5.15/2.23 YES 5.15/2.24 proof of /export/starexec/sandbox/benchmark/theBenchmark.pl 5.15/2.24 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 5.15/2.24 5.15/2.24 5.15/2.24 Left Termination of the query pattern 5.15/2.24 5.15/2.24 query(g) 5.15/2.24 5.15/2.24 w.r.t. the given Prolog program could successfully be proven: 5.15/2.24 5.15/2.24 (0) Prolog 5.15/2.24 (1) PrologToPiTRSProof [SOUND, 0 ms] 5.15/2.24 (2) PiTRS 5.15/2.24 (3) DependencyPairsProof [EQUIVALENT, 9 ms] 5.15/2.24 (4) PiDP 5.15/2.24 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 5.15/2.24 (6) AND 5.15/2.24 (7) PiDP 5.15/2.24 (8) UsableRulesProof [EQUIVALENT, 0 ms] 5.15/2.24 (9) PiDP 5.15/2.24 (10) PiDPToQDPProof [SOUND, 0 ms] 5.15/2.24 (11) QDP 5.15/2.24 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 5.15/2.24 (13) YES 5.15/2.24 (14) PiDP 5.15/2.24 (15) UsableRulesProof [EQUIVALENT, 0 ms] 5.15/2.24 (16) PiDP 5.15/2.24 (17) PiDPToQDPProof [SOUND, 0 ms] 5.15/2.24 (18) QDP 5.15/2.24 (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] 5.15/2.24 (20) YES 5.15/2.24 (21) PiDP 5.15/2.24 (22) UsableRulesProof [EQUIVALENT, 0 ms] 5.15/2.24 (23) PiDP 5.15/2.24 (24) PiDPToQDPProof [SOUND, 0 ms] 5.15/2.24 (25) QDP 5.15/2.24 (26) MRRProof [EQUIVALENT, 2 ms] 5.15/2.24 (27) QDP 5.15/2.24 (28) PisEmptyProof [EQUIVALENT, 0 ms] 5.15/2.24 (29) YES 5.15/2.24 5.15/2.24 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (0) 5.15/2.24 Obligation: 5.15/2.24 Clauses: 5.15/2.24 5.15/2.24 append(nil, XS, XS). 5.15/2.24 append(cons(X, XS), YS, cons(X, ZS)) :- append(XS, YS, ZS). 5.15/2.24 reverse(nil, nil). 5.15/2.24 reverse(cons(X, nil), cons(X, nil)). 5.15/2.24 reverse(cons(X, XS), YS) :- ','(reverse(XS, ZS), append(ZS, cons(X, nil), YS)). 5.15/2.24 shuffle(nil, nil). 5.15/2.24 shuffle(cons(X, XS), cons(X, YS)) :- ','(reverse(XS, ZS), shuffle(ZS, YS)). 5.15/2.24 query(XS) :- shuffle(cons(X, XS), YS). 5.15/2.24 5.15/2.24 5.15/2.24 Query: query(g) 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (1) PrologToPiTRSProof (SOUND) 5.15/2.24 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 5.15/2.24 5.15/2.24 query_in_1: (b) 5.15/2.24 5.15/2.24 shuffle_in_2: (b,f) 5.15/2.24 5.15/2.24 reverse_in_2: (b,f) 5.15/2.24 5.15/2.24 append_in_3: (b,b,f) 5.15/2.24 5.15/2.24 Transforming Prolog into the following Term Rewriting System: 5.15/2.24 5.15/2.24 Pi-finite rewrite system: 5.15/2.24 The TRS R consists of the following rules: 5.15/2.24 5.15/2.24 query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) 5.15/2.24 shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) 5.15/2.24 shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) 5.15/2.24 reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) 5.15/2.24 reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) 5.15/2.24 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.15/2.24 append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) 5.15/2.24 U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) 5.15/2.24 U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) 5.15/2.24 U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) 5.15/2.24 U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) 5.15/2.24 U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) 5.15/2.24 5.15/2.24 The argument filtering Pi contains the following mapping: 5.15/2.24 query_in_g(x1) = query_in_g(x1) 5.15/2.24 5.15/2.24 U6_g(x1, x2) = U6_g(x2) 5.15/2.24 5.15/2.24 shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) 5.15/2.24 5.15/2.24 cons(x1, x2) = cons(x2) 5.15/2.24 5.15/2.24 nil = nil 5.15/2.24 5.15/2.24 shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) 5.15/2.24 5.15/2.24 U4_ga(x1, x2, x3, x4) = U4_ga(x4) 5.15/2.24 5.15/2.24 