/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern plus(g,a,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND,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, 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: plus(0, Y, Y). plus(s(X), Y, Z) :- plus(X, s(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) 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, Z) -> U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z)) U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) -> plus_out_gaa(s(X), Y, 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) U1_gaa(x1, x2, x3, x4) = U1_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, Z) -> U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z)) U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) -> plus_out_gaa(s(X), Y, 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) U1_gaa(x1, x2, x3, x4) = U1_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, Z) -> U1_GAA(X, Y, Z, plus_in_gaa(X, s(Y), Z)) PLUS_IN_GAA(s(X), Y, Z) -> PLUS_IN_GAA(X, s(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, Z) -> U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z)) U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) -> plus_out_gaa(s(X), Y, 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) U1_gaa(x1, x2, x3, x4) = U1_gaa(x4) PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1) U1_GAA(x1, x2, x3, x4) = U1_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, Z) -> U1_GAA(X, Y, Z, plus_in_gaa(X, s(Y), Z)) PLUS_IN_GAA(s(X), Y, Z) -> PLUS_IN_GAA(X, s(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, Z) -> U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z)) U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) -> plus_out_gaa(s(X), Y, 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) U1_gaa(x1, x2, x3, x4) = U1_gaa(x4) PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1) U1_GAA(x1, x2, x3, x4) = U1_GAA(x4) 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: PLUS_IN_GAA(s(X), Y, Z) -> PLUS_IN_GAA(X, s(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, Z) -> U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z)) U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) -> plus_out_gaa(s(X), Y, 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) U1_gaa(x1, x2, x3, x4) = U1_gaa(x4) PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1) 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: PLUS_IN_GAA(s(X), Y, Z) -> PLUS_IN_GAA(X, s(Y), Z) R is empty. The argument filtering Pi contains the following mapping: s(x1) = s(x1) PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1) 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: PLUS_IN_GAA(s(X)) -> PLUS_IN_GAA(X) 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: *PLUS_IN_GAA(s(X)) -> PLUS_IN_GAA(X) The graph contains the following edges 1 > 1 ---------------------------------------- (12) YES