3.91/1.81 YES 3.98/1.90 proof of /export/starexec/sandbox/benchmark/theBenchmark.pl 3.98/1.90 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.98/1.90 3.98/1.90 3.98/1.90 Left Termination of the query pattern 3.98/1.90 3.98/1.90 subset(g,g) 3.98/1.90 3.98/1.90 w.r.t. the given Prolog program could successfully be proven: 3.98/1.90 3.98/1.90 (0) Prolog 3.98/1.90 (1) PrologToPiTRSProof [SOUND, 0 ms] 3.98/1.90 (2) PiTRS 3.98/1.90 (3) DependencyPairsProof [EQUIVALENT, 0 ms] 3.98/1.90 (4) PiDP 3.98/1.90 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 3.98/1.90 (6) AND 3.98/1.90 (7) PiDP 3.98/1.90 (8) UsableRulesProof [EQUIVALENT, 0 ms] 3.98/1.90 (9) PiDP 3.98/1.90 (10) PiDPToQDPProof [EQUIVALENT, 0 ms] 3.98/1.90 (11) QDP 3.98/1.90 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 3.98/1.90 (13) YES 3.98/1.90 (14) PiDP 3.98/1.90 (15) UsableRulesProof [EQUIVALENT, 0 ms] 3.98/1.90 (16) PiDP 3.98/1.90 (17) PiDPToQDPProof [SOUND, 0 ms] 3.98/1.90 (18) QDP 3.98/1.90 (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] 3.98/1.90 (20) YES 3.98/1.90 3.98/1.90 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (0) 3.98/1.90 Obligation: 3.98/1.90 Clauses: 3.98/1.90 3.98/1.90 member(X, .(Y, Xs)) :- member(X, Xs). 3.98/1.90 member(X, .(X, Xs)). 3.98/1.90 subset(.(X, Xs), Ys) :- ','(member(X, Ys), subset(Xs, Ys)). 3.98/1.90 subset([], Ys). 3.98/1.90 3.98/1.90 3.98/1.90 Query: subset(g,g) 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (1) PrologToPiTRSProof (SOUND) 3.98/1.90 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 3.98/1.90 3.98/1.90 subset_in_2: (b,b) 3.98/1.90 3.98/1.90 member_in_2: (b,b) 3.98/1.90 3.98/1.90 Transforming Prolog into the following Term Rewriting System: 3.98/1.90 3.98/1.90 Pi-finite rewrite system: 3.98/1.90 The TRS R consists of the following rules: 3.98/1.90 3.98/1.90 subset_in_gg(.(X, Xs), Ys) -> U2_gg(X, Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 member_in_gg(X, .(Y, Xs)) -> U1_gg(X, Y, Xs, member_in_gg(X, Xs)) 3.98/1.90 member_in_gg(X, .(X, Xs)) -> member_out_gg(X, .(X, Xs)) 3.98/1.90 U1_gg(X, Y, Xs, member_out_gg(X, Xs)) -> member_out_gg(X, .(Y, Xs)) 3.98/1.90 U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) -> U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys)) 3.98/1.90 subset_in_gg([], Ys) -> subset_out_gg([], Ys) 3.98/1.90 U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) -> subset_out_gg(.(X, Xs), Ys) 3.98/1.90 3.98/1.90 The argument filtering Pi contains the following mapping: 3.98/1.90 subset_in_gg(x1, x2) = subset_in_gg(x1, x2) 3.98/1.90 3.98/1.90 .(x1, x2) = .