/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern in_order(a,g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) CutEliminatorProof [SOUND, 0 ms] (2) Prolog (3) PrologToPiTRSProof [SOUND, 24 ms] (4) PiTRS (5) DependencyPairsProof [EQUIVALENT, 0 ms] (6) PiDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) PiDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) PiDP (12) PiDPToQDPProof [SOUND, 0 ms] (13) QDP (14) QDPSizeChangeProof [EQUIVALENT, 0 ms] (15) YES (16) PiDP (17) PiDPToQDPProof [SOUND, 0 ms] (18) QDP (19) MRRProof [EQUIVALENT, 58 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Clauses: in_order(void, L) :- ','(!, eq(L, [])). in_order(T, Xs) :- ','(value(T, X), ','(app(Ls, .(X, Rs), Xs), ','(left(T, L), ','(in_order(L, Ls), ','(right(T, R), in_order(R, Rs)))))). app([], X, X). app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs). left(void, void). left(node(L, X1, X2), L). right(void, void). right(node(X3, X4, R), R). value(void, X5). value(node(X6, X, X7), X). eq(X, X). Query: in_order(a,g) ---------------------------------------- (1) CutEliminatorProof (SOUND) Eliminated all cuts by simply ignoring them[PROLOG]. ---------------------------------------- (2) Obligation: Clauses: in_order(void, L) :- eq(L, []). in_order(T, Xs) :- ','(value(T, X), ','(app(Ls, .(X, Rs), Xs), ','(left(T, L), ','(in_order(L, Ls), ','(right(T, R), in_order(R, Rs)))))). app([], X, X). app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs). left(void, void). left(node(L, X1, X2), L). right(void, void). right(node(X3, X4, R), R). value(void, X5). value(node(X6, X, X7), X). eq(X, X). Query: in_order(a,g) ---------------------------------------- (3) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: in_order_in_2: (f,b) app_in_3: (f,f,b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: in_order_in_ag(void, L) -> U1_ag(L, eq_in_gg(L, [])) eq_in_gg(X, X) -> eq_out_gg(X, X) U1_ag(L, eq_out_gg(L, [])) -> in_order_out_ag(void, L) in_order_in_ag(T, Xs) -> U2_ag(T, Xs, value_in_aa(T, X)) value_in_aa(void, X5) -> value_out_aa(void, X5) value_in_aa(node(X6, X, X7), X) -> value_out_aa(node(X6, X, X7), X) U2_ag(T, Xs, value_out_aa(T, X)) -> U3_ag(T, Xs, X, app_in_aag(Ls, .(X, Rs), Xs)) app_in_aag([], X, X) -> app_out_aag([], X, X) app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U8_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs)) U8_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_ag(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> U4_ag(T, Xs, X, Ls, Rs, left_in_aa(T, L)) left_in_aa(void, void) -> left_out_aa(void, void) left_in_aa(node(L, X1, X2), L) -> left_out_aa(node(L, X1, X2), L) U4_ag(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> U5_ag(T, Xs, X, Ls, Rs, L, in_order_in_ag(L, Ls)) U5_ag(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> U6_ag(T, Xs, X, Ls, Rs, L, right_in_aa(T, R)) right_in_aa(void, void) -> right_out_aa(void, void) right_in_aa(node(X3, X4, R), R) -> right_out_aa(node(X3, X4, R), R) U6_ag(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> U7_ag(T, Xs, in_order_in_ag(R, Rs)) U7_ag(T, Xs, in_order_out_ag(R, Rs)) -> in_order_out_ag(T, Xs) The