3.94/1.78 YES 3.99/1.79 proof of /export/starexec/sandbox/benchmark/theBenchmark.pl 3.99/1.79 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.99/1.79 3.99/1.79 3.99/1.79 Left Termination of the query pattern 3.99/1.79 3.99/1.79 tree_member(a,g) 3.99/1.79 3.99/1.79 w.r.t. the given Prolog program could successfully be proven: 3.99/1.79 3.99/1.79 (0) Prolog 3.99/1.79 (1) PrologToPiTRSProof [SOUND, 0 ms] 3.99/1.79 (2) PiTRS 3.99/1.79 (3) DependencyPairsProof [EQUIVALENT, 0 ms] 3.99/1.79 (4) PiDP 3.99/1.79 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 3.99/1.79 (6) PiDP 3.99/1.79 (7) UsableRulesProof [EQUIVALENT, 0 ms] 3.99/1.79 (8) PiDP 3.99/1.79 (9) PiDPToQDPProof [SOUND, 0 ms] 3.99/1.79 (10) QDP 3.99/1.79 (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] 3.99/1.79 (12) YES 3.99/1.79 3.99/1.79 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (0) 3.99/1.79 Obligation: 3.99/1.79 Clauses: 3.99/1.79 3.99/1.79 tree_member(X, tree(X, X1, X2)). 3.99/1.79 tree_member(X, tree(X3, Left, X4)) :- tree_member(X, Left). 3.99/1.79 tree_member(X, tree(X5, X6, Right)) :- tree_member(X, Right). 3.99/1.79 3.99/1.79 3.99/1.79 Query: tree_member(a,g) 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (1) PrologToPiTRSProof (SOUND) 3.99/1.79 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 3.99/1.79 3.99/1.79 tree_member_in_2: (f,b) 3.99/1.79 3.99/1.79 Transforming Prolog into the following Term Rewriting System: 3.99/1.79 3.99/1.79 Pi-finite rewrite system: 3.99/1.79 The TRS R consists of the following rules: 3.99/1.79 3.99/1.79 tree_member_in_ag(X, tree(X, X1, X2)) -> tree_member_out_ag(X, tree(X, X1, X2)) 3.99/1.79 tree_member_in_ag(X, tree(X3, Left, X4)) -> U1_ag(X, X3, Left, X4, tree_member_in_ag(X, Left)) 3.99/1.79 tree_member_in_ag(X, tree(X5, X6, Right)) -> U2_ag(X, X5, X6, Right, tree_member_in_ag(X, Right)) 3.99/1.79 U2_ag(X, X5, X6, Right, tree_member_out_ag(X, Right)) -> tree_member_out_ag(X, tree(X5, X6, Right)) 3.99/1.79 U1_ag(X, X3, Left, X4, tree_member_out_ag(X, Left)) -> tree_member_out_ag(X, tree(X3, Left, X4)) 3.99/1.79 3.99/1.79 The argument filtering Pi contains the following mapping: 3.99/1.79 tree_member_in_ag(x1, x2) = tree_member_in_ag(x2) 3.99/1.79 3.99/1.79 tree(x1, x2, x3) = tree(x1, x2, x3) 3.99/1.79 3.99/1.79 tree_member_out_ag(x1, x2) = tree_member_out_ag(x1) 3.99/1.79 3.99/1.79 U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) 3.99/1.79 3.99/1.79 U2_ag(x1, x2, x3, x4, x5) = U2_ag(x5) 3.99/1.79 3.99/1.79 3.99/1.79 3.99/1.79 3.99/1.79 3.99/1.79 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 3.99/1.79 3.99/1.79 3.99/1.79 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (2) 3.99/1.79 Obligation: 3.99/1.79 Pi-finite rewrite system: 3.99/1.79 The TRS R consists of the following rules: 3.99/1.79 3.99/1.79 tree_member_in_ag(X, tree(X, X1, X2)) -> tree_member_out_ag(X, tree(X, X1, X2)) 3.99/1.79 tree_member_in_ag(X, tree(X3, Left, X4)) -> U1_ag(X, X3, Left, X4, tree_member_in_ag(X, Left)) 3.99/1.79 tree_member_in_ag(X, tree(X5, X6, Right)) -> U2_ag(X, X5, X6, Right, tree_member_in_ag(X, Right)) 3.99/1.79 U2_ag(X, X5, X6, Right, tree_member_out_ag(X, Right)) -> tree_member_out_ag(X, tree(X5, X6, Right)) 3.99/1.79 U1_ag(X, X3, Left, X4, tree_member_out_ag(X, Left)) -> tree_member_out_ag(X, tree(X3, Left, X4)) 3.99/1.79 3.99/1.79 The argument filtering Pi contains the following mapping: 3.99/1.79 tree_member_in_ag(x1, x2) = tree_member_in_ag(x2) 3.99/1.79 3.99/1.79 tree(x1, x2, x3) = tree(x1, x2, x3) 3.99/1.79 3.99/1.79 tree_member_out_ag(x1, x2) = tree_member_out_ag(x1) 3.99/1.79 3.99/1.79 U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) 