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 5.15/2.24 5.15/2.24 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 5.15/2.24 5.15/2.24 U2_ga(x1, x2, x3, x4) = U2_ga(x4) 5.15/2.24 5.15/2.24 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.15/2.24 5.15/2.24 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.15/2.24 5.15/2.24 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.15/2.24 5.15/2.24 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.15/2.24 5.15/2.24 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.15/2.24 5.15/2.24 query_out_g(x1) = query_out_g 5.15/2.24 5.15/2.24 5.15/2.24 5.15/2.24 5.15/2.24 5.15/2.24 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 5.15/2.24 5.15/2.24 5.15/2.24 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (2) 5.15/2.24 Obligation: 5.15/2.24 Pi-finite rewrite system: 5.15/2.24 The TRS R consists of the following rules: 5.15/2.24 5.15/2.24 query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) 5.15/2.24 shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) 5.15/2.24 shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) 5.15/2.24 reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) 5.15/2.24 reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) 5.15/2.24 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.15/2.24 append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) 5.15/2.24 U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) 5.15/2.24 U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) 5.15/2.24 U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) 5.15/2.24 U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) 5.15/2.24 U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) 5.15/2.24 5.15/2.24 The argument filtering Pi contains the following mapping: 5.15/2.24 query_in_g(x1) = query_in_g(x1) 5.15/2.24 5.15/2.24 U6_g(x1, x2) = U6_g(x2) 5.15/2.24 5.15/2.24 shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) 5.15/2.24 5.15/2.24 cons(x1, x2) = cons(x2) 5.15/2.24 5.15/2.24 nil = nil 5.15/2.24 5.15/2.24 shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) 5.15/2.24 5.15/2.24 U4_ga(x1, x2, x3, x4) = U4_ga(x4) 5.15/2.24 5.15/2.24 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 5.15/2.24 5.15/2.24 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 5.15/2.24 5.15/2.24 U2_ga(x1, x2, x3, x4) = U2_ga(x4) 5.15/2.24 5.15/2.24 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.15/2.24 5.15/2.24 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.15/2.24 5.15/2.24 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.15/2.24 5.15/2.24 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.15/2.24 5.15/2.24 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.15/2.24 5.15/2.24 query_out_g(x1) = query_out_g 5.15/2.24 5.15/2.24 5.15/2.24 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (3) DependencyPairsProof (EQUIVALENT) 5.15/2.24 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 5.15/2.24 Pi DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 QUERY_IN_G(XS) -> U6_G(XS, shuffle_in_ga(cons(X, XS), YS)) 5.15/2.24 QUERY_IN_G(XS) -> SHUFFLE_IN_GA(cons(X, XS), YS) 5.15/2.24 SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> U4_GA(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> REVERSE_IN_GA(XS, ZS) 5.15/2.24 REVERSE_IN_GA(cons(X, XS), YS) -> U2_GA(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 REVERSE_IN_GA(cons(X, XS), YS) -> REVERSE_IN_GA(XS, ZS) 5.15/2.24 U2_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_GA(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) 5.15/2.24 U2_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> APPEND_IN_GGA(ZS, cons(X, nil), YS) 5.15/2.24 APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> U1_GGA(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) 