(x1, x2) 3.98/1.90 3.98/1.90 U2_gg(x1, x2, x3, x4) = U2_gg(x2, x3, x4) 3.98/1.90 3.98/1.90 member_in_gg(x1, x2) = member_in_gg(x1, x2) 3.98/1.90 3.98/1.90 U1_gg(x1, x2, x3, x4) = U1_gg(x4) 3.98/1.90 3.98/1.90 member_out_gg(x1, x2) = member_out_gg 3.98/1.90 3.98/1.90 U3_gg(x1, x2, x3, x4) = U3_gg(x4) 3.98/1.90 3.98/1.90 [] = [] 3.98/1.90 3.98/1.90 subset_out_gg(x1, x2) = subset_out_gg 3.98/1.90 3.98/1.90 3.98/1.90 3.98/1.90 3.98/1.90 3.98/1.90 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 3.98/1.90 3.98/1.90 3.98/1.90 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (2) 3.98/1.90 Obligation: 3.98/1.90 Pi-finite rewrite system: 3.98/1.90 The TRS R consists of the following rules: 3.98/1.90 3.98/1.90 subset_in_gg(.(X, Xs), Ys) -> U2_gg(X, Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 member_in_gg(X, .(Y, Xs)) -> U1_gg(X, Y, Xs, member_in_gg(X, Xs)) 3.98/1.90 member_in_gg(X, .(X, Xs)) -> member_out_gg(X, .(X, Xs)) 3.98/1.90 U1_gg(X, Y, Xs, member_out_gg(X, Xs)) -> member_out_gg(X, .(Y, Xs)) 3.98/1.90 U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) -> U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys)) 3.98/1.90 subset_in_gg([], Ys) -> subset_out_gg([], Ys) 3.98/1.90 U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) -> subset_out_gg(.(X, Xs), Ys) 3.98/1.90 3.98/1.90 The argument filtering Pi contains the following mapping: 3.98/1.90 subset_in_gg(x1, x2) = subset_in_gg(x1, x2) 3.98/1.90 3.98/1.90 .(x1, x2) = .(x1, x2) 3.98/1.90 3.98/1.90 U2_gg(x1, x2, x3, x4) = U2_gg(x2, x3, x4) 3.98/1.90 3.98/1.90 member_in_gg(x1, x2) = member_in_gg(x1, x2) 3.98/1.90 3.98/1.90 U1_gg(x1, x2, x3, x4) = U1_gg(x4) 3.98/1.90 3.98/1.90 member_out_gg(x1, x2) = member_out_gg 3.98/1.90 3.98/1.90 U3_gg(x1, x2, x3, x4) = U3_gg(x4) 3.98/1.90 3.98/1.90 [] = [] 3.98/1.90 3.98/1.90 subset_out_gg(x1, x2) = subset_out_gg 3.98/1.90 3.98/1.90 3.98/1.90 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (3) DependencyPairsProof (EQUIVALENT) 3.98/1.90 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 3.98/1.90 Pi DP problem: 3.98/1.90 The TRS P consists of the following rules: 3.98/1.90 3.98/1.90 SUBSET_IN_GG(.(X, Xs), Ys) -> U2_GG(X, Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 SUBSET_IN_GG(.(X, Xs), Ys) -> MEMBER_IN_GG(X, Ys) 3.98/1.90 MEMBER_IN_GG(X, .(Y, Xs)) -> U1_GG(X, Y, Xs, member_in_gg(X, Xs)) 3.98/1.90 MEMBER_IN_GG(X, .(Y, Xs)) -> MEMBER_IN_GG(X, Xs) 3.98/1.90 U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) -> U3_GG(X, Xs, Ys, subset_in_gg(Xs, Ys)) 3.98/1.90 U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) -> SUBSET_IN_GG(Xs, Ys) 3.98/1.90 3.98/1.90 The TRS R consists of the following rules: 3.98/1.90 3.98/1.90 subset_in_gg(.(X, Xs), Ys) -> U2_gg(X, Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 member_in_gg(X, .(Y, Xs)) -> U1_gg(X, Y, Xs, member_in_gg(X, Xs)) 3.98/1.90 member_in_gg(X, .(X, Xs)) -> member_out_gg(X, .