argument filtering Pi contains the following mapping: in_order_in_ag(x1, x2) = in_order_in_ag(x2) U1_ag(x1, x2) = U1_ag(x2) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg [] = [] in_order_out_ag(x1, x2) = in_order_out_ag U2_ag(x1, x2, x3) = U2_ag(x2, x3) value_in_aa(x1, x2) = value_in_aa value_out_aa(x1, x2) = value_out_aa U3_ag(x1, x2, x3, x4) = U3_ag(x4) app_in_aag(x1, x2, x3) = app_in_aag(x3) .(x1, x2) = .(x2) app_out_aag(x1, x2, x3) = app_out_aag(x1, x2) U8_aag(x1, x2, x3, x4, x5) = U8_aag(x5) U4_ag(x1, x2, x3, x4, x5, x6) = U4_ag(x4, x5, x6) left_in_aa(x1, x2) = left_in_aa left_out_aa(x1, x2) = left_out_aa U5_ag(x1, x2, x3, x4, x5, x6, x7) = U5_ag(x5, x7) U6_ag(x1, x2, x3, x4, x5, x6, x7) = U6_ag(x5, x7) right_in_aa(x1, x2) = right_in_aa right_out_aa(x1, x2) = right_out_aa U7_ag(x1, x2, x3) = U7_ag(x3) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (4) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: in_order_in_ag(void, L) -> U1_ag(L, eq_in_gg(L, [])) eq_in_gg(X, X) -> eq_out_gg(X, X) U1_ag(L, eq_out_gg(L, [])) -> in_order_out_ag(void, L) in_order_in_ag(T, Xs) -> U2_ag(T, Xs, value_in_aa(T, X)) value_in_aa(void, X5) -> value_out_aa(void, X5) value_in_aa(node(X6, X, X7), X) -> value_out_aa(node(X6, X, X7), X) U2_ag(T, Xs, value_out_aa(T, X)) -> U3_ag(T, Xs, X, app_in_aag(Ls, .(X, Rs), Xs)) app_in_aag([], X, X) -> app_out_aag([], X, X) app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U8_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs)) U8_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_ag(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> U4_ag(T, Xs, X, Ls, Rs, left_in_aa(T, L)) left_in_aa(void, void) -> left_out_aa(void, void) left_in_aa(node(L, X1, X2), L) -> left_out_aa(node(L, X1, X2), L) U4_ag(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> U5_ag(T, Xs, X, Ls, Rs, L, in_order_in_ag(L, Ls)) U5_ag(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> U6_ag(T, Xs, X, Ls, Rs, L, right_in_aa(T, R)) right_in_aa(void, void) -> right_out_aa(void, void) right_in_aa(node(X3, X4, R), R) -> right_out_aa(node(X3, X4, R), R) U6_ag(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> U7_ag(T, Xs, in_order_in_ag(R, Rs)) U7_ag(T, Xs, in_order_out_ag(R, Rs)) -> in_order_out_ag(T, Xs) The argument filtering Pi contains the following mapping: in_order_in_ag(x1, x2) = in_order_in_ag(x2) U1_ag(x1, x2) = U1_ag(x2) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg [] = [] in_order_out_ag(x1, x2) = in_order_out_ag U2_ag(x1, x2, x3) = U2_ag(x2, x3) value_in_aa(x1, x2) = value_in_aa value_out_aa(x1, x2) = value_out_aa U3_ag(x1, x2, x3, x4) = U3_ag(x4) app_in_aag(x1, x2, x3) = app_in_aag(x3) .(x1, x2) = .