5.44/2.23 YES 5.44/2.25 proof of /export/starexec/sandbox/benchmark/theBenchmark.pl 5.44/2.25 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 5.44/2.25 5.44/2.25 5.44/2.25 Left Termination of the query pattern 5.44/2.25 5.44/2.25 perm(g,a) 5.44/2.25 5.44/2.25 w.r.t. the given Prolog program could successfully be proven: 5.44/2.25 5.44/2.25 (0) Prolog 5.44/2.25 (1) PrologToPiTRSProof [SOUND, 0 ms] 5.44/2.25 (2) PiTRS 5.44/2.25 (3) DependencyPairsProof [EQUIVALENT, 1 ms] 5.44/2.25 (4) PiDP 5.44/2.25 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 5.44/2.25 (6) AND 5.44/2.25 (7) PiDP 5.44/2.25 (8) UsableRulesProof [EQUIVALENT, 0 ms] 5.44/2.25 (9) PiDP 5.44/2.25 (10) PiDPToQDPProof [SOUND, 0 ms] 5.44/2.25 (11) QDP 5.44/2.25 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 5.44/2.25 (13) YES 5.44/2.25 (14) PiDP 5.44/2.25 (15) UsableRulesProof [EQUIVALENT, 0 ms] 5.44/2.25 (16) PiDP 5.44/2.25 (17) PiDPToQDPProof [SOUND, 0 ms] 5.44/2.25 (18) QDP 5.44/2.25 (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] 5.44/2.25 (20) YES 5.44/2.25 (21) PiDP 5.44/2.25 (22) UsableRulesProof [EQUIVALENT, 0 ms] 5.44/2.25 (23) PiDP 5.44/2.25 (24) PiDPToQDPProof [SOUND, 0 ms] 5.44/2.25 (25) QDP 5.44/2.25 (26) MRRProof [EQUIVALENT, 0 ms] 5.44/2.25 (27) QDP 5.44/2.25 (28) PisEmptyProof [EQUIVALENT, 0 ms] 5.44/2.25 (29) YES 5.44/2.25 5.44/2.25 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (0) 5.44/2.25 Obligation: 5.44/2.25 Clauses: 5.44/2.25 5.44/2.25 append(nil, XS, XS). 5.44/2.25 append(cons(X, XS1), XS2, cons(X, YS)) :- append(XS1, XS2, YS). 5.44/2.25 split(XS, nil, XS). 5.44/2.25 split(cons(X, XS), cons(X, YS1), YS2) :- split(XS, YS1, YS2). 5.44/2.25 perm(nil, nil). 5.44/2.25 perm(XS, cons(Y, YS)) :- ','(split(XS, YS1, cons(Y, YS2)), ','(append(YS1, YS2, ZS), perm(ZS, YS))). 5.44/2.25 5.44/2.25 5.44/2.25 Query: perm(g,a) 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (1) PrologToPiTRSProof (SOUND) 5.44/2.25 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 5.44/2.25 5.44/2.25 perm_in_2: (b,f) 5.44/2.25 5.44/2.25 split_in_3: (b,f,f) 5.44/2.25 5.44/2.25 append_in_3: (b,b,f) 5.44/2.25 5.44/2.25 Transforming Prolog into the following Term Rewriting System: 5.44/2.25 5.44/2.25 Pi-finite rewrite system: 5.44/2.25 The TRS R consists of the following rules: 5.44/2.25 5.44/2.25 perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) 5.44/2.25 perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) 5.44/2.25 split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) 5.44/2.25 U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) 5.44/2.25 U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.44/2.25 append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) 5.44/2.25 U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) 5.44/2.25 U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) 5.44/2.25 U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) 5.44/2.25 5.44/2.25 The argument filtering Pi contains the following mapping: 5.44/2.25 perm_in_ga(x1, x2) = perm_in_ga(x1) 5.44/2.25 5.44/2.25 nil = nil 5.44/2.25 5.44/2.25 perm_out_ga(x1, x2) = perm_out_ga(x2) 5.44/2.25 5.44/2.25 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.44/2.25 5.44/2.25 split_in_gaa(x1, x2, x3) = split_in_gaa(x1) 5.44/2.25 5.44/2.25 cons(x1, x2) = cons(x2) 5.44/2.25 5.44/2.25 split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) 5.44/2.25 5.44/2.25 U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) 5.44/2.25 5.44/2.25 U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 5.44/2.25 5.44/2.25 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.44/2.25 5.44/2.25 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.44/2.25 5.44/2.25 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.44/2.25 5.44/2.25 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.44/2.25 