(X, Xs)) 3.98/1.90 U1_gg(X, Y, Xs, member_out_gg(X, Xs)) -> member_out_gg(X, .(Y, Xs)) 3.98/1.90 U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) -> U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys)) 3.98/1.90 subset_in_gg([], Ys) -> subset_out_gg([], Ys) 3.98/1.90 U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) -> subset_out_gg(.(X, Xs), Ys) 3.98/1.90 3.98/1.90 The argument filtering Pi contains the following mapping: 3.98/1.90 subset_in_gg(x1, x2) = subset_in_gg(x1, x2) 3.98/1.90 3.98/1.90 .(x1, x2) = .(x1, x2) 3.98/1.90 3.98/1.90 U2_gg(x1, x2, x3, x4) = U2_gg(x2, x3, x4) 3.98/1.90 3.98/1.90 member_in_gg(x1, x2) = member_in_gg(x1, x2) 3.98/1.90 3.98/1.90 U1_gg(x1, x2, x3, x4) = U1_gg(x4) 3.98/1.90 3.98/1.90 member_out_gg(x1, x2) = member_out_gg 3.98/1.90 3.98/1.90 U3_gg(x1, x2, x3, x4) = U3_gg(x4) 3.98/1.90 3.98/1.90 [] = [] 3.98/1.90 3.98/1.90 subset_out_gg(x1, x2) = subset_out_gg 3.98/1.90 3.98/1.90 SUBSET_IN_GG(x1, x2) = SUBSET_IN_GG(x1, x2) 3.98/1.90 3.98/1.90 U2_GG(x1, x2, x3, x4) = U2_GG(x2, x3, x4) 3.98/1.90 3.98/1.90 MEMBER_IN_GG(x1, x2) = MEMBER_IN_GG(x1, x2) 3.98/1.90 3.98/1.90 U1_GG(x1, x2, x3, x4) = U1_GG(x4) 3.98/1.90 3.98/1.90 U3_GG(x1, x2, x3, x4) = U3_GG(x4) 3.98/1.90 3.98/1.90 3.98/1.90 We have to consider all (P,R,Pi)-chains 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (4) 3.98/1.90 Obligation: 3.98/1.90 Pi DP problem: 3.98/1.90 The TRS P consists of the following rules: 3.98/1.90 3.98/1.90 SUBSET_IN_GG(.(X, Xs), Ys) -> U2_GG(X, Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 SUBSET_IN_GG(.(X, Xs), Ys) -> MEMBER_IN_GG(X, Ys) 3.98/1.90 MEMBER_IN_GG(X, .(Y, Xs)) -> U1_GG(X, Y, Xs, member_in_gg(X, Xs)) 3.98/1.90 MEMBER_IN_GG(X, .(Y, Xs)) -> MEMBER_IN_GG(X, Xs) 3.98/1.90 U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) -> U3_GG(X, Xs, Ys, subset_in_gg(Xs, Ys)) 3.98/1.90 U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) -> SUBSET_IN_GG(Xs, Ys) 3.98/1.90 3.98/1.90 The TRS R consists of the following rules: 3.98/1.90 3.98/1.90 subset_in_gg(.(X, Xs), Ys) -> U2_gg(X, Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 member_in_gg(X, .(Y, Xs)) -> U1_gg(X, Y, Xs, member_in_gg(X, Xs)) 3.98/1.90 member_in_gg(X, .(X, Xs)) -> member_out_gg(X, .(X, Xs)) 3.98/1.90 U1_gg(X, Y, Xs, member_out_gg(X, Xs)) -> member_out_gg(X, .(Y, Xs)) 3.98/1.90 U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) -> U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys)) 3.98/1.90 subset_in_gg([], Ys) -> subset_out_gg([], Ys) 3.98/1.90 U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) -> subset_out_gg(.(X, Xs), Ys) 3.98/1.90 3.98/1.90 The argument filtering Pi contains the following mapping: 3.98/1.90 subset_in_gg(x1, x2) = subset_in_gg(x1, x2) 3.98/1.90 3.98/1.90 .(x1, x2) = .