(x2) app_out_aag(x1, x2, x3) = app_out_aag(x1, x2) U8_aag(x1, x2, x3, x4, x5) = U8_aag(x5) U4_ag(x1, x2, x3, x4, x5, x6) = U4_ag(x4, x5, x6) left_in_aa(x1, x2) = left_in_aa left_out_aa(x1, x2) = left_out_aa U5_ag(x1, x2, x3, x4, x5, x6, x7) = U5_ag(x5, x7) U6_ag(x1, x2, x3, x4, x5, x6, x7) = U6_ag(x5, x7) right_in_aa(x1, x2) = right_in_aa right_out_aa(x1, x2) = right_out_aa U7_ag(x1, x2, x3) = U7_ag(x3) ---------------------------------------- (5) 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: IN_ORDER_IN_AG(void, L) -> U1_AG(L, eq_in_gg(L, [])) IN_ORDER_IN_AG(void, L) -> EQ_IN_GG(L, []) IN_ORDER_IN_AG(T, Xs) -> U2_AG(T, Xs, value_in_aa(T, X)) IN_ORDER_IN_AG(T, Xs) -> VALUE_IN_AA(T, X) U2_AG(T, Xs, value_out_aa(T, X)) -> U3_AG(T, Xs, X, app_in_aag(Ls, .(X, Rs), Xs)) U2_AG(T, Xs, value_out_aa(T, X)) -> APP_IN_AAG(Ls, .(X, Rs), Xs) APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U8_AAG(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs)) APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs) U3_AG(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> U4_AG(T, Xs, X, Ls, Rs, left_in_aa(T, L)) U3_AG(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> LEFT_IN_AA(T, L) U4_AG(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> U5_AG(T, Xs, X, Ls, Rs, L, in_order_in_ag(L, Ls)) U4_AG(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> IN_ORDER_IN_AG(L, Ls) U5_AG(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> U6_AG(T, Xs, X, Ls, Rs, L, right_in_aa(T, R)) U5_AG(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> RIGHT_IN_AA(T, R) U6_AG(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> U7_AG(T, Xs, in_order_in_ag(R, Rs)) U6_AG(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> IN_ORDER_IN_AG(R, Rs) The TRS R consists of the following rules: in_order_in_ag(void, L) -> U1_ag(L, eq_in_gg(L, [])) eq_in_gg(X, X) -> eq_out_gg(X, X) U1_ag(L, eq_out_gg(L, [])) -> in_order_out_ag(void, L) in_order_in_ag(T, Xs) -> U2_ag(T, Xs, value_in_aa(T, X)) value_in_aa(void, X5) -> value_out_aa(void, X5) value_in_aa(node(X6, X, X7), X) -> value_out_aa(node(X6, X, X7), X) U2_ag(T, Xs, value_out_aa(T, X)) -> U3_ag(T, Xs, X, app_in_aag(Ls, .(X, Rs), Xs)) app_in_aag([], X, X) -> app_out_aag([], X, X) app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U8_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs)) U8_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_ag(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> U4_ag(T, Xs, X, Ls, Rs, left_in_aa(T, L)) left_in_aa(void, void) -> left_out_aa(void, void) left_in_aa(node(L, X1, X2), L) -> left_out_aa(node(L, X1, X2), L) U4_ag(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> U5_ag(T, Xs, X, Ls, Rs, L, in_order_in_ag(L, Ls)) U5_ag(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> U6_ag(T, Xs, X, Ls, Rs, L, right_in_aa(T, R)) right_in_aa(void, void) -> right_out_aa(void, void) right_in_aa(node(X3, X4, R), R) -> right_out_aa(node(X3, X4, R), R) U6_ag(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> U7_ag(T, Xs, in_order_in_ag(R, Rs)) U7_ag(T, Xs, in_order_out_ag(R, Rs)) -> in_order_out_ag(T, Xs) The argument filtering Pi contains the following mapping: in_order_in_ag(x1, x2) = in_order_in_ag(x2) U1_ag(x1, x2) = U1_ag(x2) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg [] = [] in_order_out_ag(x1, x2) = in_order_out_ag U2_ag(x1, x2, x3) = U2_ag(x2, x3) value_in_aa(x1, x2) = value_in_aa value_out_aa(x1, x2) = value_out_aa U3_ag(x1, x2, x3, x4) = U3_ag(x4) app_in_aag(x1, x2, x3) = app_in_aag(x3) .