/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern conf(g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 12 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) PiDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) PiDP (10) PiDPToQDPProof [SOUND, 10 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, 3 ms] (27) QDP (28) PisEmptyProof [EQUIVALENT, 0 ms] (29) YES ---------------------------------------- (0) Obligation: Clauses: conf(X) :- ','(del2(X, Z), ','(del(U, Y, Z), conf(Y))). del2(X, Y) :- ','(del(U, X, Z), del(V, Z, Y)). del(X, cons(X, T), T). del(X, cons(H, T), cons(H, T1)) :- del(X, T, T1). s2l(s(X), cons(Y, Xs)) :- s2l(X, Xs). s2l(0, nil). goal(X) :- ','(s2l(X, XS), conf(XS)). Query: conf(g) ---------------------------------------- (1) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: conf_in_1: (b) del2_in_2: (b,f) del_in_3: (f,b,f) (f,f,b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z)) del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z)) del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T) del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1)) U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1)) U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y)) U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y) U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z)) del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T) del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1)) U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1)) U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y)) U3_g(X, conf_out_g(Y)) -> conf_out_g(X) The argument filtering Pi contains the following mapping: conf_in_g(x1) = conf_in_g(x1) U1_g(x1, x2) = U1_g(x2) del2_in_ga(x1, x2) = del2_in_ga(x1) U4_ga(x1, x2, x3) = U4_ga(x3) del_in_aga(x1, x2, x3) = del_in_aga(x2) cons(x1, x2) = cons(x2) del_out_aga(x1, x2, x3) = del_out_aga(x3) U6_aga(x1, x2, x3, x4, x5) = U6_aga(x5) U5_ga(x1, x2, x3) = U5_ga(x3) del2_out_ga(x1, x2) = del2_out_ga(x2) U2_g(x1, x2) = U2_g(x2) del_in_aag(x1, x2, x3) = del_in_aag(x3) del_out_aag(x1, x2, x3) = del_out_aag(x2) U6_aag(x1, x2, x3, x4, x5) = U6_aag(x5) U3_g(x1, x2) = U3_g(x2) conf_out_g(x1) = conf_out_g Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z)) del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z)) del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T) del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1)) U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1)) U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y)) U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y) U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z)) del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T) del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1)) U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1)) U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y)) U3_g(X, conf_out_g(Y)) -> conf_out_g(X) The argument filtering Pi contains the following mapping: conf_in_g(x1) = conf_in_g(x1) U1_g(x1, x2) = U1_g(x2) del2_in_ga(x1, x2) = del2_in_ga(x1) U4_ga(x1, x2, x3) = U4_ga(x3) del_in_aga(x1, x2, x3) = del_in_aga(x2) cons(x1, x2) = cons(x2) del_out_aga(x1, x2, x3) = del_out_aga(x3) U6_aga(x1, x2, x3, x4, x5) = U6_aga(x5) U5_ga(x1, x2, x3) = U5_ga(x3) del2_out_ga(x1, x2) = del2_out_ga(x2) U2_g(x1, x2) = U2_g(x2) del_in_aag(x1, x2, x3) = del_in_aag(x3) del_out_aag(x1, x2, x3) = del_out_aag(x2) U6_aag(x1, x2, x3, x4, x5) = U6_aag(x5) U3_g(x1, x2) = U3_g(x2) conf_out_g(x1) = conf_out_g ---------------------------------------- (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: CONF_IN_G(X) -> U1_G(X, del2_in_ga(X, Z)) CONF_IN_G(X) -> DEL2_IN_GA(X, Z) DEL2_IN_GA(X, Y) -> U4_GA(X, Y, del_in_aga(U, X, Z)) DEL2_IN_GA(X, Y) -> DEL_IN_AGA(U, X, Z) DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> U6_AGA(X, H, T, T1, del_in_aga(X, T, T1)) DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> DEL_IN_AGA(X, T, T1) U4_GA(X, Y, del_out_aga(U, X, Z)) -> U5_GA(X, Y, del_in_aga(V, Z, Y)) U4_GA(X, Y, del_out_aga(U, X, Z)) -> DEL_IN_AGA(V, Z, Y) U1_G(X, del2_out_ga(X, Z)) -> U2_G(X, del_in_aag(U, Y, Z)) U1_G(X, del2_out_ga(X, Z)) -> DEL_IN_AAG(U, Y, Z) DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> U6_AAG(X, H, T, T1, del_in_aag(X, T, T1)) DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> DEL_IN_AAG(X, T, T1) U2_G(X, del_out_aag(U, Y, Z)) -> U3_G(X, conf_in_g(Y)) U2_G(X, del_out_aag(U, Y, Z)) -> CONF_IN_G(Y) The TRS R consists of the following rules: conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z)) del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z)) del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T) del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1)) U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1)) U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y)) U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y) U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z)) del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T) del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1)) U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1)) U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y)) U3_g(X, conf_out_g(Y)) -> conf_out_g(X) The argument filtering Pi contains the following mapping: conf_in_g(x1) = conf_in_g(x1) U1_g(x1, x2) = U1_g(x2) del2_in_ga(x1, x2) = del2_in_ga(x1) U4_ga(x1, x2, x3) = U4_ga(x3) del_in_aga(x1, x2, x3) = del_in_aga(x2) cons(x1, x2) = cons(x2) del_out_aga(x1, x2, x3) = del_out_aga(x3) U6_aga(x1, x2, x3, x4, x5) = U6_aga(x5) U5_ga(x1, x2, x3) = U5_ga(x3) del2_out_ga(x1, x2) = del2_out_ga(x2) U2_g(x1, x2) = U2_g(x2) del_in_aag(x1, x2, x3) = del_in_aag(x3) del_out_aag(x1, x2, x3) = del_out_aag(x2) U6_aag(x1, x2, x3, x4, x5) = U6_aag(x5) U3_g(x1, x2) = U3_g(x2) conf_out_g(x1) = conf_out_g CONF_IN_G(x1) = CONF_IN_G(x1) U1_G(x1, x2) = U1_G(x2) DEL2_IN_GA(x1, x2) = DEL2_IN_GA(x1) U4_GA(x1, x2, x3) = U4_GA(x3) DEL_IN_AGA(x1, x2, x3) = DEL_IN_AGA(x2) U6_AGA(x1, x2, x3, x4, x5) = U6_AGA(x5) U5_GA(x1, x2, x3) = U5_GA(x3) U2_G(x1, x2) = U2_G(x2) DEL_IN_AAG(x1, x2, x3) = DEL_IN_AAG(x3) U6_AAG(x1, x2, x3, x4, x5) = U6_AAG(x5) U3_G(x1, x2) = U3_G(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: CONF_IN_G(X) -> U1_G(X, del2_in_ga(X, Z)) CONF_IN_G(X) -> DEL2_IN_GA(X, Z) DEL2_IN_GA(X, Y) -> U4_GA(X, Y, del_in_aga(U, X, Z)) DEL2_IN_GA(X, Y) -> DEL_IN_AGA(U, X, Z) DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> U6_AGA(X, H, T, T1, del_in_aga(X, T, T1)) DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> DEL_IN_AGA(X, T, T1) U4_GA(X, Y, del_out_aga(U, X, Z)) -> U5_GA(X, Y, del_in_aga(V, Z, Y)) U4_GA(X, Y, del_out_aga(U, X, Z)) -> DEL_IN_AGA(V, Z, Y) U1_G(X, del2_out_ga(X, Z)) -> U2_G(X, del_in_aag(U, Y, Z)) U1_G(X, del2_out_ga(X, Z)) -> DEL_IN_AAG(U, Y, Z) DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> U6_AAG(X, H, T, T1, del_in_aag(X, T, T1)) DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> DEL_IN_AAG(X, T, T1) U2_G(X, del_out_aag(U, Y, Z)) -> U3_G(X, conf_in_g(Y)) U2_G(X, del_out_aag(U, Y, Z)) -> CONF_IN_G(Y) The TRS R consists of the following rules: conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z)) del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z)) del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T) del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1)) U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1)) U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y)) U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y) U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z)) del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T) del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1)) U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1)) U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y)) U3_g(X, conf_out_g(Y)) -> conf_out_g(X) The argument filtering Pi contains the following mapping: conf_in_g(x1) = conf_in_g(x1) U1_g(x1, x2) = U1_g(x2) del2_in_ga(x1, x2) = del2_in_ga(x1) U4_ga(x1, x2, x3) = U4_ga(x3) del_in_aga(x1, x2, x3) = del_in_aga(x2) cons(x1, x2) = cons(x2) del_out_aga(x1, x2, x3) = del_out_aga(x3) U6_aga(x1, x2, x3, x4, x5) = U6_aga(x5) U5_ga(x1, x2, x3) = U5_ga(x3) del2_out_ga(x1, x2) = del2_out_ga(x2) U2_g(x1, x2) = U2_g(x2) del_in_aag(x1, x2, x3) = del_in_aag(x3) del_out_aag(x1, x2, x3) = del_out_aag(x2) U6_aag(x1, x2, x3, x4, x5) = U6_aag(x5) U3_g(x1, x2) = U3_g(x2) conf_out_g(x1) = conf_out_g CONF_IN_G(x1) = CONF_IN_G(x1) U1_G(x1, x2) = U1_G(x2) DEL2_IN_GA(x1, x2) = DEL2_IN_GA(x1) U4_GA(x1, x2, x3) = U4_GA(x3) DEL_IN_AGA(x1, x2, x3) = DEL_IN_AGA(x2) U6_AGA(x1, x2, x3, x4, x5) = U6_AGA(x5) U5_GA(x1, x2, x3) = U5_GA(x3) U2_G(x1, x2) = U2_G(x2) DEL_IN_AAG(x1, x2, x3) = DEL_IN_AAG(x3) U6_AAG(x1, x2, x3, x4, x5) = U6_AAG(x5) U3_G(x1, x2) = U3_G(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 9 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> DEL_IN_AAG(X, T, T1) The TRS R consists of the following rules: conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z)) del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z)) del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T) del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1)) U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1)) U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y)) U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y) U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z)) del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T) del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1)) U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1)) U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y)) U3_g(X, conf_out_g(Y)) -> conf_out_g(X) The argument filtering Pi contains the following mapping: conf_in_g(x1) = conf_in_g(x1) U1_g(x1, x2) = U1_g(x2) del2_in_ga(x1, x2) = del2_in_ga(x1) U4_ga(x1, x2, x3) = U4_ga(x3) del_in_aga(x1, x2, x3) = del_in_aga(x2) cons(x1, x2) = cons(x2) del_out_aga(x1, x2, x3) = del_out_aga(x3) U6_aga(x1, x2, x3, x4, x5) = U6_aga(x5) U5_ga(x1, x2, x3) = U5_ga(x3) del2_out_ga(x1, x2) = del2_out_ga(x2) U2_g(x1, x2) = U2_g(x2) del_in_aag(x1, x2, x3) = del_in_aag(x3) del_out_aag(x1, x2, x3) = del_out_aag(x2) U6_aag(x1, x2, x3, x4, x5) = U6_aag(x5) U3_g(x1, x2) = U3_g(x2) conf_out_g(x1) = conf_out_g DEL_IN_AAG(x1, x2, x3) = DEL_IN_AAG(x3) 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: DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> DEL_IN_AAG(X, T, T1) R is empty. The argument filtering Pi contains the following mapping: cons(x1, x2) = cons(x2) DEL_IN_AAG(x1, x2, x3) = DEL_IN_AAG(x3) 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: DEL_IN_AAG(cons(T1)) -> DEL_IN_AAG(T1) 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: *DEL_IN_AAG(cons(T1)) -> DEL_IN_AAG(T1) The graph contains the following edges 1 > 1 