3.54/1.62 YES 3.64/1.64 proof of /export/starexec/sandbox/benchmark/theBenchmark.pl 3.64/1.64 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.64/1.64 3.64/1.64 3.64/1.64 Left Termination of the query pattern 3.64/1.64 3.64/1.64 member(a,g) 3.64/1.64 3.64/1.64 w.r.t. the given Prolog program could successfully be proven: 3.64/1.64 3.64/1.64 (0) Prolog 3.64/1.64 (1) PrologToPiTRSProof [SOUND, 0 ms] 3.64/1.64 (2) PiTRS 3.64/1.64 (3) DependencyPairsProof [EQUIVALENT, 0 ms] 3.64/1.64 (4) PiDP 3.64/1.64 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 3.64/1.64 (6) PiDP 3.64/1.64 (7) UsableRulesProof [EQUIVALENT, 0 ms] 3.64/1.64 (8) PiDP 3.64/1.64 (9) PiDPToQDPProof [SOUND, 0 ms] 3.64/1.64 (10) QDP 3.64/1.64 (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] 3.64/1.64 (12) YES 3.64/1.64 3.64/1.64 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (0) 3.64/1.64 Obligation: 3.64/1.64 Clauses: 3.64/1.64 3.64/1.64 member(X, .(Y, Xs)) :- member(X, Xs). 3.64/1.64 member(X, .(X, Xs)). 3.64/1.64 3.64/1.64 3.64/1.64 Query: member(a,g) 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (1) PrologToPiTRSProof (SOUND) 3.64/1.64 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 3.64/1.64 3.64/1.64 member_in_2: (f,b) 3.64/1.64 3.64/1.64 Transforming Prolog into the following Term Rewriting System: 3.64/1.64 3.64/1.64 Pi-finite rewrite system: 3.64/1.64 The TRS R consists of the following rules: 3.64/1.64 3.64/1.64 member_in_ag(X, .(Y, Xs)) -> U1_ag(X, Y, Xs, member_in_ag(X, Xs)) 3.64/1.64 member_in_ag(X, .(X, Xs)) -> member_out_ag(X, .(X, Xs)) 3.64/1.64 U1_ag(X, Y, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(Y, Xs)) 3.64/1.64 3.64/1.64 The argument filtering Pi contains the following mapping: 3.64/1.64 member_in_ag(x1, x2) = member_in_ag(x2) 3.64/1.64 3.64/1.64 .(x1, x2) = .(x1, x2) 3.64/1.64 3.64/1.64 U1_ag(x1, x2, x3, x4) = U1_ag(x2, x3, x4) 3.64/1.64 3.64/1.64 member_out_ag(x1, x2) = member_out_ag(x1, x2) 3.64/1.64 3.64/1.64 3.64/1.64 3.64/1.64 3.64/1.64 3.64/1.64 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 3.64/1.64 3.64/1.64 3.64/1.64 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (2) 3.64/1.64 Obligation: 3.64/1.64 Pi-finite rewrite system: 3.64/1.64 The TRS R consists of the following rules: 3.64/1.64 3.64/1.64 member_in_ag(X, .(Y, Xs)) -> U1_ag(X, Y, Xs, member_in_ag(X, Xs)) 3.64/1.64 member_in_ag(X, .(X, Xs)) -> member_out_ag(X, .(X, Xs)) 3.64/1.64 U1_ag(X, Y, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(Y, Xs)) 3.64/1.64 3.64/1.64 The argument filtering Pi contains the following mapping: 3.64/1.64 member_in_ag(x1, x2) = member_in_ag(x2) 3.64/1.64 3.64/1.64 .(x1, x2) = .(x1, x2) 3.64/1.64 3.64/1.64 U1_ag(x1, x2, x3, x4) = U1_ag(x2, x3, x4) 3.64/1.64 3.64/1.64 member_out_ag(x1, x2) = member_out_ag(x1, x2) 3.64/1.64 3.64/1.64 3.64/1.64 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (3) DependencyPairsProof (EQUIVALENT) 3.64/1.64 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 3.64/1.64 Pi DP problem: 3.64/1.64 The TRS P consists of the following rules: 3.64/1.64 3.64/1.64 MEMBER_IN_AG(X, .(Y, Xs)) -> U1_AG(X, Y, Xs, member_in_ag(X, Xs)) 3.64/1.64 MEMBER_IN_AG(X, .(Y, Xs)) -> MEMBER_IN_AG(X, Xs) 3.64/1.64 3.64/1.64 The TRS R consists of the following rules: 3.64/1.64 3.64/1.64 member_in_ag(X, .(Y, Xs)) -> U1_ag(X, Y, Xs, member_in_ag(X, Xs)) 3.64/1.64 member_in_ag(X, .(X, Xs)) -> member_out_ag(X, .(X, Xs)) 3.64/1.64 U1_ag(X, Y, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(Y, Xs)) 3.64/1.64 3.64/1.64 The argument filtering Pi contains the following mapping: 3.64/1.64 member_in_ag(x1, x2) = member_in_ag(x2) 3.64/1.64 3.64/1.64 .(x1, x2) = .