(x1, x2) = .(x2) app_out_aag(x1, x2, x3) = app_out_aag(x1, x2) U8_aag(x1, x2, x3, x4, x5) = U8_aag(x5) U4_ag(x1, x2, x3, x4, x5, x6) = U4_ag(x4, x5, x6) left_in_aa(x1, x2) = left_in_aa left_out_aa(x1, x2) = left_out_aa U5_ag(x1, x2, x3, x4, x5, x6, x7) = U5_ag(x5, x7) U6_ag(x1, x2, x3, x4, x5, x6, x7) = U6_ag(x5, x7) right_in_aa(x1, x2) = right_in_aa right_out_aa(x1, x2) = right_out_aa U7_ag(x1, x2, x3) = U7_ag(x3) IN_ORDER_IN_AG(x1, x2) = IN_ORDER_IN_AG(x2) U1_AG(x1, x2) = U1_AG(x2) EQ_IN_GG(x1, x2) = EQ_IN_GG(x1, x2) U2_AG(x1, x2, x3) = U2_AG(x2, x3) VALUE_IN_AA(x1, x2) = VALUE_IN_AA U3_AG(x1, x2, x3, x4) = U3_AG(x4) APP_IN_AAG(x1, x2, x3) = APP_IN_AAG(x3) U8_AAG(x1, x2, x3, x4, x5) = U8_AAG(x5) U4_AG(x1, x2, x3, x4, x5, x6) = U4_AG(x4, x5, x6) LEFT_IN_AA(x1, x2) = LEFT_IN_AA U5_AG(x1, x2, x3, x4, x5, x6, x7) = U5_AG(x5, x7) U6_AG(x1, x2, x3, x4, x5, x6, x7) = U6_AG(x5, x7) RIGHT_IN_AA(x1, x2) = RIGHT_IN_AA U7_AG(x1, x2, x3) = U7_AG(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: IN_ORDER_IN_AG(void, L) -> U1_AG(L, eq_in_gg(L, [])) IN_ORDER_IN_AG(void, L) -> EQ_IN_GG(L, []) IN_ORDER_IN_AG(T, Xs) -> U2_AG(T, Xs, value_in_aa(T, X)) IN_ORDER_IN_AG(T, Xs) -> VALUE_IN_AA(T, X) U2_AG(T, Xs, value_out_aa(T, X)) -> U3_AG(T, Xs, X, app_in_aag(Ls, .(X, Rs), Xs)) U2_AG(T, Xs, value_out_aa(T, X)) -> APP_IN_AAG(Ls, .(X, Rs), Xs) APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U8_AAG(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs)) APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs) U3_AG(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> U4_AG(T, Xs, X, Ls, Rs, left_in_aa(T, L)) U3_AG(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> LEFT_IN_AA(T, L) U4_AG(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> U5_AG(T, Xs, X, Ls, Rs, L, in_order_in_ag(L, Ls)) U4_AG(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> IN_ORDER_IN_AG(L, Ls) U5_AG(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> U6_AG(T, Xs, X, Ls, Rs, L, right_in_aa(T, R)) U5_AG(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> RIGHT_IN_AA(T, R) U6_AG(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> U7_AG(T, Xs, in_order_in_ag(R, Rs)) U6_AG(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> IN_ORDER_IN_AG(R, Rs) The TRS R consists of the following rules: in_order_in_ag(void, L) -> U1_ag(L, eq_in_gg(L, [])) eq_in_gg(X, X) -> eq_out_gg(X, X) U1_ag(L, eq_out_gg(L, [])) -> in_order_out_ag(void, L) in_order_in_ag(T, Xs) -> U2_ag(T, Xs, value_in_aa(T, X)) value_in_aa(void, X5) -> value_out_aa(void, X5) value_in_aa(node(X6, X, X7), X) -> value_out_aa(node(X6, X, X7), X) U2_ag(T, Xs, value_out_aa(T, X)) -> U3_ag(T, Xs, X, app_in_aag(Ls, .