(x1, x2) 3.98/1.90 3.98/1.90 U2_gg(x1, x2, x3, x4) = U2_gg(x2, x3, x4) 3.98/1.90 3.98/1.90 member_in_gg(x1, x2) = member_in_gg(x1, x2) 3.98/1.90 3.98/1.90 U1_gg(x1, x2, x3, x4) = U1_gg(x4) 3.98/1.90 3.98/1.90 member_out_gg(x1, x2) = member_out_gg 3.98/1.90 3.98/1.90 U3_gg(x1, x2, x3, x4) = U3_gg(x4) 3.98/1.90 3.98/1.90 [] = [] 3.98/1.90 3.98/1.90 subset_out_gg(x1, x2) = subset_out_gg 3.98/1.90 3.98/1.90 SUBSET_IN_GG(x1, x2) = SUBSET_IN_GG(x1, x2) 3.98/1.90 3.98/1.90 U2_GG(x1, x2, x3, x4) = U2_GG(x2, x3, x4) 3.98/1.90 3.98/1.90 MEMBER_IN_GG(x1, x2) = MEMBER_IN_GG(x1, x2) 3.98/1.90 3.98/1.90 U1_GG(x1, x2, x3, x4) = U1_GG(x4) 3.98/1.90 3.98/1.90 U3_GG(x1, x2, x3, x4) = U3_GG(x4) 3.98/1.90 3.98/1.90 3.98/1.90 We have to consider all (P,R,Pi)-chains 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (5) DependencyGraphProof (EQUIVALENT) 3.98/1.90 The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes. 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (6) 3.98/1.90 Complex Obligation (AND) 3.98/1.90 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (7) 3.98/1.90 Obligation: 3.98/1.90 Pi DP problem: 3.98/1.90 The TRS P consists of the following rules: 3.98/1.90 3.98/1.90 MEMBER_IN_GG(X, .(Y, Xs)) -> MEMBER_IN_GG(X, Xs) 3.98/1.90 3.98/1.90 The TRS R consists of the following rules: 3.98/1.90 3.98/1.90 subset_in_gg(.(X, Xs), Ys) -> U2_gg(X, Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 member_in_gg(X, .(Y, Xs)) -> U1_gg(X, Y, Xs, member_in_gg(X, Xs)) 3.98/1.90 member_in_gg(X, .(X, Xs)) -> member_out_gg(X, .(X, Xs)) 3.98/1.90 U1_gg(X, Y, Xs, member_out_gg(X, Xs)) -> member_out_gg(X, .(Y, Xs)) 3.98/1.90 U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) -> U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys)) 3.98/1.90 subset_in_gg([], Ys) -> subset_out_gg([], Ys) 3.98/1.90 U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) -> subset_out_gg(.(X, Xs), Ys) 3.98/1.90 3.98/1.90 The argument filtering Pi contains the following mapping: 3.98/1.90 subset_in_gg(x1, x2) = subset_in_gg(x1, x2) 3.98/1.90 3.98/1.90 .(x1, x2) = .(x1, x2) 3.98/1.90 3.98/1.90 U2_gg(x1, x2, x3, x4) = U2_gg(x2, x3, x4) 3.98/1.90 3.98/1.90 member_in_gg(x1, x2) = member_in_gg(x1, x2) 3.98/1.90 3.98/1.90 U1_gg(x1, x2, x3, x4) = U1_gg(x4) 3.98/1.90 3.98/1.90 member_out_gg(x1, x2) = member_out_gg 3.98/1.90 3.98/1.90 U3_gg(x1, x2, x3, x4) = U3_gg(x4) 3.98/1.90 3.98/1.90 [] = [] 3.98/1.90 3.98/1.90 subset_out_gg(x1, x2) = subset_out_gg 3.98/1.90 3.98/1.90 MEMBER_IN_GG(x1, x2) = MEMBER_IN_GG(x1, x2) 3.98/1.90 3.98/1.90 3.98/1.90 We have to consider all (P,R,Pi)-chains 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (8) UsableRulesProof (EQUIVALENT) 3.98/1.90 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (9) 3.98/1.90 Obligation: 3.98/1.90 Pi DP problem: 3.98/1.90 The TRS P consists of the following rules: 3.98/1.90 3.98/1.90 MEMBER_IN_GG(X, .