5.44/2.25 5.44/2.25 5.44/2.25 5.44/2.25 5.44/2.25 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 5.44/2.25 5.44/2.25 5.44/2.25 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (2) 5.44/2.25 Obligation: 5.44/2.25 Pi-finite rewrite system: 5.44/2.25 The TRS R consists of the following rules: 5.44/2.25 5.44/2.25 perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) 5.44/2.25 perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) 5.44/2.25 split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) 5.44/2.25 U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) 5.44/2.25 U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.44/2.25 append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) 5.44/2.25 U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) 5.44/2.25 U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) 5.44/2.25 U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) 5.44/2.25 5.44/2.25 The argument filtering Pi contains the following mapping: 5.44/2.25 perm_in_ga(x1, x2) = perm_in_ga(x1) 5.44/2.25 5.44/2.25 nil = nil 5.44/2.25 5.44/2.25 perm_out_ga(x1, x2) = perm_out_ga(x2) 5.44/2.25 5.44/2.25 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.44/2.25 5.44/2.25 split_in_gaa(x1, x2, x3) = split_in_gaa(x1) 5.44/2.25 5.44/2.25 cons(x1, x2) = cons(x2) 5.44/2.25 5.44/2.25 split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) 5.44/2.25 5.44/2.25 U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) 5.44/2.25 5.44/2.25 U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 5.44/2.25 5.44/2.25 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.44/2.25 5.44/2.25 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.44/2.25 5.44/2.25 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.44/2.25 5.44/2.25 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.44/2.25 5.44/2.25 5.44/2.25 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (3) DependencyPairsProof (EQUIVALENT) 5.44/2.25 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 5.44/2.25 Pi DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 PERM_IN_GA(XS, cons(Y, YS)) -> U3_GA(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 PERM_IN_GA(XS, cons(Y, YS)) -> SPLIT_IN_GAA(XS, YS1, cons(Y, YS2)) 5.44/2.25 SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> U2_GAA(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) 5.44/2.25 SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> SPLIT_IN_GAA(XS, YS1, YS2) 5.44/2.25 U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_GA(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> APPEND_IN_GGA(YS1, YS2, ZS) 5.44/2.25 APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> U1_GGA(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) 5.44/2.25 APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> APPEND_IN_GGA(XS1, XS2, YS) 5.44/2.25 U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_GA(XS, Y, YS, perm_in_ga(ZS, YS)) 5.44/2.25 U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> PERM_IN_GA(ZS, YS) 5.44/2.25 5.44/2.25 The TRS R consists of the following rules: 5.44/2.25 5.44/2.25 perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) 5.44/2.25 perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) 5.44/2.25 split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) 5.44/2.25 U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) 5.44/2.25 U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.44/2.25 append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) 5.44/2.25 U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) 5.44/2.25 U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) 5.44/2.25 U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) 5.44/2.25 5.44/2.25 The argument filtering Pi contains the following mapping: 5.44/2.25 perm_in_ga(x1, x2) = perm_in_ga(x1) 5.44/2.25 