(X, Rs), Xs)) app_in_aag([], X, X) -> app_out_aag([], X, X) app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U8_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs)) U8_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_ag(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> U4_ag(T, Xs, X, Ls, Rs, left_in_aa(T, L)) left_in_aa(void, void) -> left_out_aa(void, void) left_in_aa(node(L, X1, X2), L) -> left_out_aa(node(L, X1, X2), L) U4_ag(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> U5_ag(T, Xs, X, Ls, Rs, L, in_order_in_ag(L, Ls)) U5_ag(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> U6_ag(T, Xs, X, Ls, Rs, L, right_in_aa(T, R)) right_in_aa(void, void) -> right_out_aa(void, void) right_in_aa(node(X3, X4, R), R) -> right_out_aa(node(X3, X4, R), R) U6_ag(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> U7_ag(T, Xs, in_order_in_ag(R, Rs)) U7_ag(T, Xs, in_order_out_ag(R, Rs)) -> in_order_out_ag(T, Xs) The argument filtering Pi contains the following mapping: in_order_in_ag(x1, x2) = in_order_in_ag(x2) U1_ag(x1, x2) = U1_ag(x2) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg [] = [] in_order_out_ag(x1, x2) = in_order_out_ag U2_ag(x1, x2, x3) = U2_ag(x2, x3) value_in_aa(x1, x2) = value_in_aa value_out_aa(x1, x2) = value_out_aa U3_ag(x1, x2, x3, x4) = U3_ag(x4) app_in_aag(x1, x2, x3) = app_in_aag(x3) .(x1, x2) = .(x2) app_out_aag(x1, x2, x3) = app_out_aag(x1, x2) U8_aag(x1, x2, x3, x4, x5) = U8_aag(x5) U4_ag(x1, x2, x3, x4, x5, x6) = U4_ag(x4, x5, x6) left_in_aa(x1, x2) = left_in_aa left_out_aa(x1, x2) = left_out_aa U5_ag(x1, x2, x3, x4, x5, x6, x7) = U5_ag(x5, x7) U6_ag(x1, x2, x3, x4, x5, x6, x7) = U6_ag(x5, x7) right_in_aa(x1, x2) = right_in_aa right_out_aa(x1, x2) = right_out_aa U7_ag(x1, x2, x3) = U7_ag(x3) IN_ORDER_IN_AG(x1, x2) = IN_ORDER_IN_AG(x2) U1_AG(x1, x2) = U1_AG(x2) EQ_IN_GG(x1, x2) = EQ_IN_GG(x1, x2) U2_AG(x1, x2, x3) = U2_AG(x2, x3) VALUE_IN_AA(x1, x2) = VALUE_IN_AA U3_AG(x1, x2, x3, x4) = U3_AG(x4) APP_IN_AAG(x1, x2, x3) = APP_IN_AAG(x3) U8_AAG(x1, x2, x3, x4, x5) = U8_AAG(x5) U4_AG(x1, x2, x3, x4, x5, x6) = U4_AG(x4, x5, x6) LEFT_IN_AA(x1, x2) = LEFT_IN_AA U5_AG(x1, x2, x3, x4, x5, x6, x7) = U5_AG(x5, x7) U6_AG(x1, x2, x3, x4, x5, x6, x7) = U6_AG(x5, x7) RIGHT_IN_AA(x1, x2) = RIGHT_IN_AA U7_AG(x1, x2, x3) = U7_AG(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 8 less nodes. ---------------------------------------- (8) Complex Obligation (AND) ---------------------------------------- (9) Obligation: Pi DP problem: The TRS P consists of the following rules: APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs) The TRS R consists of the following rules: in_order_in_ag(void, L) -> U1_ag(L, eq_in_gg(L, [])) eq_in_gg(X, X) -> eq_out_gg(X, X) U1_ag(L, eq_out_gg(L, [])) -> in_order_out_ag(void, L) in_order_in_ag(T, Xs) -> U2_ag(T, Xs, value_in_aa(T, X)) value_in_aa(void, X5) -> value_out_aa(void, X5) value_in_aa(node(X6, X, X7), X) -> value_out_aa(node(X6, X, X7), X) U2_ag(T, Xs, value_out_aa(T, X)) -> U3_ag(T, Xs, X, app_in_aag(Ls, .