(Y, Xs)) -> MEMBER_IN_GG(X, Xs) 3.98/1.90 3.98/1.90 R is empty. 3.98/1.90 Pi is empty. 3.98/1.90 We have to consider all (P,R,Pi)-chains 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (10) PiDPToQDPProof (EQUIVALENT) 3.98/1.90 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (11) 3.98/1.90 Obligation: 3.98/1.90 Q DP problem: 3.98/1.90 The TRS P consists of the following rules: 3.98/1.90 3.98/1.90 MEMBER_IN_GG(X, .(Y, Xs)) -> MEMBER_IN_GG(X, Xs) 3.98/1.90 3.98/1.90 R is empty. 3.98/1.90 Q is empty. 3.98/1.90 We have to consider all (P,Q,R)-chains. 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (12) QDPSizeChangeProof (EQUIVALENT) 3.98/1.90 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. 3.98/1.90 3.98/1.90 From the DPs we obtained the following set of size-change graphs: 3.98/1.90 *MEMBER_IN_GG(X, .(Y, Xs)) -> MEMBER_IN_GG(X, Xs) 3.98/1.90 The graph contains the following edges 1 >= 1, 2 > 2 3.98/1.90 3.98/1.90 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (13) 3.98/1.90 YES 3.98/1.90 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (14) 3.98/1.90 Obligation: 3.98/1.90 Pi DP problem: 3.98/1.90 The TRS P consists of the following rules: 3.98/1.90 3.98/1.90 U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) -> SUBSET_IN_GG(Xs, Ys) 3.98/1.90 SUBSET_IN_GG(.(X, Xs), Ys) -> U2_GG(X, Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 3.98/1.90 The TRS R consists of the following rules: 3.98/1.90 3.98/1.90 subset_in_gg(.(X, Xs), Ys) -> U2_gg(X, Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 member_in_gg(X, .(Y, Xs)) -> U1_gg(X, Y, Xs, member_in_gg(X, Xs)) 3.98/1.90 member_in_gg(X, .(X, Xs)) -> member_out_gg(X, .(X, Xs)) 3.98/1.90 U1_gg(X, Y, Xs, member_out_gg(X, Xs)) -> member_out_gg(X, .(Y, Xs)) 3.98/1.90 U2_gg(X, Xs, Ys, member_out_gg(X, Ys)) -> U3_gg(X, Xs, Ys, subset_in_gg(Xs, Ys)) 3.98/1.90 subset_in_gg([], Ys) -> subset_out_gg([], Ys) 3.98/1.90 U3_gg(X, Xs, Ys, subset_out_gg(Xs, Ys)) -> subset_out_gg(.(X, Xs), Ys) 3.98/1.90 3.98/1.90 The argument filtering Pi contains the following mapping: 3.98/1.90 subset_in_gg(x1, x2) = subset_in_gg(x1, x2) 3.98/1.90 3.98/1.90 .(x1, x2) = .(x1, x2) 3.98/1.90 3.98/1.90 U2_gg(x1, x2, x3, x4) = U2_gg(x2, x3, x4) 3.98/1.90 3.98/1.90 member_in_gg(x1, x2) = member_in_gg(x1, x2) 3.98/1.90 3.98/1.90 U1_gg(x1, x2, x3, x4) = U1_gg(x4) 3.98/1.90 3.98/1.90 member_out_gg(x1, x2) = member_out_gg 3.98/1.90 3.98/1.90 U3_gg(x1, x2, x3, x4) = U3_gg(x4) 3.98/1.90 3.98/1.90 [] = [] 3.98/1.90 3.98/1.90 subset_out_gg(x1, x2) = subset_out_gg 3.98/1.90 3.98/1.90 SUBSET_IN_GG(x1, x2) = SUBSET_IN_GG(x1, x2) 3.98/1.90 3.98/1.90 U2_GG(x1, x2, x3, x4) = U2_GG(x2, x3, x4) 