5.44/2.25 nil = nil 5.44/2.25 5.44/2.25 perm_out_ga(x1, x2) = perm_out_ga(x2) 5.44/2.25 5.44/2.25 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.44/2.25 5.44/2.25 split_in_gaa(x1, x2, x3) = split_in_gaa(x1) 5.44/2.25 5.44/2.25 cons(x1, x2) = cons(x2) 5.44/2.25 5.44/2.25 split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) 5.44/2.25 5.44/2.25 U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) 5.44/2.25 5.44/2.25 U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 5.44/2.25 5.44/2.25 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.44/2.25 5.44/2.25 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.44/2.25 5.44/2.25 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.44/2.25 5.44/2.25 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.44/2.25 5.44/2.25 PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) 5.44/2.25 5.44/2.25 U3_GA(x1, x2, x3, x4) = U3_GA(x4) 5.44/2.25 5.44/2.25 SPLIT_IN_GAA(x1, x2, x3) = SPLIT_IN_GAA(x1) 5.44/2.25 5.44/2.25 U2_GAA(x1, x2, x3, x4, x5) = U2_GAA(x5) 5.44/2.25 5.44/2.25 U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) 5.44/2.25 5.44/2.25 APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2) 5.44/2.25 5.44/2.25 U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x5) 5.44/2.25 5.44/2.25 U5_GA(x1, x2, x3, x4) = U5_GA(x4) 5.44/2.25 5.44/2.25 5.44/2.25 We have to consider all (P,R,Pi)-chains 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (4) 5.44/2.25 Obligation: 5.44/2.25 Pi DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 PERM_IN_GA(XS, cons(Y, YS)) -> U3_GA(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 PERM_IN_GA(XS, cons(Y, YS)) -> SPLIT_IN_GAA(XS, YS1, cons(Y, YS2)) 5.44/2.25 SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> U2_GAA(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) 5.44/2.25 SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> SPLIT_IN_GAA(XS, YS1, YS2) 5.44/2.25 U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_GA(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> APPEND_IN_GGA(YS1, YS2, ZS) 5.44/2.25 APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> U1_GGA(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) 5.44/2.25 APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> APPEND_IN_GGA(XS1, XS2, YS) 5.44/2.25 U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_GA(XS, Y, YS, perm_in_ga(ZS, YS)) 5.44/2.25 U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> PERM_IN_GA(ZS, YS) 5.44/2.25 5.44/2.25 The TRS R consists of the following rules: 5.44/2.25 5.44/2.25 perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) 5.44/2.25 perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) 5.44/2.25 split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) 5.44/2.25 U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) 5.44/2.25 U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.44/2.25 append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) 5.44/2.25 U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) 5.44/2.25 U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) 5.44/2.25 U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) 5.44/2.25 5.44/2.25 The argument filtering Pi contains the following mapping: 5.44/2.25 perm_in_ga(x1, x2) = perm_in_ga(x1) 5.44/2.25 5.44/2.25 nil = nil 5.44/2.25 5.44/2.25 perm_out_ga(x1, x2) = perm_out_ga(x2) 5.44/2.25 5.44/2.25 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.44/2.25 5.44/2.25 split_in_gaa(x1, x2, x3) = split_in_gaa(x1) 5.44/2.25 5.44/2.25 cons(x1, x2) = cons(x2) 5.44/2.25 5.44/2.25 split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) 5.44/2.25 5.44/2.25 U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) 5.44/2.25 5.44/2.25 U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 5.44/2.25 5.44/2.25 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.44/2.25 