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Pi DP problem: The TRS P consists of the following rules: DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> DEL_IN_AGA(X, T, T1) The TRS R consists of the following rules: conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z)) del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z)) del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T) del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1)) U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1)) U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y)) U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y) U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z)) del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T) del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1)) U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1)) U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y)) U3_g(X, conf_out_g(Y)) -> conf_out_g(X) The argument filtering Pi contains the following mapping: conf_in_g(x1) = conf_in_g(x1) U1_g(x1, x2) = U1_g(x2) del2_in_ga(x1, x2) = del2_in_ga(x1) U4_ga(x1, x2, x3) = U4_ga(x3) del_in_aga(x1, x2, x3) = del_in_aga(x2) cons(x1, x2) = cons(x2) del_out_aga(x1, x2, x3) = del_out_aga(x3) U6_aga(x1, x2, x3, x4, x5) = U6_aga(x5) U5_ga(x1, x2, x3) = U5_ga(x3) del2_out_ga(x1, x2) = del2_out_ga(x2) U2_g(x1, x2) = U2_g(x2) del_in_aag(x1, x2, x3) = del_in_aag(x3) del_out_aag(x1, x2, x3) = del_out_aag(x2) U6_aag(x1, x2, x3, x4, x5) = U6_aag(x5) U3_g(x1, x2) = U3_g(x2) conf_out_g(x1) = conf_out_g DEL_IN_AGA(x1, x2, x3) = DEL_IN_AGA(x2) 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: DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> DEL_IN_AGA(X, T, T1) R is empty. The argument filtering Pi contains the following mapping: cons(x1, x2) = cons(x2) DEL_IN_AGA(x1, x2, x3) = DEL_IN_AGA(x2) 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: DEL_IN_AGA(cons(T)) -> DEL_IN_AGA(T) 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: *DEL_IN_AGA(cons(T)) -> DEL_IN_AGA(T) The graph contains the following edges 1 > 1 ---------------------------------------- (20) YES ---------------------------------------- (21) Obligation: Pi DP problem: The TRS P consists of the following rules: U1_G(X, del2_out_ga(X, Z)) -> U2_G(X, del_in_aag(U, Y, Z)) U2_G(X, del_out_aag(U, Y, Z)) -> CONF_IN_G(Y) CONF_IN_G(X) -> U1_G(X, del2_in_ga(X, Z)) The TRS R consists of the following rules: conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z)) del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z)) del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T) del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1)) U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1)) U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y)) U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y) U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z)) del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T) del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1)) U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1)) U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y)) U3_g(X, conf_out_g(Y)) -> conf_out_g(X) The argument filtering Pi contains the following mapping: conf_in_g(x1) = conf_in_g(x1) U1_g(x1, x2) = U1_g(x2) del2_in_ga(x1, x2) = del2_in_ga(x1) U4_ga(x1, x2, x3) = U4_ga(x3) del_in_aga(x1, x2, x3) = del_in_aga(x2) cons(x1, x2) = cons(x2) del_out_aga(x1, x2, x3) = del_out_aga(x3) U6_aga(x1, x2, x3, x4, x5) = U6_aga(x5) U5_ga(x1, x2, x3) = U5_ga(x3) del2_out_ga(x1, x2) = del2_out_ga(x2) U2_g(x1, x2) = U2_g(x2) del_in_aag(x1, x2, x3) = del_in_aag(x3) del_out_aag(x1, x2, x3) = del_out_aag(x2) U6_aag(x1, x2, x3, x4, x5) = U6_aag(x5) U3_g(x1, x2) = U3_g(x2) conf_out_g(x1) = conf_out_g