(X, Rs), Xs)) app_in_aag([], X, X) -> app_out_aag([], X, X) app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U8_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs)) U8_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_ag(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> U4_ag(T, Xs, X, Ls, Rs, left_in_aa(T, L)) left_in_aa(void, void) -> left_out_aa(void, void) left_in_aa(node(L, X1, X2), L) -> left_out_aa(node(L, X1, X2), L) U4_ag(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> U5_ag(T, Xs, X, Ls, Rs, L, in_order_in_ag(L, Ls)) U5_ag(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> U6_ag(T, Xs, X, Ls, Rs, L, right_in_aa(T, R)) right_in_aa(void, void) -> right_out_aa(void, void) right_in_aa(node(X3, X4, R), R) -> right_out_aa(node(X3, X4, R), R) U6_ag(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> U7_ag(T, Xs, in_order_in_ag(R, Rs)) U7_ag(T, Xs, in_order_out_ag(R, Rs)) -> in_order_out_ag(T, Xs) The argument filtering Pi contains the following mapping: in_order_in_ag(x1, x2) = in_order_in_ag(x2) U1_ag(x1, x2) = U1_ag(x2) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg [] = [] in_order_out_ag(x1, x2) = in_order_out_ag U2_ag(x1, x2, x3) = U2_ag(x2, x3) value_in_aa(x1, x2) = value_in_aa value_out_aa(x1, x2) = value_out_aa U3_ag(x1, x2, x3, x4) = U3_ag(x4) app_in_aag(x1, x2, x3) = app_in_aag(x3) .(x1, x2) = .(x2) app_out_aag(x1, x2, x3) = app_out_aag(x1, x2) U8_aag(x1, x2, x3, x4, x5) = U8_aag(x5) U4_ag(x1, x2, x3, x4, x5, x6) = U4_ag(x4, x5, x6) left_in_aa(x1, x2) = left_in_aa left_out_aa(x1, x2) = left_out_aa U5_ag(x1, x2, x3, x4, x5, x6, x7) = U5_ag(x5, x7) U6_ag(x1, x2, x3, x4, x5, x6, x7) = U6_ag(x5, x7) right_in_aa(x1, x2) = right_in_aa right_out_aa(x1, x2) = right_out_aa U7_ag(x1, x2, x3) = U7_ag(x3) APP_IN_AAG(x1, x2, x3) = APP_IN_AAG(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (10) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (11) Obligation: Pi DP problem: The TRS P consists of the following rules: APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x2) APP_IN_AAG(x1, x2, x3) = APP_IN_AAG(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (12) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: APP_IN_AAG(.(Zs)) -> APP_IN_AAG(Zs) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (14) 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: *APP_IN_AAG(.(Zs)) -> APP_IN_AAG(Zs) The graph contains the following edges 1 > 1 ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Pi DP problem: The TRS P consists of the following rules: IN_ORDER_IN_AG(T, Xs) -> U2_AG(T, Xs, value_in_aa(T, X)) U2_AG(T, Xs, value_out_aa(T, X)) -> U3_AG(T, Xs, X, app_in_aag(Ls, .(X, Rs), Xs)) U3_AG(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> U4_AG(T, Xs, X, Ls, Rs, left_in_aa(T, L)) U4_AG(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> U5_AG(T, Xs, X, Ls, Rs, L, in_order_in_ag(L, Ls)) U5_AG(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> U6_AG(T, Xs, X, Ls, Rs, L, right_in_aa(T, R)) U6_AG(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> IN_ORDER_IN_AG(R, Rs) U4_AG(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> IN_ORDER_IN_AG(L, Ls) The TRS R consists of the following rules: in_order_in_ag(void, L) -> U1_ag(L, eq_in_gg(L, [])) eq_in_gg(X, X) -> eq_out_gg(X, X) U1_ag(L, eq_out_gg(L, [])) -> in_order_out_ag(void, L) in_order_in_ag(T, Xs) -> U2_ag(T, Xs, value_in_aa(T, X)) value_in_aa(void, X5) -> value_out_aa(void, X5) value_in_aa(node(X6, X, X7), X) -> value_out_aa(node(X6, X, X7), X) U2_ag(T, Xs, value_out_aa(T, X)) -> U3_ag(T, Xs, X, app_in_aag(Ls, .