5.44/2.25 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.44/2.25 5.44/2.25 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.44/2.25 5.44/2.25 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.44/2.25 5.44/2.25 PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) 5.44/2.25 5.44/2.25 U3_GA(x1, x2, x3, x4) = U3_GA(x4) 5.44/2.25 5.44/2.25 SPLIT_IN_GAA(x1, x2, x3) = SPLIT_IN_GAA(x1) 5.44/2.25 5.44/2.25 U2_GAA(x1, x2, x3, x4, x5) = U2_GAA(x5) 5.44/2.25 5.44/2.25 U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) 5.44/2.25 5.44/2.25 APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2) 5.44/2.25 5.44/2.25 U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x5) 5.44/2.25 5.44/2.25 U5_GA(x1, x2, x3, x4) = U5_GA(x4) 5.44/2.25 5.44/2.25 5.44/2.25 We have to consider all (P,R,Pi)-chains 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (5) DependencyGraphProof (EQUIVALENT) 5.44/2.25 The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 5 less nodes. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (6) 5.44/2.25 Complex Obligation (AND) 5.44/2.25 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (7) 5.44/2.25 Obligation: 5.44/2.25 Pi DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> APPEND_IN_GGA(XS1, XS2, YS) 5.44/2.25 5.44/2.25 The TRS R consists of the following rules: 5.44/2.25 5.44/2.25 perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) 5.44/2.25 perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) 5.44/2.25 split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) 5.44/2.25 U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) 5.44/2.25 U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.44/2.25 append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) 5.44/2.25 U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) 5.44/2.25 U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) 5.44/2.25 U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) 5.44/2.25 5.44/2.25 The argument filtering Pi contains the following mapping: 5.44/2.25 perm_in_ga(x1, x2) = perm_in_ga(x1) 5.44/2.25 5.44/2.25 nil = nil 5.44/2.25 5.44/2.25 perm_out_ga(x1, x2) = perm_out_ga(x2) 5.44/2.25 5.44/2.25 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.44/2.25 5.44/2.25 split_in_gaa(x1, x2, x3) = split_in_gaa(x1) 5.44/2.25 5.44/2.25 cons(x1, x2) = cons(x2) 5.44/2.25 5.44/2.25 split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) 5.44/2.25 5.44/2.25 U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) 5.44/2.25 5.44/2.25 U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 5.44/2.25 5.44/2.25 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.44/2.25 5.44/2.25 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.44/2.25 5.44/2.25 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.44/2.25 5.44/2.25 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.44/2.25 5.44/2.25 APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2) 5.44/2.25 5.44/2.25 5.44/2.25 We have to consider all (P,R,Pi)-chains 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (8) UsableRulesProof (EQUIVALENT) 5.44/2.25 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (9) 5.44/2.25 Obligation: 5.44/2.25 Pi DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 APPEND_IN_GGA(cons(X, XS1), XS2, cons(X, YS)) -> APPEND_IN_GGA(XS1, XS2, YS) 5.44/2.25 5.44/2.25 R is empty. 5.44/2.25 The argument filtering Pi contains the following mapping: 5.44/2.25 cons(x1, x2) = cons(x2) 5.44/2.25 5.44/2.25 APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2) 5.44/2.25 5.44/2.25 5.44/2.25 We have to consider all (P,R,Pi)-chains 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (10) PiDPToQDPProof (SOUND) 5.44/2.25 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (11) 5.44/2.25 Obligation: 5.44/2.25 Q DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 APPEND_IN_GGA(cons(XS1), XS2) -> APPEND_IN_GGA(XS1, XS2) 5.44/2.25 5.44/2.25 R is empty. 