.(X, Z)) -> APPEND3_IN_AAG(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: *APPEND3_IN_AAG(.(X, Z)) -> APPEND3_IN_AAG(Z) The graph contains the following edges 1 > 1 ---------------------------------------- (12) YES