3.99/1.79 3.99/1.79 U2_ag(x1, x2, x3, x4, x5) = U2_ag(x5) 3.99/1.79 3.99/1.79 3.99/1.79 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (3) DependencyPairsProof (EQUIVALENT) 3.99/1.79 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 3.99/1.79 Pi DP problem: 3.99/1.79 The TRS P consists of the following rules: 3.99/1.79 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) -> U1_AG(X, X3, Left, X4, tree_member_in_ag(X, Left)) 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) -> TREE_MEMBER_IN_AG(X, Left) 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) -> U2_AG(X, X5, X6, Right, tree_member_in_ag(X, Right)) 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) -> TREE_MEMBER_IN_AG(X, Right) 3.99/1.79 3.99/1.79 The TRS R consists of the following rules: 3.99/1.79 3.99/1.79 tree_member_in_ag(X, tree(X, X1, X2)) -> tree_member_out_ag(X, tree(X, X1, X2)) 3.99/1.79 tree_member_in_ag(X, tree(X3, Left, X4)) -> U1_ag(X, X3, Left, X4, tree_member_in_ag(X, Left)) 3.99/1.79 tree_member_in_ag(X, tree(X5, X6, Right)) -> U2_ag(X, X5, X6, Right, tree_member_in_ag(X, Right)) 3.99/1.79 U2_ag(X, X5, X6, Right, tree_member_out_ag(X, Right)) -> tree_member_out_ag(X, tree(X5, X6, Right)) 3.99/1.79 U1_ag(X, X3, Left, X4, tree_member_out_ag(X, Left)) -> tree_member_out_ag(X, tree(X3, Left, X4)) 3.99/1.79 3.99/1.79 The argument filtering Pi contains the following mapping: 3.99/1.79 tree_member_in_ag(x1, x2) = tree_member_in_ag(x2) 3.99/1.79 3.99/1.79 tree(x1, x2, x3) = tree(x1, x2, x3) 3.99/1.79 3.99/1.79 tree_member_out_ag(x1, x2) = tree_member_out_ag(x1) 3.99/1.79 3.99/1.79 U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) 3.99/1.79 3.99/1.79 U2_ag(x1, x2, x3, x4, x5) = U2_ag(x5) 3.99/1.79 3.99/1.79 TREE_MEMBER_IN_AG(x1, x2) = TREE_MEMBER_IN_AG(x2) 3.99/1.79 3.99/1.79 U1_AG(x1, x2, x3, x4, x5) = U1_AG(x5) 3.99/1.79 3.99/1.79 U2_AG(x1, x2, x3, x4, x5) = U2_AG(x5) 3.99/1.79 3.99/1.79 3.99/1.79 We have to consider all (P,R,Pi)-chains 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (4) 3.99/1.79 Obligation: 3.99/1.79 Pi DP problem: 3.99/1.79 The TRS P consists of the following rules: 3.99/1.79 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) -> U1_AG(X, X3, Left, X4, tree_member_in_ag(X, Left)) 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) -> TREE_MEMBER_IN_AG(X, Left) 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) -> U2_AG(X, X5, X6, Right, tree_member_in_ag(X, Right)) 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) -> TREE_MEMBER_IN_AG(X, Right) 3.99/1.79 3.99/1.79 The TRS R consists of the following rules: 3.99/1.79 3.99/1.79 tree_member_in_ag(X, tree(X, X1, X2)) -> tree_member_out_ag(X, tree(X, X1, X2)) 3.99/1.79 tree_member_in_ag(X, tree(X3, Left, X4)) -> U1_ag(X, X3, Left, X4, tree_member_in_ag(X, Left)) 3.99/1.79 tree_member_in_ag(X, tree(X5, X6, Right)) -> U2_ag(X, X5, X6, Right, tree_member_in_ag(X, Right)) 3.99/1.79 U2_ag(X, X5, X6, Right, tree_member_out_ag(X, Right)) -> tree_member_out_ag(X, tree(X5, X6, Right)) 3.99/1.79 U1_ag(X, X3, Left, X4, tree_member_out_ag(X, Left)) -> tree_member_out_ag(X, tree(X3, Left, X4)) 3.99/1.79 3.99/1.79 The argument filtering Pi contains the following mapping: 3.99/1.79 tree_member_in_ag(x1, x2) = tree_member_in_ag(x2) 3.99/1.79 3.99/1.79 tree(x1, x2, x3) = tree(x1, x2, x3) 3.99/1.79 3.99/1.79 tree_member_out_ag(x1, x2) = tree_member_out_ag(x1) 3.99/1.79 3.99/1.79 U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) 3.99/1.79 3.99/1.79 U2_ag(x1, x2, x3, x4, x5) = U2_ag(x5) 3.99/1.79 3.99/1.79 TREE_MEMBER_IN_AG(x1, x2) = TREE_MEMBER_IN_AG(x2) 3.99/1.79 3.99/1.79 U1_AG(x1, x2, x3, x4, x5) = U1_AG(x5) 3.99/1.79 3.99/1.79 U2_AG(x1, x2, x3, x4, x5) = U2_AG(x5) 3.99/1.79 3.99/1.79 3.99/1.79 We have to consider all (P,R,Pi)-chains 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (5) DependencyGraphProof (EQUIVALENT) 3.99/1.79 The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes. 