5.44/2.25 Q is empty. 5.44/2.25 We have to consider all (P,Q,R)-chains. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (12) QDPSizeChangeProof (EQUIVALENT) 5.44/2.25 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. 5.44/2.25 5.44/2.25 From the DPs we obtained the following set of size-change graphs: 5.44/2.25 *APPEND_IN_GGA(cons(XS1), XS2) -> APPEND_IN_GGA(XS1, XS2) 5.44/2.25 The graph contains the following edges 1 > 1, 2 >= 2 5.44/2.25 5.44/2.25 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (13) 5.44/2.25 YES 5.44/2.25 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (14) 5.44/2.25 Obligation: 5.44/2.25 Pi DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> SPLIT_IN_GAA(XS, YS1, YS2) 5.44/2.25 5.44/2.25 The TRS R consists of the following rules: 5.44/2.25 5.44/2.25 perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) 5.44/2.25 perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) 5.44/2.25 split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) 5.44/2.25 U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) 5.44/2.25 U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.44/2.25 append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) 5.44/2.25 U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) 5.44/2.25 U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) 5.44/2.25 U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) 5.44/2.25 5.44/2.25 The argument filtering Pi contains the following mapping: 5.44/2.25 perm_in_ga(x1, x2) = perm_in_ga(x1) 5.44/2.25 5.44/2.25 nil = nil 5.44/2.25 5.44/2.25 perm_out_ga(x1, x2) = perm_out_ga(x2) 5.44/2.25 5.44/2.25 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.44/2.25 5.44/2.25 split_in_gaa(x1, x2, x3) = split_in_gaa(x1) 5.44/2.25 5.44/2.25 cons(x1, x2) = cons(x2) 5.44/2.25 5.44/2.25 split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) 5.44/2.25 5.44/2.25 U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) 5.44/2.25 5.44/2.25 U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 5.44/2.25 5.44/2.25 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.44/2.25 5.44/2.25 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.44/2.25 5.44/2.25 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.44/2.25 5.44/2.25 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.44/2.25 5.44/2.25 SPLIT_IN_GAA(x1, x2, x3) = SPLIT_IN_GAA(x1) 5.44/2.25 5.44/2.25 5.44/2.25 We have to consider all (P,R,Pi)-chains 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (15) UsableRulesProof (EQUIVALENT) 5.44/2.25 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (16) 5.44/2.25 Obligation: 5.44/2.25 Pi DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 SPLIT_IN_GAA(cons(X, XS), cons(X, YS1), YS2) -> SPLIT_IN_GAA(XS, YS1, YS2) 5.44/2.25 5.44/2.25 R is empty. 5.44/2.25 The argument filtering Pi contains the following mapping: 5.44/2.25 cons(x1, x2) = cons(x2) 5.44/2.25 5.44/2.25 SPLIT_IN_GAA(x1, x2, x3) = SPLIT_IN_GAA(x1) 5.44/2.25 5.44/2.25 5.44/2.25 We have to consider all (P,R,Pi)-chains 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (17) PiDPToQDPProof (SOUND) 5.44/2.25 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (18) 5.44/2.25 Obligation: 5.44/2.25 Q DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 SPLIT_IN_GAA(cons(XS)) -> SPLIT_IN_GAA(XS) 5.44/2.25 5.44/2.25 R is empty. 5.44/2.25 Q is empty. 5.44/2.25 We have to consider all (P,Q,R)-chains. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (19) QDPSizeChangeProof (EQUIVALENT) 5.44/2.25 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. 