5.15/2.24 APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> APPEND_IN_GGA(XS, YS, ZS) 5.15/2.24 U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_GA(X, XS, YS, shuffle_in_ga(ZS, YS)) 5.15/2.24 U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> SHUFFLE_IN_GA(ZS, YS) 5.15/2.24 5.15/2.24 The TRS R consists of the following rules: 5.15/2.24 5.15/2.24 query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) 5.15/2.24 shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) 5.15/2.24 shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) 5.15/2.24 reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) 5.15/2.24 reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) 5.15/2.24 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.15/2.24 append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) 5.15/2.24 U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) 5.15/2.24 U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) 5.15/2.24 U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) 5.15/2.24 U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) 5.15/2.24 U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) 5.15/2.24 5.15/2.24 The argument filtering Pi contains the following mapping: 5.15/2.24 query_in_g(x1) = query_in_g(x1) 5.15/2.24 5.15/2.24 U6_g(x1, x2) = U6_g(x2) 5.15/2.24 5.15/2.24 shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) 5.15/2.24 5.15/2.24 cons(x1, x2) = cons(x2) 5.15/2.24 5.15/2.24 nil = nil 5.15/2.24 5.15/2.24 shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) 5.15/2.24 5.15/2.24 U4_ga(x1, x2, x3, x4) = U4_ga(x4) 5.15/2.24 5.15/2.24 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 5.15/2.24 5.15/2.24 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 5.15/2.24 5.15/2.24 U2_ga(x1, x2, x3, x4) = U2_ga(x4) 5.15/2.24 5.15/2.24 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.15/2.24 5.15/2.24 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.15/2.24 5.15/2.24 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.15/2.24 5.15/2.24 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.15/2.24 5.15/2.24 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.15/2.24 5.15/2.24 query_out_g(x1) = query_out_g 5.15/2.24 5.15/2.24 QUERY_IN_G(x1) = QUERY_IN_G(x1) 5.15/2.24 5.15/2.24 U6_G(x1, x2) = U6_G(x2) 5.15/2.24 5.15/2.24 SHUFFLE_IN_GA(x1, x2) = SHUFFLE_IN_GA(x1) 5.15/2.24 5.15/2.24 U4_GA(x1, x2, x3, x4) = U4_GA(x4) 5.15/2.24 5.15/2.24 REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1) 5.15/2.24 5.15/2.24 U2_GA(x1, x2, x3, x4) = U2_GA(x4) 5.15/2.24 5.15/2.24 U3_GA(x1, x2, x3, x4) = U3_GA(x4) 5.15/2.24 5.15/2.24 APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2) 5.15/2.24 5.15/2.24 U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x5) 5.15/2.24 5.15/2.24 U5_GA(x1, x2, x3, x4) = U5_GA(x4) 5.15/2.24 5.15/2.24 5.15/2.24 We have to consider all (P,R,Pi)-chains 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (4) 5.15/2.24 Obligation: 5.15/2.24 Pi DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 QUERY_IN_G(XS) -> U6_G(XS, shuffle_in_ga(cons(X, XS), YS)) 5.15/2.24 QUERY_IN_G(XS) -> SHUFFLE_IN_GA(cons(X, XS), YS) 5.15/2.24 SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> U4_GA(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> REVERSE_IN_GA(XS, ZS) 5.15/2.24 REVERSE_IN_GA(cons(X, XS), YS) -> U2_GA(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 REVERSE_IN_GA(cons(X, XS), YS) -> REVERSE_IN_GA(XS, ZS) 5.15/2.24 U2_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_GA(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) 5.15/2.24 U2_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> APPEND_IN_GGA(ZS, cons(X, nil), YS) 