3.49/1.79 YES 3.49/1.80 proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl 3.49/1.80 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.49/1.80 3.49/1.80 3.49/1.80 Left Termination of the query pattern 3.49/1.80 3.49/1.80 member(a,g) 3.49/1.80 3.49/1.80 w.r.t. the given Prolog program could successfully be proven: 3.49/1.80 3.49/1.80 (0) Prolog 3.49/1.80 (1) PrologToPiTRSProof [SOUND, 0 ms] 3.49/1.80 (2) PiTRS 3.49/1.80 (3) DependencyPairsProof [EQUIVALENT, 0 ms] 3.49/1.80 (4) PiDP 3.49/1.80 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 3.49/1.80 (6) PiDP 3.49/1.80 (7) UsableRulesProof [EQUIVALENT, 0 ms] 3.49/1.80 (8) PiDP 3.49/1.80 (9) PiDPToQDPProof [SOUND, 0 ms] 3.49/1.80 (10) QDP 3.49/1.80 (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] 3.49/1.80 (12) YES 3.49/1.80 3.49/1.80 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (0) 3.49/1.80 Obligation: 3.49/1.80 Clauses: 3.49/1.80 3.49/1.80 member(X, .(X, X1)). 3.49/1.80 member(X, .(X2, Xs)) :- member(X, Xs). 3.49/1.80 3.49/1.80 3.49/1.80 Query: member(a,g) 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (1) PrologToPiTRSProof (SOUND) 3.49/1.80 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 3.49/1.80 3.49/1.80 member_in_2: (f,b) 3.49/1.80 3.49/1.80 Transforming Prolog into the following Term Rewriting System: 3.49/1.80 3.49/1.80 Pi-finite rewrite system: 3.49/1.80 The TRS R consists of the following rules: 3.49/1.80 3.49/1.80 member_in_ag(X, .(X, X1)) -> member_out_ag(X, .(X, X1)) 3.49/1.80 member_in_ag(X, .(X2, Xs)) -> U1_ag(X, X2, Xs, member_in_ag(X, Xs)) 3.49/1.80 U1_ag(X, X2, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(X2, Xs)) 3.49/1.80 3.49/1.80 The argument filtering Pi contains the following mapping: 3.49/1.80 member_in_ag(x1, x2) = member_in_ag(x2) 3.49/1.80 3.49/1.80 .(x1, x2) = .(x1, x2) 3.49/1.80 3.49/1.80 member_out_ag(x1, x2) = member_out_ag(x1) 3.49/1.80 3.49/1.80 U1_ag(x1, x2, x3, x4) = U1_ag(x4) 3.49/1.80 3.49/1.80 3.49/1.80 3.49/1.80 3.49/1.80 3.49/1.80 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 3.49/1.80 3.49/1.80 3.49/1.80 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (2) 3.49/1.80 Obligation: 3.49/1.80 Pi-finite rewrite system: 3.49/1.80 The TRS R consists of the following rules: 3.49/1.80 3.49/1.80 member_in_ag(X, .(X, X1)) -> member_out_ag(X, .(X, X1)) 3.49/1.80 member_in_ag(X, .(X2, Xs)) -> U1_ag(X, X2, Xs, member_in_ag(X, Xs)) 3.49/1.80 U1_ag(X, X2, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(X2, Xs)) 3.49/1.80 3.49/1.80 The argument filtering Pi contains the following mapping: 3.49/1.80 member_in_ag(x1, x2) = member_in_ag(x2) 3.49/1.80 3.49/1.80 .(x1, x2) = .(x1, x2) 3.49/1.80 3.49/1.80 member_out_ag(x1, x2) = member_out_ag(x1) 3.49/1.80 3.49/1.80 U1_ag(x1, x2, x3, x4) = U1_ag(x4) 3.49/1.80 3.49/1.80 3.49/1.80 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (3) DependencyPairsProof (EQUIVALENT) 3.49/1.80 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 3.49/1.80 Pi DP problem: 3.49/1.80 The TRS P consists of the following rules: 3.49/1.80 3.49/1.80 MEMBER_IN_AG(X, .(X2, Xs)) -> U1_AG(X, X2, Xs, member_in_ag(X, Xs)) 3.49/1.80 MEMBER_IN_AG(X, .(X2, Xs)) -> MEMBER_IN_AG(X, Xs) 3.49/1.80 3.49/1.80 The TRS R consists of the following rules: 3.49/1.80 3.49/1.80 member_in_ag(X, .(X, X1)) -> member_out_ag(X, .(X, X1)) 3.49/1.80 member_in_ag(X, .(X2, Xs)) -> U1_ag(X, X2, Xs, member_in_ag(X, Xs)) 3.49/1.80 U1_ag(X, X2, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(X2, Xs)) 3.49/1.80 3.49/1.80 The argument filtering Pi contains the following mapping: 3.49/1.80 member_in_ag(x1, x2) = member_in_ag(x2) 3.49/1.80 3.49/1.80 .(x1, x2) = .