5.44/2.25 5.44/2.25 From the DPs we obtained the following set of size-change graphs: 5.44/2.25 *SPLIT_IN_GAA(cons(XS)) -> SPLIT_IN_GAA(XS) 5.44/2.25 The graph contains the following edges 1 > 1 5.44/2.25 5.44/2.25 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (20) 5.44/2.25 YES 5.44/2.25 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (21) 5.44/2.25 Obligation: 5.44/2.25 Pi DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_GA(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> PERM_IN_GA(ZS, YS) 5.44/2.25 PERM_IN_GA(XS, cons(Y, YS)) -> U3_GA(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 5.44/2.25 The TRS R consists of the following rules: 5.44/2.25 5.44/2.25 perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) 5.44/2.25 perm_in_ga(XS, cons(Y, YS)) -> U3_ga(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) 5.44/2.25 split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) 5.44/2.25 U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) 5.44/2.25 U3_ga(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_ga(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.44/2.25 append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) 5.44/2.25 U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) 5.44/2.25 U4_ga(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> U5_ga(XS, Y, YS, perm_in_ga(ZS, YS)) 5.44/2.25 U5_ga(XS, Y, YS, perm_out_ga(ZS, YS)) -> perm_out_ga(XS, cons(Y, YS)) 5.44/2.25 5.44/2.25 The argument filtering Pi contains the following mapping: 5.44/2.25 perm_in_ga(x1, x2) = perm_in_ga(x1) 5.44/2.25 5.44/2.25 nil = nil 5.44/2.25 5.44/2.25 perm_out_ga(x1, x2) = perm_out_ga(x2) 5.44/2.25 5.44/2.25 U3_ga(x1, x2, x3, x4) = U3_ga(x4) 5.44/2.25 5.44/2.25 split_in_gaa(x1, x2, x3) = split_in_gaa(x1) 5.44/2.25 5.44/2.25 cons(x1, x2) = cons(x2) 5.44/2.25 5.44/2.25 split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) 5.44/2.25 5.44/2.25 U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) 5.44/2.25 5.44/2.25 U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) 5.44/2.25 5.44/2.25 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.44/2.25 5.44/2.25 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.44/2.25 5.44/2.25 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.44/2.25 5.44/2.25 U5_ga(x1, x2, x3, x4) = U5_ga(x4) 5.44/2.25 5.44/2.25 PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) 5.44/2.25 5.44/2.25 U3_GA(x1, x2, x3, x4) = U3_GA(x4) 5.44/2.25 5.44/2.25 U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) 5.44/2.25 5.44/2.25 5.44/2.25 We have to consider all (P,R,Pi)-chains 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (22) UsableRulesProof (EQUIVALENT) 5.44/2.25 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (23) 5.44/2.25 Obligation: 5.44/2.25 Pi DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 U3_GA(XS, Y, YS, split_out_gaa(XS, YS1, cons(Y, YS2))) -> U4_GA(XS, Y, YS, YS1, YS2, append_in_gga(YS1, YS2, ZS)) 5.44/2.25 U4_GA(XS, Y, YS, YS1, YS2, append_out_gga(YS1, YS2, ZS)) -> PERM_IN_GA(ZS, YS) 5.44/2.25 PERM_IN_GA(XS, cons(Y, YS)) -> U3_GA(XS, Y, YS, split_in_gaa(XS, YS1, cons(Y, YS2))) 5.44/2.25 5.44/2.25 The TRS R consists of the following rules: 5.44/2.25 5.44/2.25 append_in_gga(nil, XS, XS) -> append_out_gga(nil, XS, XS) 5.44/2.25 append_in_gga(cons(X, XS1), XS2, cons(X, YS)) -> U1_gga(X, XS1, XS2, YS, append_in_gga(XS1, XS2, YS)) 5.44/2.25 split_in_gaa(XS, nil, XS) -> split_out_gaa(XS, nil, XS) 5.44/2.25 split_in_gaa(cons(X, XS), cons(X, YS1), YS2) -> U2_gaa(X, XS, YS1, YS2, split_in_gaa(XS, YS1, YS2)) 5.44/2.25 U1_gga(X, XS1, XS2, YS, append_out_gga(XS1, XS2, YS)) -> append_out_gga(cons(X, XS1), XS2, cons(X, YS)) 5.44/2.25 U2_gaa(X, XS, YS1, YS2, split_out_gaa(XS, YS1, YS2)) -> split_out_gaa(cons(X, XS), cons(X, YS1), YS2) 5.44/2.25 5.44/2.25 The argument filtering Pi contains the following mapping: 5.44/2.25 nil = nil 5.44/2.25 5.44/2.25 split_in_gaa(x1, x2, x3) = split_in_gaa(x1) 5.44/2.25 