5.15/2.24 APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> U1_GGA(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) 5.15/2.24 APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> APPEND_IN_GGA(XS, YS, ZS) 5.15/2.24 U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_GA(X, XS, YS, shuffle_in_ga(ZS, YS)) 5.15/2.24 U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> SHUFFLE_IN_GA(ZS, YS) 5.15/2.24 5.15/2.24 The TRS R consists of the following rules: 5.15/2.24 5.15/2.24 query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) 5.15/2.24 shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) 5.15/2.24 shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) 5.15/2.24 reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) 5.15/2.24 reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) 5.15/2.24 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.15/2.24 append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) 5.15/2.24 U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) 5.15/2.24 U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) 5.15/2.24 U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) 5.15/2.24 U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) 5.15/2.24 U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) 5.15/2.24 5.15/2.24 The argument filtering Pi contains the following mapping: 5.15/2.24 query_in_g(x1) = query_in_g(x1) 5.15/2.24 5.15/2.24 U6_g(x1, x2) = U6_g(x2) 5.15/2.24 5.15/2.24 shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) 5.15/2.24 5.15/2.24 cons(x1, x2) = cons(x2) 5.15/2.24 5.15/2.24 nil = nil 5.15/2.24 5.15/2.24 shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) 5.15/2.24 5.15/2.24 U4_ga(x1, x2, x3, x4) = U4_ga(x4) 5.15/2.24 5.15/2.24 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 5.15/2.24 5.15/2.24 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 5.15/2.24 5.15/2.24 U2_ga(x1, x2, x3, x4) = U2_ga(x4) 5.15/2.24 5.15/2.24 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.15/2.24 5.15/2.24 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.15/2.24 5.15/2.24 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.15/2.24 5.15/2.24 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.15/2.24 5.15/2.24 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.15/2.24 5.15/2.24 query_out_g(x1) = query_out_g 5.15/2.24 5.15/2.24 QUERY_IN_G(x1) = QUERY_IN_G(x1) 5.15/2.24 5.15/2.24 U6_G(x1, x2) = U6_G(x2) 5.15/2.24 5.15/2.24 SHUFFLE_IN_GA(x1, x2) = SHUFFLE_IN_GA(x1) 5.15/2.24 5.15/2.24 U4_GA(x1, x2, x3, x4) = U4_GA(x4) 5.15/2.24 5.15/2.24 REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1) 5.15/2.24 5.15/2.24 U2_GA(x1, x2, x3, x4) = U2_GA(x4) 5.15/2.24 5.15/2.24 U3_GA(x1, x2, x3, x4) = U3_GA(x4) 5.15/2.24 5.15/2.24 APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2) 5.15/2.24 5.15/2.24 U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x5) 5.15/2.24 5.15/2.24 U5_GA(x1, x2, x3, x4) = U5_GA(x4) 5.15/2.24 5.15/2.24 5.15/2.24 We have to consider all (P,R,Pi)-chains 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (5) DependencyGraphProof (EQUIVALENT) 5.15/2.24 The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 8 less nodes. 