(X, Rs), Xs)) app_in_aag([], X, X) -> app_out_aag([], X, X) app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U8_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs)) U8_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs)) U3_ag(T, Xs, X, app_out_aag(Ls, .(X, Rs), Xs)) -> U4_ag(T, Xs, X, Ls, Rs, left_in_aa(T, L)) left_in_aa(void, void) -> left_out_aa(void, void) left_in_aa(node(L, X1, X2), L) -> left_out_aa(node(L, X1, X2), L) U4_ag(T, Xs, X, Ls, Rs, left_out_aa(T, L)) -> U5_ag(T, Xs, X, Ls, Rs, L, in_order_in_ag(L, Ls)) U5_ag(T, Xs, X, Ls, Rs, L, in_order_out_ag(L, Ls)) -> U6_ag(T, Xs, X, Ls, Rs, L, right_in_aa(T, R)) right_in_aa(void, void) -> right_out_aa(void, void) right_in_aa(node(X3, X4, R), R) -> right_out_aa(node(X3, X4, R), R) U6_ag(T, Xs, X, Ls, Rs, L, right_out_aa(T, R)) -> U7_ag(T, Xs, in_order_in_ag(R, Rs)) U7_ag(T, Xs, in_order_out_ag(R, Rs)) -> in_order_out_ag(T, Xs) The argument filtering Pi contains the following mapping: in_order_in_ag(x1, x2) = in_order_in_ag(x2) U1_ag(x1, x2) = U1_ag(x2) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg [] = [] in_order_out_ag(x1, x2) = in_order_out_ag U2_ag(x1, x2, x3) = U2_ag(x2, x3) value_in_aa(x1, x2) = value_in_aa value_out_aa(x1, x2) = value_out_aa U3_ag(x1, x2, x3, x4) = U3_ag(x4) app_in_aag(x1, x2, x3) = app_in_aag(x3) .(x1, x2) = .(x2) app_out_aag(x1, x2, x3) = app_out_aag(x1, x2) U8_aag(x1, x2, x3, x4, x5) = U8_aag(x5) U4_ag(x1, x2, x3, x4, x5, x6) = U4_ag(x4, x5, x6) left_in_aa(x1, x2) = left_in_aa left_out_aa(x1, x2) = left_out_aa U5_ag(x1, x2, x3, x4, x5, x6, x7) = U5_ag(x5, x7) U6_ag(x1, x2, x3, x4, x5, x6, x7) = U6_ag(x5, x7) right_in_aa(x1, x2) = right_in_aa right_out_aa(x1, x2) = right_out_aa U7_ag(x1, x2, x3) = U7_ag(x3) IN_ORDER_IN_AG(x1, x2) = IN_ORDER_IN_AG(x2) U2_AG(x1, x2, x3) = U2_AG(x2, x3) U3_AG(x1, x2, x3, x4) = U3_AG(x4) U4_AG(x1, x2, x3, x4, x5, x6) = U4_AG(x4, x5, x6) U5_AG(x1, x2, x3, x4, x5, x6, x7) = U5_AG(x5, x7) U6_AG(x1, x2, x3, x4, x5, x6, x7) = U6_AG(x5, x7) 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: IN_ORDER_IN_AG(Xs) -> U2_AG(Xs, value_in_aa) U2_AG(Xs, value_out_aa) -> U3_AG(app_in_aag(Xs)) U3_AG(app_out_aag(Ls, .(Rs))) -> U4_AG(Ls, Rs, left_in_aa) U4_AG(Ls, Rs, left_out_aa) -> U5_AG(Rs, in_order_in_ag(Ls)) U5_AG(Rs, in_order_out_ag) -> U6_AG(Rs, right_in_aa) U6_AG(Rs, right_out_aa) -> IN_ORDER_IN_AG(Rs) U4_AG(Ls, Rs, left_out_aa) -> IN_ORDER_IN_AG(Ls) The TRS R consists of the following rules: in_order_in_ag(L) -> U1_ag(eq_in_gg(L, [])) eq_in_gg(X, X) -> eq_out_gg U1_ag(eq_out_gg) -> in_order_out_ag in_order_in_ag(Xs) -> U2_ag(Xs, value_in_aa) value_in_aa -> value_out_aa U2_ag(Xs, value_out_aa) -> U3_ag(app_in_aag(Xs)) app_in_aag(X) -> app_out_aag([], X) app_in_aag(.