5.44/2.25 cons(x1, x2) = cons(x2) 5.44/2.25 5.44/2.25 split_out_gaa(x1, x2, x3) = split_out_gaa(x2, x3) 5.44/2.25 5.44/2.25 U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5) 5.44/2.25 5.44/2.25 append_in_gga(x1, x2, x3) = append_in_gga(x1, x2) 5.44/2.25 5.44/2.25 append_out_gga(x1, x2, x3) = append_out_gga(x3) 5.44/2.25 5.44/2.25 U1_gga(x1, x2, x3, x4, x5) = U1_gga(x5) 5.44/2.25 5.44/2.25 PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) 5.44/2.25 5.44/2.25 U3_GA(x1, x2, x3, x4) = U3_GA(x4) 5.44/2.25 5.44/2.25 U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) 5.44/2.25 5.44/2.25 5.44/2.25 We have to consider all (P,R,Pi)-chains 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (24) PiDPToQDPProof (SOUND) 5.44/2.25 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (25) 5.44/2.25 Obligation: 5.44/2.25 Q DP problem: 5.44/2.25 The TRS P consists of the following rules: 5.44/2.25 5.44/2.25 U3_GA(split_out_gaa(YS1, cons(YS2))) -> U4_GA(append_in_gga(YS1, YS2)) 5.44/2.25 U4_GA(append_out_gga(ZS)) -> PERM_IN_GA(ZS) 5.44/2.25 PERM_IN_GA(XS) -> U3_GA(split_in_gaa(XS)) 5.44/2.25 5.44/2.25 The TRS R consists of the following rules: 5.44/2.25 5.44/2.25 append_in_gga(nil, XS) -> append_out_gga(XS) 5.44/2.25 append_in_gga(cons(XS1), XS2) -> U1_gga(append_in_gga(XS1, XS2)) 5.44/2.25 split_in_gaa(XS) -> split_out_gaa(nil, XS) 5.44/2.25 split_in_gaa(cons(XS)) -> U2_gaa(split_in_gaa(XS)) 5.44/2.25 U1_gga(append_out_gga(YS)) -> append_out_gga(cons(YS)) 5.44/2.25 U2_gaa(split_out_gaa(YS1, YS2)) -> split_out_gaa(cons(YS1), YS2) 5.44/2.25 5.44/2.25 The set Q consists of the following terms: 5.44/2.25 5.44/2.25 append_in_gga(x0, x1) 5.44/2.25 split_in_gaa(x0) 5.44/2.25 U1_gga(x0) 5.44/2.25 U2_gaa(x0) 5.44/2.25 5.44/2.25 We have to consider all (P,Q,R)-chains. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (26) MRRProof (EQUIVALENT) 5.44/2.25 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. 5.44/2.25 5.44/2.25 Strictly oriented dependency pairs: 5.44/2.25 5.44/2.25 U3_GA(split_out_gaa(YS1, cons(YS2))) -> U4_GA(append_in_gga(YS1, YS2)) 5.44/2.25 U4_GA(append_out_gga(ZS)) -> PERM_IN_GA(ZS) 5.44/2.25 PERM_IN_GA(XS) -> U3_GA(split_in_gaa(XS)) 5.44/2.25 5.44/2.25 Strictly oriented rules of the TRS R: 5.44/2.25 5.44/2.25 append_in_gga(nil, XS) -> append_out_gga(XS) 5.44/2.25 append_in_gga(cons(XS1), XS2) -> U1_gga(append_in_gga(XS1, XS2)) 5.44/2.25 split_in_gaa(XS) -> split_out_gaa(nil, XS) 5.44/2.25 split_in_gaa(cons(XS)) -> U2_gaa(split_in_gaa(XS)) 5.44/2.25 U1_gga(append_out_gga(YS)) -> append_out_gga(cons(YS)) 5.44/2.25 U2_gaa(split_out_gaa(YS1, YS2)) -> split_out_gaa(cons(YS1), YS2) 5.44/2.25 5.44/2.25 Used ordering: Knuth-Bendix order [KBO] with precedence:append_in_gga_2 > nil > PERM_IN_GA_1 > cons_1 > U4_GA_1 > split_in_gaa_1 > U3_GA_1 > U2_gaa_1 > split_out_gaa_2 > U1_gga_1 > append_out_gga_1 5.44/2.25 5.44/2.25 and weight map: 5.44/2.25 5.44/2.25 nil=2 5.44/2.25 append_out_gga_1=6 5.44/2.25 cons_1=7 5.44/2.25 U1_gga_1=7 5.44/2.25 split_in_gaa_1=2 5.44/2.25 U2_gaa_1=7 5.44/2.25 U3_GA_1=3 5.44/2.25 U4_GA_1=1 5.44/2.25 PERM_IN_GA_1=6 5.44/2.25 append_in_gga_2=8 5.44/2.25 split_out_gaa_2=0 5.44/2.25 5.44/2.25 The variable weight is 1 5.44/2.25 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (27) 5.44/2.25 Obligation: 5.44/2.25 Q DP problem: 5.44/2.25 P is empty. 5.44/2.25 R is empty. 5.44/2.25 The set Q consists of the following terms: 5.44/2.25 5.44/2.25 append_in_gga(x0, x1) 5.44/2.25 split_in_gaa(x0) 5.44/2.25 U1_gga(x0) 5.44/2.25 U2_gaa(x0) 5.44/2.25 5.44/2.25 We have to consider all (P,Q,R)-chains. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (28) PisEmptyProof (EQUIVALENT) 5.44/2.25 The TRS P is empty. Hence, there is no (P,Q,R) chain. 5.44/2.25 ---------------------------------------- 5.44/2.25 5.44/2.25 (29) 5.44/2.25 YES 5.44/2.28 EOF