3.98/1.90 3.98/1.90 3.98/1.90 We have to consider all (P,R,Pi)-chains 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (15) UsableRulesProof (EQUIVALENT) 3.98/1.90 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (16) 3.98/1.90 Obligation: 3.98/1.90 Pi DP problem: 3.98/1.90 The TRS P consists of the following rules: 3.98/1.90 3.98/1.90 U2_GG(X, Xs, Ys, member_out_gg(X, Ys)) -> SUBSET_IN_GG(Xs, Ys) 3.98/1.90 SUBSET_IN_GG(.(X, Xs), Ys) -> U2_GG(X, Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 3.98/1.90 The TRS R consists of the following rules: 3.98/1.90 3.98/1.90 member_in_gg(X, .(Y, Xs)) -> U1_gg(X, Y, Xs, member_in_gg(X, Xs)) 3.98/1.90 member_in_gg(X, .(X, Xs)) -> member_out_gg(X, .(X, Xs)) 3.98/1.90 U1_gg(X, Y, Xs, member_out_gg(X, Xs)) -> member_out_gg(X, .(Y, Xs)) 3.98/1.90 3.98/1.90 The argument filtering Pi contains the following mapping: 3.98/1.90 .(x1, x2) = .(x1, x2) 3.98/1.90 3.98/1.90 member_in_gg(x1, x2) = member_in_gg(x1, x2) 3.98/1.90 3.98/1.90 U1_gg(x1, x2, x3, x4) = U1_gg(x4) 3.98/1.90 3.98/1.90 member_out_gg(x1, x2) = member_out_gg 3.98/1.90 3.98/1.90 SUBSET_IN_GG(x1, x2) = SUBSET_IN_GG(x1, x2) 3.98/1.90 3.98/1.90 U2_GG(x1, x2, x3, x4) = U2_GG(x2, x3, x4) 3.98/1.90 3.98/1.90 3.98/1.90 We have to consider all (P,R,Pi)-chains 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (17) PiDPToQDPProof (SOUND) 3.98/1.90 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (18) 3.98/1.90 Obligation: 3.98/1.90 Q DP problem: 3.98/1.90 The TRS P consists of the following rules: 3.98/1.90 3.98/1.90 U2_GG(Xs, Ys, member_out_gg) -> SUBSET_IN_GG(Xs, Ys) 3.98/1.90 SUBSET_IN_GG(.(X, Xs), Ys) -> U2_GG(Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 3.98/1.90 The TRS R consists of the following rules: 3.98/1.90 3.98/1.90 member_in_gg(X, .(Y, Xs)) -> U1_gg(member_in_gg(X, Xs)) 3.98/1.90 member_in_gg(X, .(X, Xs)) -> member_out_gg 3.98/1.90 U1_gg(member_out_gg) -> member_out_gg 3.98/1.90 3.98/1.90 The set Q consists of the following terms: 3.98/1.90 3.98/1.90 member_in_gg(x0, x1) 3.98/1.90 U1_gg(x0) 3.98/1.90 3.98/1.90 We have to consider all (P,Q,R)-chains. 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (19) QDPSizeChangeProof (EQUIVALENT) 3.98/1.90 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. 3.98/1.90 3.98/1.90 From the DPs we obtained the following set of size-change graphs: 3.98/1.90 *SUBSET_IN_GG(.(X, Xs), Ys) -> U2_GG(Xs, Ys, member_in_gg(X, Ys)) 3.98/1.90 The graph contains the following edges 1 > 1, 2 >= 2 3.98/1.90 3.98/1.90 3.98/1.90 *U2_GG(Xs, Ys, member_out_gg) -> SUBSET_IN_GG(Xs, Ys) 3.98/1.90 The graph contains the following edges 1 >= 1, 2 >= 2 3.98/1.90 3.98/1.90 3.98/1.90 ---------------------------------------- 3.98/1.90 3.98/1.90 (20) 3.98/1.90 YES 4.24/1.92 EOF