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (6) 5.15/2.24 Complex Obligation (AND) 5.15/2.24 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (7) 5.15/2.24 Obligation: 5.15/2.24 Pi DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> APPEND_IN_GGA(XS, YS, ZS) 5.15/2.24 5.15/2.24 The TRS R consists of the following rules: 5.15/2.24 5.15/2.24 query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) 5.15/2.24 shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) 5.15/2.24 shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) 5.15/2.24 reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) 5.15/2.24 reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) 5.15/2.24 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.15/2.24 append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) 5.15/2.24 U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) 5.15/2.24 U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) 5.15/2.24 U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) 5.15/2.24 U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) 5.15/2.24 U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) 5.15/2.24 5.15/2.24 The argument filtering Pi contains the following mapping: 5.15/2.24 query_in_g(x1) = query_in_g(x1) 5.15/2.24 5.15/2.24 U6_g(x1, x2) = U6_g(x2) 5.15/2.24 5.15/2.24 shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) 5.15/2.24 5.15/2.24 cons(x1, x2) = cons(x2) 5.15/2.24 5.15/2.24 nil = nil 5.15/2.24 5.15/2.24 shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) 5.15/2.24 5.15/2.24 U4_ga(x1, x2, x3, x4) = U4_ga(x4) 5.15/2.24 5.15/2.24 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 5.15/2.24 5.15/2.24 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 5.15/2.24 5.15/2.24 U2_ga(x1, x2, x3, x4) = U2_ga(x4) 5.15/2.24 5.15/2.24 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.15/2.24 5.15/2.24 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.15/2.24 5.15/2.24 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.15/2.24 5.15/2.24 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.15/2.24 5.15/2.24 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.15/2.24 5.15/2.24 query_out_g(x1) = query_out_g 5.15/2.24 5.15/2.24 APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2) 5.15/2.24 5.15/2.24 5.15/2.24 We have to consider all (P,R,Pi)-chains 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (8) UsableRulesProof (EQUIVALENT) 5.15/2.24 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (9) 5.15/2.24 Obligation: 5.15/2.24 Pi DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 APPEND_IN_GGA(cons(X, XS), YS, cons(X, ZS)) -> APPEND_IN_GGA(XS, YS, ZS) 5.15/2.24 5.15/2.24 R is empty. 5.15/2.24 The argument filtering Pi contains the following mapping: 5.15/2.24 cons(x1, x2) = cons(x2) 5.15/2.24 5.15/2.24 APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2) 5.15/2.24 5.15/2.24 5.15/2.24 We have to consider all (P,R,Pi)-chains 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (10) PiDPToQDPProof (SOUND) 5.15/2.24 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (11) 5.15/2.24 Obligation: 5.15/2.24 Q DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 APPEND_IN_GGA(cons(XS), YS) -> APPEND_IN_GGA(XS, YS) 5.15/2.24 5.15/2.24 R is empty. 5.15/2.24 Q is empty. 5.15/2.24 We have to consider all (P,Q,R)-chains. 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (12) QDPSizeChangeProof (EQUIVALENT) 5.15/2.24 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. 5.15/2.24 5.15/2.24 From the DPs we obtained the following set of size-change graphs: 5.15/2.24 *APPEND_IN_GGA(cons(XS), YS) -> APPEND_IN_GGA(XS, YS) 5.15/2.24 The graph contains the following edges 1 > 1, 2 >= 2 5.15/2.24 5.15/2.24 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (13) 5.15/2.24 YES 5.15/2.24 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (14) 5.15/2.24 Obligation: 5.15/2.24 Pi DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 REVERSE_IN_GA(cons(X, XS), YS) -> REVERSE_IN_GA(XS, ZS) 5.15/2.24 5.15/2.24 The TRS R consists of the following rules: 5.15/2.24 5.15/2.24 query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) 5.15/2.24 shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) 5.15/2.24 shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) 5.15/2.24 reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) 5.15/2.24 reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) 5.15/2.24 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.15/2.24 append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) 5.15/2.24 U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) 5.15/2.24 U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) 5.15/2.24 U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) 5.15/2.24 U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) 5.15/2.24 U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) 5.15/2.24 5.15/2.24 The argument filtering Pi contains the following mapping: 5.15/2.24 query_in_g(x1) = query_in_g(x1) 5.15/2.24 5.15/2.24 U6_g(x1, x2) = U6_g(x2) 5.15/2.24 5.15/2.24 shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) 5.15/2.24 5.15/2.24 cons(x1, x2) = cons(x2) 5.15/2.24 5.15/2.24 nil = nil 5.15/2.24 5.15/2.24 shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) 5.15/2.24 5.15/2.24 U4_ga(x1, x2, x3, x4) = U4_ga(x4) 5.15/2.24 5.15/2.24 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 5.15/2.24 5.15/2.24 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 5.15/2.24 5.15/2.24 U2_ga(x1, x2, x3, x4) = U2_ga(x4) 5.15/2.24 5.15/2.24 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.15/2.24 5.15/2.24 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.15/2.24 5.15/2.24 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.15/2.24 5.15/2.24 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.15/2.24 5.15/2.24 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.15/2.24 5.15/2.24 query_out_g(x1) = query_out_g 5.15/2.24 5.15/2.24 REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1) 5.15/2.24 5.15/2.24 5.15/2.24 We have to consider all (P,R,Pi)-chains 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (15) UsableRulesProof (EQUIVALENT) 5.15/2.24 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (16) 5.15/2.24 Obligation: 5.15/2.24 Pi DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 REVERSE_IN_GA(cons(X, XS), YS) -> REVERSE_IN_GA(XS, ZS) 5.15/2.24 5.15/2.24 R is empty. 5.15/2.24 The argument filtering Pi contains the following mapping: 5.15/2.24 cons(x1, x2) = cons(x2) 5.15/2.24 5.15/2.24 REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1) 5.15/2.24 5.15/2.24 5.15/2.24 We have to consider all (P,R,Pi)-chains 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (17) PiDPToQDPProof (SOUND) 5.15/2.24 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (18) 5.15/2.24 Obligation: 5.15/2.24 Q DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 REVERSE_IN_GA(cons(XS)) -> REVERSE_IN_GA(XS) 5.15/2.24 5.15/2.24 R is empty. 5.15/2.24 Q is empty. 5.15/2.24 We have to consider all (P,Q,R)-chains. 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (19) QDPSizeChangeProof (EQUIVALENT) 5.15/2.24 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. 5.15/2.24 5.15/2.24 From the DPs we obtained the following set of size-change graphs: 5.15/2.24 *REVERSE_IN_GA(cons(XS)) -> REVERSE_IN_GA(XS) 5.15/2.24 The graph contains the following edges 1 > 1 5.15/2.24 5.15/2.24 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (20) 5.15/2.24 YES 5.15/2.24 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (21) 5.15/2.24 Obligation: 5.15/2.24 Pi DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> SHUFFLE_IN_GA(ZS, YS) 5.15/2.24 SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> U4_GA(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 5.15/2.24 The TRS R consists of the following rules: 5.15/2.24 5.15/2.24 query_in_g(XS) -> U6_g(XS, shuffle_in_ga(cons(X, XS), YS)) 5.15/2.24 