(x1, x2) 3.64/1.64 3.64/1.64 U1_ag(x1, x2, x3, x4) = U1_ag(x2, x3, x4) 3.64/1.64 3.64/1.64 member_out_ag(x1, x2) = member_out_ag(x1, x2) 3.64/1.64 3.64/1.64 MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2) 3.64/1.64 3.64/1.64 U1_AG(x1, x2, x3, x4) = U1_AG(x2, x3, x4) 3.64/1.64 3.64/1.64 3.64/1.64 We have to consider all (P,R,Pi)-chains 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (4) 3.64/1.64 Obligation: 3.64/1.64 Pi DP problem: 3.64/1.64 The TRS P consists of the following rules: 3.64/1.64 3.64/1.64 MEMBER_IN_AG(X, .(Y, Xs)) -> U1_AG(X, Y, Xs, member_in_ag(X, Xs)) 3.64/1.64 MEMBER_IN_AG(X, .(Y, Xs)) -> MEMBER_IN_AG(X, Xs) 3.64/1.64 3.64/1.64 The TRS R consists of the following rules: 3.64/1.64 3.64/1.64 member_in_ag(X, .(Y, Xs)) -> U1_ag(X, Y, Xs, member_in_ag(X, Xs)) 3.64/1.64 member_in_ag(X, .(X, Xs)) -> member_out_ag(X, .(X, Xs)) 3.64/1.64 U1_ag(X, Y, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(Y, Xs)) 3.64/1.64 3.64/1.64 The argument filtering Pi contains the following mapping: 3.64/1.64 member_in_ag(x1, x2) = member_in_ag(x2) 3.64/1.64 3.64/1.64 .(x1, x2) = .(x1, x2) 3.64/1.64 3.64/1.64 U1_ag(x1, x2, x3, x4) = U1_ag(x2, x3, x4) 3.64/1.64 3.64/1.64 member_out_ag(x1, x2) = member_out_ag(x1, x2) 3.64/1.64 3.64/1.64 MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2) 3.64/1.64 3.64/1.64 U1_AG(x1, x2, x3, x4) = U1_AG(x2, x3, x4) 3.64/1.64 3.64/1.64 3.64/1.64 We have to consider all (P,R,Pi)-chains 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (5) DependencyGraphProof (EQUIVALENT) 3.64/1.64 The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node. 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (6) 3.64/1.64 Obligation: 3.64/1.64 Pi DP problem: 3.64/1.64 The TRS P consists of the following rules: 3.64/1.64 3.64/1.64 MEMBER_IN_AG(X, .(Y, Xs)) -> MEMBER_IN_AG(X, Xs) 3.64/1.64 3.64/1.64 The TRS R consists of the following rules: 3.64/1.64 3.64/1.64 member_in_ag(X, .(Y, Xs)) -> U1_ag(X, Y, Xs, member_in_ag(X, Xs)) 3.64/1.64 member_in_ag(X, .(X, Xs)) -> member_out_ag(X, .(X, Xs)) 3.64/1.64 U1_ag(X, Y, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(Y, Xs)) 3.64/1.64 3.64/1.64 The argument filtering Pi contains the following mapping: 3.64/1.64 member_in_ag(x1, x2) = member_in_ag(x2) 3.64/1.64 3.64/1.64 .(x1, x2) = .(x1, x2) 3.64/1.64 3.64/1.64 U1_ag(x1, x2, x3, x4) = U1_ag(x2, x3, x4) 3.64/1.64 3.64/1.64 member_out_ag(x1, x2) = member_out_ag(x1, x2) 3.64/1.64 3.64/1.64 MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2) 3.64/1.64 3.64/1.64 3.64/1.64 We have to consider all (P,R,Pi)-chains 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (7) UsableRulesProof (EQUIVALENT) 3.64/1.64 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (8) 3.64/1.64 Obligation: 3.64/1.64 Pi DP problem: 3.64/1.64 The TRS P consists of the following rules: 3.64/1.64 3.64/1.64 MEMBER_IN_AG(X, .(Y, Xs)) -> MEMBER_IN_AG(X, Xs) 3.64/1.64 3.64/1.64 R is empty. 3.64/1.64 The argument filtering Pi contains the following mapping: 3.64/1.64 .(x1, x2) = .(x1, x2) 3.64/1.64 3.64/1.64 MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2) 3.64/1.64 3.64/1.64 3.64/1.64 We have to consider all (P,R,Pi)-chains 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (9) PiDPToQDPProof (SOUND) 3.64/1.64 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (10) 3.64/1.64 Obligation: 3.64/1.64 Q DP problem: 3.64/1.64 The TRS P consists of the following rules: 3.64/1.64 3.64/1.64 MEMBER_IN_AG(.(Y, Xs)) -> MEMBER_IN_AG(Xs) 3.64/1.64 3.64/1.64 R is empty. 3.64/1.64 Q is empty. 3.64/1.64 We have to consider all (P,Q,R)-chains. 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (11) QDPSizeChangeProof (EQUIVALENT) 3.64/1.64 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. 3.64/1.64 3.64/1.64 From the DPs we obtained the following set of size-change graphs: 3.64/1.64 *MEMBER_IN_AG(.(Y, Xs)) -> MEMBER_IN_AG(Xs) 3.64/1.64 The graph contains the following edges 1 > 1 3.64/1.64 3.64/1.64 3.64/1.64 ---------------------------------------- 3.64/1.64 3.64/1.64 (12) 3.64/1.64 YES 3.64/1.67 EOF