/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.pl /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.pl # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern perm(g,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) PiDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) PiDP (10) PiDPToQDPProof [SOUND, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) PiDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) PiDP (17) PiDPToQDPProof [SOUND, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES (21) PiDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) PiDP (24) PiDPToQDPProof [SOUND, 0 ms] (25) QDP (26) MRRProof [EQUIVALENT, 18 ms] (27) QDP (28) DependencyGraphProof [EQUIVALENT, 0 ms] (29) TRUE ---------------------------------------- (0) Obligation: Clauses: ap1(nil, X, X). ap1(cons(H, X), Y, cons(H, Z)) :- ap1(X, Y, Z). ap2(nil, X, X). ap2(cons(H, X), Y, cons(H, Z)) :- ap2(X, Y, Z). perm(nil, nil). perm(Xs, cons(X, Ys)) :- ','(ap1(X1s, cons(X, X2s), Xs), ','(ap2(X1s, X2s, Zs), perm(Zs, Ys))). Query: perm(g,a) ---------------------------------------- (1) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: perm_in_2: (b,f) ap1_in_3: (f,f,b) ap2_in_3: (b,b,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X) ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z)) U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z)) U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X) ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z)) U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z)) U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys)) U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) ap1_in_aag(x1, x2, x3) = ap1_in_aag(x3) cons(x1, x2) = cons(x2) ap1_out_aag(x1, x2, x3) = ap1_out_aag(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) ap2_in_gga(x1, x2, x3) = ap2_in_gga(x1, x2) ap2_out_gga(x1, x2, x3) = ap2_out_gga(x3) U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X) ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z)) U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z)) U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X) ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z)) U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z)) U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys)) U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) ap1_in_aag(x1, x2, x3) = ap1_in_aag(x3) cons(x1, x2) = cons(x2) ap1_out_aag(x1, x2, x3) = ap1_out_aag(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) ap2_in_gga(x1, x2, x3) = ap2_in_gga(x1, x2) ap2_out_gga(x1, x2, x3) = ap2_out_gga(x3) U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: Pi DP problem: The TRS P consists of the following rules: PERM_IN_GA(Xs, cons(X, Ys)) -> U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) PERM_IN_GA(Xs, cons(X, Ys)) -> AP1_IN_AAG(X1s, cons(X, X2s), Xs) AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> U1_AAG(H, X, Y, Z, ap1_in_aag(X, Y, Z)) AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> AP1_IN_AAG(X, Y, Z) U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> AP2_IN_GGA(X1s, X2s, Zs) AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> U2_GGA(H, X, Y, Z, ap2_in_gga(X, Y, Z)) AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> AP2_IN_GGA(X, Y, Z) U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_GA(Xs, X, Ys, perm_in_ga(Zs, Ys)) U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> PERM_IN_GA(Zs, Ys) The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X) ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z)) U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z)) U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X) ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z)) U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z)) U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys)) U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) ap1_in_aag(x1, x2, x3) = ap1_in_aag(x3) cons(x1, x2) = cons(x2) ap1_out_aag(x1, x2, x3) = ap1_out_aag(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) ap2_in_gga(x1, x2, x3) = ap2_in_gga(x1, x2) ap2_out_gga(x1, x2, x3) = ap2_out_gga(x3) U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) AP1_IN_AAG(x1, x2, x3) = AP1_IN_AAG(x3) U1_AAG(x1, x2, x3, x4, x5) = U1_AAG(x5) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) AP2_IN_GGA(x1, x2, x3) = AP2_IN_GGA(x1, x2) U2_GGA(x1, x2, x3, x4, x5) = U2_GGA(x5) U5_GA(x1, x2, x3, x4) = U5_GA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: PERM_IN_GA(Xs, cons(X, Ys)) -> U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) PERM_IN_GA(Xs, cons(X, Ys)) -> AP1_IN_AAG(X1s, cons(X, X2s), Xs) AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> U1_AAG(H, X, Y, Z, ap1_in_aag(X, Y, Z)) AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> AP1_IN_AAG(X, Y, Z) U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> AP2_IN_GGA(X1s, X2s, Zs) AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> U2_GGA(H, X, Y, Z, ap2_in_gga(X, Y, Z)) AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> AP2_IN_GGA(X, Y, Z) U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_GA(Xs, X, Ys, perm_in_ga(Zs, Ys)) U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> PERM_IN_GA(Zs, Ys) The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X) ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z)) U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z)) U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X) ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z)) U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z)) U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys)) U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) ap1_in_aag(x1, x2, x3) = ap1_in_aag(x3) cons(x1, x2) = cons(x2) ap1_out_aag(x1, x2, x3) = ap1_out_aag(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) ap2_in_gga(x1, x2, x3) = ap2_in_gga(x1, x2) ap2_out_gga(x1, x2, x3) = ap2_out_gga(x3) U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) AP1_IN_AAG(x1, x2, x3) = AP1_IN_AAG(x3) U1_AAG(x1, x2, x3, x4, x5) = U1_AAG(x5) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) AP2_IN_GGA(x1, x2, x3) = AP2_IN_GGA(x1, x2) U2_GGA(x1, x2, x3, x4, x5) = U2_GGA(x5) U5_GA(x1, x2, x3, x4) = U5_GA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 5 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> AP2_IN_GGA(X, Y, Z) The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X) ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z)) U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z)) U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X) ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z)) U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z)) U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys)) U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) ap1_in_aag(x1, x2, x3) = ap1_in_aag(x3) cons(x1, x2) = cons(x2) ap1_out_aag(x1, x2, x3) = ap1_out_aag(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) ap2_in_gga(x1, x2, x3) = ap2_in_gga(x1, x2) ap2_out_gga(x1, x2, x3) = ap2_out_gga(x3) U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) AP2_IN_GGA(x1, x2, x3) = AP2_IN_GGA(x1, x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (8) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (9) Obligation: Pi DP problem: The TRS P consists of the following rules: AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> AP2_IN_GGA(X, Y, Z) R is empty. The argument filtering Pi contains the following mapping: cons(x1, x2) = cons(x2) AP2_IN_GGA(x1, x2, x3) = AP2_IN_GGA(x1, x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (10) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: AP2_IN_GGA(cons(X), Y) -> AP2_IN_GGA(X, Y) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (12) QDPSizeChangeProof (EQUIVALENT) 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. From the DPs we obtained the following set of size-change graphs: *AP2_IN_GGA(cons(X), Y) -> AP2_IN_GGA(X, Y) The graph contains the following edges 1 > 1, 2 >= 2 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Pi DP problem: The TRS P consists of the following rules: AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> AP1_IN_AAG(X, Y, Z) The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X) ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z)) U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z)) U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X) ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z)) U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z)) U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys)) U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) ap1_in_aag(x1, x2, x3) = ap1_in_aag(x3) cons(x1, x2) = cons(x2) ap1_out_aag(x1, x2, x3) = ap1_out_aag(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) ap2_in_gga(x1, x2, x3) = ap2_in_gga(x1, x2) ap2_out_gga(x1, x2, x3) = ap2_out_gga(x3) U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) AP1_IN_AAG(x1, x2, x3) = AP1_IN_AAG(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (15) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (16) Obligation: Pi DP problem: The TRS P consists of the following rules: AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> AP1_IN_AAG(X, Y, Z) R is empty. The argument filtering Pi contains the following mapping: cons(x1, x2) = cons(x2) AP1_IN_AAG(x1, x2, x3) = AP1_IN_AAG(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (17) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: AP1_IN_AAG(cons(Z)) -> AP1_IN_AAG(Z) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (19) QDPSizeChangeProof (EQUIVALENT) 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. From the DPs we obtained the following set of size-change graphs: *AP1_IN_AAG(cons(Z)) -> AP1_IN_AAG(Z) The graph contains the following edges 1 > 1 ---------------------------------------- (20) YES ---------------------------------------- (21) Obligation: Pi DP problem: The TRS P consists of the following rules: U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> PERM_IN_GA(Zs, Ys) PERM_IN_GA(Xs, cons(X, Ys)) -> U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) The TRS R consists of the following rules: perm_in_ga(nil, nil) -> perm_out_ga(nil, nil) perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X) ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z)) U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z)) U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X) ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z)) U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z)) U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys)) U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) nil = nil perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) ap1_in_aag(x1, x2, x3) = ap1_in_aag(x3) cons(x1, x2) = cons(x2) ap1_out_aag(x1, x2, x3) = ap1_out_aag(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) ap2_in_gga(x1, x2, x3) = ap2_in_gga(x1, x2) ap2_out_gga(x1, x2, x3) = ap2_out_gga(x3) U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) We have to consider all (P,R,Pi)-chains ---------------------------------------- (22) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (23) Obligation: Pi DP problem: The TRS P consists of the following rules: U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs)) U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> PERM_IN_GA(Zs, Ys) PERM_IN_GA(Xs, cons(X, Ys)) -> U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs)) The TRS R consists of the following rules: ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X) ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z)) ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X) ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z)) U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z)) U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z)) The argument filtering Pi contains the following mapping: nil = nil ap1_in_aag(x1, x2, x3) = ap1_in_aag(x3) cons(x1, x2) = cons(x2) ap1_out_aag(x1, x2, x3) = ap1_out_aag(x1, x2) U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5) ap2_in_gga(x1, x2, x3) = ap2_in_gga(x1, x2) ap2_out_gga(x1, x2, x3) = ap2_out_gga(x3) U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) We have to consider all (P,R,Pi)-chains ---------------------------------------- (24) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: U3_GA(ap1_out_aag(X1s, cons(X2s))) -> U4_GA(ap2_in_gga(X1s, X2s)) U4_GA(ap2_out_gga(Zs)) -> PERM_IN_GA(Zs) PERM_IN_GA(Xs) -> U3_GA(ap1_in_aag(Xs)) The TRS R consists of the following rules: ap2_in_gga(nil, X) -> ap2_out_gga(X) ap2_in_gga(cons(X), Y) -> U2_gga(ap2_in_gga(X, Y)) ap1_in_aag(X) -> ap1_out_aag(nil, X) ap1_in_aag(cons(Z)) -> U1_aag(ap1_in_aag(Z)) U2_gga(ap2_out_gga(Z)) -> ap2_out_gga(cons(Z)) U1_aag(ap1_out_aag(X, Y)) -> ap1_out_aag(cons(X), Y) The set Q consists of the following terms: ap2_in_gga(x0, x1) ap1_in_aag(x0) U2_gga(x0) U1_aag(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (26) MRRProof (EQUIVALENT) 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. Strictly oriented dependency pairs: U3_GA(ap1_out_aag(X1s, cons(X2s))) -> U4_GA(ap2_in_gga(X1s, X2s)) U4_GA(ap2_out_gga(Zs)) -> PERM_IN_GA(Zs) Used ordering: Polynomial interpretation [POLO]: POL(PERM_IN_GA(x_1)) = 2*x_1 POL(U1_aag(x_1)) = 2 + x_1 POL(U2_gga(x_1)) = 2 + x_1 POL(U3_GA(x_1)) = 2*x_1 POL(U4_GA(x_1)) = 1 + 2*x_1 POL(ap1_in_aag(x_1)) = x_1 POL(ap1_out_aag(x_1, x_2)) = x_1 + x_2 POL(ap2_in_gga(x_1, x_2)) = x_1 + x_2 POL(ap2_out_gga(x_1)) = x_1 POL(cons(x_1)) = 2 + x_1 POL(nil) = 0 ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: PERM_IN_GA(Xs) -> U3_GA(ap1_in_aag(Xs)) The TRS R consists of the following rules: ap2_in_gga(nil, X) -> ap2_out_gga(X) ap2_in_gga(cons(X), Y) -> U2_gga(ap2_in_gga(X, Y)) ap1_in_aag(X) -> ap1_out_aag(nil, X) ap1_in_aag(cons(Z)) -> U1_aag(ap1_in_aag(Z)) U2_gga(ap2_out_gga(Z)) -> ap2_out_gga(cons(Z)) U1_aag(ap1_out_aag(X, Y)) -> ap1_out_aag(cons(X), Y) The set Q consists of the following terms: ap2_in_gga(x0, x1) ap1_in_aag(x0) U2_gga(x0) U1_aag(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (28) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (29) TRUE