shuffle_in_ga(nil, nil) -> shuffle_out_ga(nil, nil) 5.15/2.24 shuffle_in_ga(cons(X, XS), cons(X, YS)) -> U4_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) 5.15/2.24 reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) 5.15/2.24 reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) 5.15/2.24 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.15/2.24 append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) 5.15/2.24 U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) 5.15/2.24 U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) 5.15/2.24 U4_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U5_ga(X, XS, YS, shuffle_in_ga(ZS, YS)) 5.15/2.24 U5_ga(X, XS, YS, shuffle_out_ga(ZS, YS)) -> shuffle_out_ga(cons(X, XS), cons(X, YS)) 5.15/2.24 U6_g(XS, shuffle_out_ga(cons(X, XS), YS)) -> query_out_g(XS) 5.15/2.24 5.15/2.24 The argument filtering Pi contains the following mapping: 5.15/2.24 query_in_g(x1) = query_in_g(x1) 5.15/2.24 5.15/2.24 U6_g(x1, x2) = U6_g(x2) 5.15/2.24 5.15/2.24 shuffle_in_ga(x1, x2) = shuffle_in_ga(x1) 5.15/2.24 5.15/2.24 cons(x1, x2) = cons(x2) 5.15/2.24 5.15/2.24 nil = nil 5.15/2.24 5.15/2.24 shuffle_out_ga(x1, x2) = shuffle_out_ga(x2) 5.15/2.24 5.15/2.24 U4_ga(x1, x2, x3, x4) = U4_ga(x4) 5.15/2.24 5.15/2.24 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 5.15/2.24 5.15/2.24 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 5.15/2.24 5.15/2.24 U2_ga(x1, x2, x3, x4) = U2_ga(x4) 5.15/2.24 5.15/2.24 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.15/2.24 5.15/2.24 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.15/2.24 5.15/2.24 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.15/2.24 5.15/2.24 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.15/2.24 5.15/2.24 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.15/2.24 5.15/2.24 query_out_g(x1) = query_out_g 5.15/2.24 5.15/2.24 SHUFFLE_IN_GA(x1, x2) = SHUFFLE_IN_GA(x1) 5.15/2.24 5.15/2.24 U4_GA(x1, x2, x3, x4) = U4_GA(x4) 5.15/2.24 5.15/2.24 5.15/2.24 We have to consider all (P,R,Pi)-chains 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (22) UsableRulesProof (EQUIVALENT) 5.15/2.24 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (23) 5.15/2.24 Obligation: 5.15/2.24 Pi DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 U4_GA(X, XS, YS, reverse_out_ga(XS, ZS)) -> SHUFFLE_IN_GA(ZS, YS) 5.15/2.24 SHUFFLE_IN_GA(cons(X, XS), cons(X, YS)) -> U4_GA(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 5.15/2.24 The TRS R consists of the following rules: 5.15/2.24 5.15/2.24 reverse_in_ga(nil, nil) -> reverse_out_ga(nil, nil) 5.15/2.24 reverse_in_ga(cons(X, nil), cons(X, nil)) -> reverse_out_ga(cons(X, nil), cons(X, nil)) 5.15/2.24 reverse_in_ga(cons(X, XS), YS) -> U2_ga(X, XS, YS, reverse_in_ga(XS, ZS)) 5.15/2.24 U2_ga(X, XS, YS, reverse_out_ga(XS, ZS)) -> U3_ga(X, XS, YS, append_in_gga(ZS, cons(X, nil), YS)) 5.15/2.24 U3_ga(X, XS, YS, append_out_gga(ZS, cons(X, nil), YS)) -> reverse_out_ga(cons(X, XS), YS) 5.15/2.24 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.15/2.24 append_in_gga(cons(X, XS), YS, cons(X, ZS)) -> U1_gga(X, XS, YS, ZS, append_in_gga(XS, YS, ZS)) 5.15/2.24 U1_gga(X, XS, YS, ZS, append_out_gga(XS, YS, ZS)) -> append_out_gga(cons(X, XS), YS, cons(X, ZS)) 5.15/2.24 5.15/2.24 The argument filtering Pi contains the following mapping: 5.15/2.24 cons(x1, x2) = cons(x2) 5.15/2.24 5.15/2.24 nil = nil 5.15/2.24 5.15/2.24 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 5.15/2.24 5.15/2.24 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 5.15/2.24 5.15/2.24 U2_ga(x1, x2, x3, x4) = U2_ga(x4) 5.15/2.24 5.15/2.24 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.15/2.24 5.15/2.24 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.15/2.24 