(x1, x2) 3.49/1.80 3.49/1.80 member_out_ag(x1, x2) = member_out_ag(x1) 3.49/1.80 3.49/1.80 U1_ag(x1, x2, x3, x4) = U1_ag(x4) 3.49/1.80 3.49/1.80 MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2) 3.49/1.80 3.49/1.80 U1_AG(x1, x2, x3, x4) = U1_AG(x4) 3.49/1.80 3.49/1.80 3.49/1.80 We have to consider all (P,R,Pi)-chains 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (4) 3.49/1.80 Obligation: 3.49/1.80 Pi DP problem: 3.49/1.80 The TRS P consists of the following rules: 3.49/1.80 3.49/1.80 MEMBER_IN_AG(X, .(X2, Xs)) -> U1_AG(X, X2, Xs, member_in_ag(X, Xs)) 3.49/1.80 MEMBER_IN_AG(X, .(X2, Xs)) -> MEMBER_IN_AG(X, Xs) 3.49/1.80 3.49/1.80 The TRS R consists of the following rules: 3.49/1.80 3.49/1.80 member_in_ag(X, .(X, X1)) -> member_out_ag(X, .(X, X1)) 3.49/1.80 member_in_ag(X, .(X2, Xs)) -> U1_ag(X, X2, Xs, member_in_ag(X, Xs)) 3.49/1.80 U1_ag(X, X2, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(X2, Xs)) 3.49/1.80 3.49/1.80 The argument filtering Pi contains the following mapping: 3.49/1.80 member_in_ag(x1, x2) = member_in_ag(x2) 3.49/1.80 3.49/1.80 .(x1, x2) = .(x1, x2) 3.49/1.80 3.49/1.80 member_out_ag(x1, x2) = member_out_ag(x1) 3.49/1.80 3.49/1.80 U1_ag(x1, x2, x3, x4) = U1_ag(x4) 3.49/1.80 3.49/1.80 MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2) 3.49/1.80 3.49/1.80 U1_AG(x1, x2, x3, x4) = U1_AG(x4) 3.49/1.80 3.49/1.80 3.49/1.80 We have to consider all (P,R,Pi)-chains 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (5) DependencyGraphProof (EQUIVALENT) 3.49/1.80 The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node. 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (6) 3.49/1.80 Obligation: 3.49/1.80 Pi DP problem: 3.49/1.80 The TRS P consists of the following rules: 3.49/1.80 3.49/1.80 MEMBER_IN_AG(X, .(X2, Xs)) -> MEMBER_IN_AG(X, Xs) 3.49/1.80 3.49/1.80 The TRS R consists of the following rules: 3.49/1.80 3.49/1.80 member_in_ag(X, .(X, X1)) -> member_out_ag(X, .(X, X1)) 3.49/1.80 member_in_ag(X, .(X2, Xs)) -> U1_ag(X, X2, Xs, member_in_ag(X, Xs)) 3.49/1.80 U1_ag(X, X2, Xs, member_out_ag(X, Xs)) -> member_out_ag(X, .(X2, Xs)) 3.49/1.80 3.49/1.80 The argument filtering Pi contains the following mapping: 3.49/1.80 member_in_ag(x1, x2) = member_in_ag(x2) 3.49/1.80 3.49/1.80 .(x1, x2) = .(x1, x2) 3.49/1.80 3.49/1.80 member_out_ag(x1, x2) = member_out_ag(x1) 3.49/1.80 3.49/1.80 U1_ag(x1, x2, x3, x4) = U1_ag(x4) 3.49/1.80 3.49/1.80 MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2) 3.49/1.80 3.49/1.80 3.49/1.80 We have to consider all (P,R,Pi)-chains 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (7) UsableRulesProof (EQUIVALENT) 3.49/1.80 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (8) 3.49/1.80 Obligation: 3.49/1.80 Pi DP problem: 3.49/1.80 The TRS P consists of the following rules: 3.49/1.80 3.49/1.80 MEMBER_IN_AG(X, .(X2, Xs)) -> MEMBER_IN_AG(X, Xs) 3.49/1.80 3.49/1.80 R is empty. 3.49/1.80 The argument filtering Pi contains the following mapping: 3.49/1.80 .(x1, x2) = .(x1, x2) 3.49/1.80 3.49/1.80 MEMBER_IN_AG(x1, x2) = MEMBER_IN_AG(x2) 3.49/1.80 3.49/1.80 3.49/1.80 We have to consider all (P,R,Pi)-chains 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (9) PiDPToQDPProof (SOUND) 3.49/1.80 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (10) 3.49/1.80 Obligation: 3.49/1.80 Q DP problem: 3.49/1.80 The TRS P consists of the following rules: 3.49/1.80 3.49/1.80 MEMBER_IN_AG(.(X2, Xs)) -> MEMBER_IN_AG(Xs) 3.49/1.80 3.49/1.80 R is empty. 3.49/1.80 Q is empty. 3.49/1.80 We have to consider all (P,Q,R)-chains. 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (11) QDPSizeChangeProof (EQUIVALENT) 3.49/1.80 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. 3.49/1.80 3.49/1.80 From the DPs we obtained the following set of size-change graphs: 3.49/1.80 *MEMBER_IN_AG(.(X2, Xs)) -> MEMBER_IN_AG(Xs) 3.49/1.80 The graph contains the following edges 1 > 1 3.49/1.80 3.49/1.80 3.49/1.80 ---------------------------------------- 3.49/1.80 3.49/1.80 (12) 3.49/1.80 YES 3.67/1.83 EOF