(Zs)) -> U8_aag(app_in_aag(Zs)) U8_aag(app_out_aag(Xs, Ys)) -> app_out_aag(.(Xs), Ys) U3_ag(app_out_aag(Ls, .(Rs))) -> U4_ag(Ls, Rs, left_in_aa) left_in_aa -> left_out_aa U4_ag(Ls, Rs, left_out_aa) -> U5_ag(Rs, in_order_in_ag(Ls)) U5_ag(Rs, in_order_out_ag) -> U6_ag(Rs, right_in_aa) right_in_aa -> right_out_aa U6_ag(Rs, right_out_aa) -> U7_ag(in_order_in_ag(Rs)) U7_ag(in_order_out_ag) -> in_order_out_ag The set Q consists of the following terms: in_order_in_ag(x0) eq_in_gg(x0, x1) U1_ag(x0) value_in_aa U2_ag(x0, x1) app_in_aag(x0) U8_aag(x0) U3_ag(x0) left_in_aa U4_ag(x0, x1, x2) U5_ag(x0, x1) right_in_aa U6_ag(x0, x1) U7_ag(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (19) 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: IN_ORDER_IN_AG(Xs) -> U2_AG(Xs, value_in_aa) U2_AG(Xs, value_out_aa) -> U3_AG(app_in_aag(Xs)) U3_AG(app_out_aag(Ls, .(Rs))) -> U4_AG(Ls, Rs, left_in_aa) U4_AG(Ls, Rs, left_out_aa) -> U5_AG(Rs, in_order_in_ag(Ls)) U5_AG(Rs, in_order_out_ag) -> U6_AG(Rs, right_in_aa) U6_AG(Rs, right_out_aa) -> IN_ORDER_IN_AG(Rs) U4_AG(Ls, Rs, left_out_aa) -> IN_ORDER_IN_AG(Ls) Strictly oriented rules of the TRS R: in_order_in_ag(L) -> U1_ag(eq_in_gg(L, [])) eq_in_gg(X, X) -> eq_out_gg U1_ag(eq_out_gg) -> in_order_out_ag in_order_in_ag(Xs) -> U2_ag(Xs, value_in_aa) value_in_aa -> value_out_aa U2_ag(Xs, value_out_aa) -> U3_ag(app_in_aag(Xs)) app_in_aag(X) -> app_out_aag([], X) app_in_aag(.(Zs)) -> U8_aag(app_in_aag(Zs)) U8_aag(app_out_aag(Xs, Ys)) -> app_out_aag(.(Xs), Ys) U3_ag(app_out_aag(Ls, .(Rs))) -> U4_ag(Ls, Rs, left_in_aa) left_in_aa -> left_out_aa U4_ag(Ls, Rs, left_out_aa) -> U5_ag(Rs, in_order_in_ag(Ls)) U5_ag(Rs, in_order_out_ag) -> U6_ag(Rs, right_in_aa) right_in_aa -> right_out_aa U6_ag(Rs, right_out_aa) -> U7_ag(in_order_in_ag(Rs)) U7_ag(in_order_out_ag) -> in_order_out_ag Used ordering: Knuth-Bendix order [KBO] with precedence:U4_AG_3 > left_in_aa > U1_ag_1 > right_out_aa > right_in_aa > IN_ORDER_IN_AG_1 > left_out_aa > in_order_in_ag_1 > U2_ag_2 > in_order_out_ag > eq_out_gg > app_in_aag_1 > U8_aag_1 > [] > app_out_aag_2 > U2_AG_2 > value_in_aa > U3_ag_1 > U7_ag_1 > U4_ag_3 > U5_ag_2 > U6_ag_2 > U5_AG_2 > U6_AG_2 > U3_AG_1 > ._1 > value_out_aa > eq_in_gg_2 and weight map: []=2 eq_out_gg=10 in_order_out_ag=11 value_in_aa=2 value_out_aa=1 left_in_aa=19 left_out_aa=18 right_in_aa=10 right_out_aa=9 in_order_in_ag_1=13 U1_ag_1=1 U3_ag_1=1 app_in_aag_1=11 ._1=10 U8_aag_1=10 U7_ag_1=1 IN_ORDER_IN_AG_1=14 U3_AG_1=2 eq_in_gg_2=9 U2_ag_2=11 app_out_aag_2=8 U4_ag_3=0 U5_ag_2=5 U6_ag_2=6 U2_AG_2=12 U4_AG_3=0 U5_AG_2=5 U6_AG_2=6 The variable weight is 1 ---------------------------------------- (20) Obligation: Q DP problem: P is empty. R is empty. The set Q consists of the following terms: in_order_in_ag(x0) eq_in_gg(x0, x1) U1_ag(x0) value_in_aa U2_ag(x0, x1) app_in_aag(x0) U8_aag(x0) U3_ag(x0) left_in_aa U4_ag(x0, x1, x2) U5_ag(x0, x1) right_in_aa U6_ag(x0, x1) U7_ag(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (21) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (22) YES