5.15/2.24 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.15/2.24 5.15/2.24 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.15/2.24 5.15/2.24 SHUFFLE_IN_GA(x1, x2) = SHUFFLE_IN_GA(x1) 5.15/2.24 5.15/2.24 U4_GA(x1, x2, x3, x4) = U4_GA(x4) 5.15/2.24 5.15/2.24 5.15/2.24 We have to consider all (P,R,Pi)-chains 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (24) PiDPToQDPProof (SOUND) 5.15/2.24 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (25) 5.15/2.24 Obligation: 5.15/2.24 Q DP problem: 5.15/2.24 The TRS P consists of the following rules: 5.15/2.24 5.15/2.24 U4_GA(reverse_out_ga(ZS)) -> SHUFFLE_IN_GA(ZS) 5.15/2.24 SHUFFLE_IN_GA(cons(XS)) -> U4_GA(reverse_in_ga(XS)) 5.15/2.24 5.15/2.24 The TRS R consists of the following rules: 5.15/2.24 5.15/2.24 reverse_in_ga(nil) -> reverse_out_ga(nil) 5.15/2.24 reverse_in_ga(cons(nil)) -> reverse_out_ga(cons(nil)) 5.15/2.24 reverse_in_ga(cons(XS)) -> U2_ga(reverse_in_ga(XS)) 5.15/2.24 U2_ga(reverse_out_ga(ZS)) -> U3_ga(append_in_gga(ZS, cons(nil))) 5.15/2.24 U3_ga(append_out_gga(YS)) -> reverse_out_ga(YS) 5.15/2.24 append_in_gga(nil, XS) -> append_out_gga(XS) 5.15/2.24 append_in_gga(cons(XS), YS) -> U1_gga(append_in_gga(XS, YS)) 5.15/2.24 U1_gga(append_out_gga(ZS)) -> append_out_gga(cons(ZS)) 5.15/2.24 5.15/2.24 The set Q consists of the following terms: 5.15/2.24 5.15/2.24 reverse_in_ga(x0) 5.15/2.24 U2_ga(x0) 5.15/2.24 U3_ga(x0) 5.15/2.24 append_in_gga(x0, x1) 5.15/2.24 U1_gga(x0) 5.15/2.24 5.15/2.24 We have to consider all (P,Q,R)-chains. 5.15/2.24 ---------------------------------------- 5.15/2.24 5.15/2.24 (26) MRRProof (EQUIVALENT) 5.15/2.24 By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. 5.15/2.24 5.15/2.24 Strictly oriented dependency pairs: 5.15/2.24 5.15/2.24 U4_GA(reverse_out_ga(ZS)) -> SHUFFLE_IN_GA(ZS) 5.15/2.24 SHUFFLE_IN_GA(cons(XS)) -> U4_GA(reverse_in_ga(XS)) 5.15/2.24 5.15/2.24 Strictly oriented rules of the TRS R: 5.15/2.24 5.15/2.24 reverse_in_ga(nil) -> reverse_out_ga(nil) 5.15/2.24 reverse_in_ga(cons(nil)) -> reverse_out_ga(cons(nil)) 5.15/2.24 reverse_in_ga(cons(XS)) -> U2_ga(reverse_in_ga(XS)) 5.15/2.24 U2_ga(reverse_out_ga(ZS)) -> U3_ga(append_in_gga(ZS, cons(nil))) 5.15/2.24 U3_ga(append_out_gga(YS)) -> reverse_out_ga(YS) 5.15/2.24 append_in_gga(nil, XS) -> append_out_gga(XS) 5.15/2.25 append_in_gga(cons(XS), YS) -> U1_gga(append_in_gga(XS, YS)) 5.15/2.25 U1_gga(append_out_gga(ZS)) -> append_out_gga(cons(ZS)) 5.15/2.25 5.15/2.25 Used ordering: Knuth-Bendix order [KBO] with precedence:SHUFFLE_IN_GA_1 > append_in_gga_2 > U1_gga_1 > nil > U4_GA_1 > append_out_gga_1 > reverse_in_ga_1 > U2_ga_1 > U3_ga_1 > cons_1 > reverse_out_ga_1 5.15/2.25 5.15/2.25 and weight map: 5.15/2.25 5.15/2.25 nil=1 5.15/2.25 reverse_in_ga_1=4 5.15/2.25 reverse_out_ga_1=4 5.15/2.25 cons_1=3 5.15/2.25 U2_ga_1=3 5.15/2.25 U3_ga_1=2 5.15/2.25 append_out_gga_1=2 5.15/2.25 U1_gga_1=3 5.15/2.25 U4_GA_1=1 5.15/2.25 SHUFFLE_IN_GA_1=2 5.15/2.25 append_in_gga_2=1 5.15/2.25 5.15/2.25 The variable weight is 1 5.15/2.25 5.15/2.25 ---------------------------------------- 5.15/2.25 5.15/2.25 (27) 5.15/2.25 Obligation: 5.15/2.25 Q DP problem: 5.15/2.25 P is empty. 5.15/2.25 R is empty. 5.15/2.25 The set Q consists of the following terms: 5.15/2.25 5.15/2.25 reverse_in_ga(x0) 5.15/2.25 U2_ga(x0) 5.15/2.25 U3_ga(x0) 5.15/2.25 append_in_gga(x0, x1) 5.15/2.25 U1_gga(x0) 5.15/2.25 5.15/2.25 We have to consider all (P,Q,R)-chains. 5.15/2.25 ---------------------------------------- 5.15/2.25 5.15/2.25 (28) PisEmptyProof (EQUIVALENT) 5.15/2.25 The TRS P is empty. Hence, there is no (P,Q,R) chain. 5.15/2.25 ---------------------------------------- 5.15/2.25 5.15/2.25 (29) 5.15/2.25 YES 5.45/2.28 EOF