CONF_IN_G(x1) = CONF_IN_G(x1) U1_G(x1, x2) = U1_G(x2) U2_G(x1, x2) = U2_G(x2) 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: U1_G(X, del2_out_ga(X, Z)) -> U2_G(X, del_in_aag(U, Y, Z)) U2_G(X, del_out_aag(U, Y, Z)) -> CONF_IN_G(Y) CONF_IN_G(X) -> U1_G(X, del2_in_ga(X, Z)) The TRS R consists of the following rules: del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T) del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1)) del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z)) U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1)) U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y)) del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T) del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1)) U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y) U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1)) The argument filtering Pi contains the following mapping: del2_in_ga(x1, x2) = del2_in_ga(x1) U4_ga(x1, x2, x3) = U4_ga(x3) del_in_aga(x1, x2, x3) = del_in_aga(x2) cons(x1, x2) = cons(x2) del_out_aga(x1, x2, x3) = del_out_aga(x3) U6_aga(x1, x2, x3, x4, x5) = U6_aga(x5) U5_ga(x1, x2, x3) = U5_ga(x3) del2_out_ga(x1, x2) = del2_out_ga(x2) del_in_aag(x1, x2, x3) = del_in_aag(x3) del_out_aag(x1, x2, x3) = del_out_aag(x2) U6_aag(x1, x2, x3, x4, x5) = U6_aag(x5) CONF_IN_G(x1) = CONF_IN_G(x1) U1_G(x1, x2) = U1_G(x2) U2_G(x1, x2) = U2_G(x2) 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: U1_G(del2_out_ga(Z)) -> U2_G(del_in_aag(Z)) U2_G(del_out_aag(Y)) -> CONF_IN_G(Y) CONF_IN_G(X) -> U1_G(del2_in_ga(X)) The TRS R consists of the following rules: del_in_aag(T) -> del_out_aag(cons(T)) del_in_aag(cons(T1)) -> U6_aag(del_in_aag(T1)) del2_in_ga(X) -> U4_ga(del_in_aga(X)) U6_aag(del_out_aag(T)) -> del_out_aag(cons(T)) U4_ga(del_out_aga(Z)) -> U5_ga(del_in_aga(Z)) del_in_aga(cons(T)) -> del_out_aga(T) del_in_aga(cons(T)) -> U6_aga(del_in_aga(T)) U5_ga(del_out_aga(Y)) -> del2_out_ga(Y) U6_aga(del_out_aga(T1)) -> del_out_aga(cons(T1)) The set Q consists of the following terms: del_in_aag(x0) del2_in_ga(x0) U6_aag(x0) U4_ga(x0) del_in_aga(x0) U5_ga(x0) U6_aga(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: U1_G(del2_out_ga(Z)) -> U2_G(del_in_aag(Z)) U2_G(del_out_aag(Y)) -> CONF_IN_G(Y) CONF_IN_G(X) -> U1_G(del2_in_ga(X)) Strictly oriented rules of the TRS R: del_in_aag(T) -> del_out_aag(cons(T)) del_in_aag(cons(T1)) -> U6_aag(del_in_aag(T1)) del2_in_ga(X) -> U4_ga(del_in_aga(X)) U6_aag(del_out_aag(T)) -> del_out_aag(cons(T)) U4_ga(del_out_aga(Z)) -> U5_ga(del_in_aga(Z)) del_in_aga(cons(T)) -> del_out_aga(T) del_in_aga(cons(T)) -> U6_aga(del_in_aga(T)) U5_ga(del_out_aga(Y)) -> del2_out_ga(Y) U6_aga(del_out_aga(T1)) -> del_out_aga(cons(T1)) Used ordering: Knuth-Bendix order [KBO] with precedence:del2_in_ga_1 > U4_ga_1 > U1_G_1 > U2_G_1 > CONF_IN_G_1 > del_in_aag_1 > del_in_aga_1 > U6_aga_1 > del2_out_ga_1 > U5_ga_1 > del_out_aga_1 > U6_aag_1 > del_out_aag_1 > cons_1 and weight map: del_in_aag_1=3 del_out_aag_1=1 cons_1=2 U6_aag_1=2 del2_in_ga_1=4 U4_ga_1=1 del_in_aga_1=3 del_out_aga_1=5 U5_ga_1=3 U6_aga_1=2 del2_out_ga_1=7 U1_G_1=1 U2_G_1=5 CONF_IN_G_1=6 The variable weight is 1 ---------------------------------------- (27) Obligation: Q DP problem: P is empty. R is empty. The set Q consists of the following terms: del_in_aag(x0) del2_in_ga(x0) U6_aag(x0) U4_ga(x0) del_in_aga(x0) U5_ga(x0) U6_aga(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (28) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (29) YES