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (6) 3.99/1.79 Obligation: 3.99/1.79 Pi DP problem: 3.99/1.79 The TRS P consists of the following rules: 3.99/1.79 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) -> TREE_MEMBER_IN_AG(X, Right) 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) -> TREE_MEMBER_IN_AG(X, Left) 3.99/1.79 3.99/1.79 The TRS R consists of the following rules: 3.99/1.79 3.99/1.79 tree_member_in_ag(X, tree(X, X1, X2)) -> tree_member_out_ag(X, tree(X, X1, X2)) 3.99/1.79 tree_member_in_ag(X, tree(X3, Left, X4)) -> U1_ag(X, X3, Left, X4, tree_member_in_ag(X, Left)) 3.99/1.79 tree_member_in_ag(X, tree(X5, X6, Right)) -> U2_ag(X, X5, X6, Right, tree_member_in_ag(X, Right)) 3.99/1.79 U2_ag(X, X5, X6, Right, tree_member_out_ag(X, Right)) -> tree_member_out_ag(X, tree(X5, X6, Right)) 3.99/1.79 U1_ag(X, X3, Left, X4, tree_member_out_ag(X, Left)) -> tree_member_out_ag(X, tree(X3, Left, X4)) 3.99/1.79 3.99/1.79 The argument filtering Pi contains the following mapping: 3.99/1.79 tree_member_in_ag(x1, x2) = tree_member_in_ag(x2) 3.99/1.79 3.99/1.79 tree(x1, x2, x3) = tree(x1, x2, x3) 3.99/1.79 3.99/1.79 tree_member_out_ag(x1, x2) = tree_member_out_ag(x1) 3.99/1.79 3.99/1.79 U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) 3.99/1.79 3.99/1.79 U2_ag(x1, x2, x3, x4, x5) = U2_ag(x5) 3.99/1.79 3.99/1.79 TREE_MEMBER_IN_AG(x1, x2) = TREE_MEMBER_IN_AG(x2) 3.99/1.79 3.99/1.79 3.99/1.79 We have to consider all (P,R,Pi)-chains 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (7) UsableRulesProof (EQUIVALENT) 3.99/1.79 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (8) 3.99/1.79 Obligation: 3.99/1.79 Pi DP problem: 3.99/1.79 The TRS P consists of the following rules: 3.99/1.79 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) -> TREE_MEMBER_IN_AG(X, Right) 3.99/1.79 TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) -> TREE_MEMBER_IN_AG(X, Left) 3.99/1.79 3.99/1.79 R is empty. 3.99/1.79 The argument filtering Pi contains the following mapping: 3.99/1.79 tree(x1, x2, x3) = tree(x1, x2, x3) 3.99/1.79 3.99/1.79 TREE_MEMBER_IN_AG(x1, x2) = TREE_MEMBER_IN_AG(x2) 3.99/1.79 3.99/1.79 3.99/1.79 We have to consider all (P,R,Pi)-chains 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (9) PiDPToQDPProof (SOUND) 3.99/1.79 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (10) 3.99/1.79 Obligation: 3.99/1.79 Q DP problem: 3.99/1.79 The TRS P consists of the following rules: 3.99/1.79 3.99/1.79 TREE_MEMBER_IN_AG(tree(X5, X6, Right)) -> TREE_MEMBER_IN_AG(Right) 3.99/1.79 TREE_MEMBER_IN_AG(tree(X3, Left, X4)) -> TREE_MEMBER_IN_AG(Left) 3.99/1.79 3.99/1.79 R is empty. 3.99/1.79 Q is empty. 3.99/1.79 We have to consider all (P,Q,R)-chains. 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (11) QDPSizeChangeProof (EQUIVALENT) 3.99/1.79 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. 3.99/1.79 3.99/1.79 From the DPs we obtained the following set of size-change graphs: 3.99/1.79 *TREE_MEMBER_IN_AG(tree(X5, X6, Right)) -> TREE_MEMBER_IN_AG(Right) 3.99/1.79 The graph contains the following edges 1 > 1 3.99/1.79 3.99/1.79 3.99/1.79 *TREE_MEMBER_IN_AG(tree(X3, Left, X4)) -> TREE_MEMBER_IN_AG(Left) 3.99/1.79 The graph contains the following edges 1 > 1 3.99/1.79 3.99/1.79 3.99/1.79 ---------------------------------------- 3.99/1.79 3.99/1.79 (12) 3.99/1.79 YES 3.99/1.82 EOF