/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.pl /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern balance(g,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof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, 21 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) PiDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) PiDP (9) PiDPToQDPProof [SOUND, 18 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Clauses: balance(T, TB) :- balance(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])). balance(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)). balance(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) :- ','(balance(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)), balance(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))). Query: balance(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: balance_in_2: (b,f) balance_in_5: (b,f,f,f,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: balance_in_ga(T, TB) -> U1_ga(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) balance_in_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) U1_ga(T, TB, balance_out_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) -> balance_out_ga(T, TB) The argument filtering Pi contains the following mapping: balance_in_ga(x1, x2) = balance_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x1, x3) balance_in_gaaaa(x1, x2, x3, x4, x5) = balance_in_gaaaa(x1) nil = nil balance_out_gaaaa(x1, x2, x3, x4, x5) = balance_out_gaaaa(x1) tree(x1, x2, x3) = tree(x1, x2, x3) U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_gaaaa(x1, x2, x3, x18) U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_gaaaa(x1, x2, x3, x18) balance_out_ga(x1, x2) = balance_out_ga(x1) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: balance_in_ga(T, TB) -> U1_ga(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) balance_in_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) U1_ga(T, TB, balance_out_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) -> balance_out_ga(T, TB) The argument filtering Pi contains the following mapping: balance_in_ga(x1, x2) = balance_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x1, x3) balance_in_gaaaa(x1, x2, x3, x4, x5) = balance_in_gaaaa(x1) nil = nil balance_out_gaaaa(x1, x2, x3, x4, x5) = balance_out_gaaaa(x1) tree(x1, x2, x3) = tree(x1, x2, x3) U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_gaaaa(x1, x2, x3, x18) U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_gaaaa(x1, x2, x3, x18) balance_out_ga(x1, x2) = balance_out_ga(x1) ---------------------------------------- (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: BALANCE_IN_GA(T, TB) -> U1_GA(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) BALANCE_IN_GA(T, TB) -> BALANCE_IN_GAAAA(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])) BALANCE_IN_GAAAA(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) BALANCE_IN_GAAAA(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> BALANCE_IN_GAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)) U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> BALANCE_IN_GAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)) The TRS R consists of the following rules: balance_in_ga(T, TB) -> U1_ga(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) balance_in_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) U1_ga(T, TB, balance_out_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) -> balance_out_ga(T, TB) The argument filtering Pi contains the following mapping: balance_in_ga(x1, x2) = balance_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x1, x3) balance_in_gaaaa(x1, x2, x3, x4, x5) = balance_in_gaaaa(x1) nil = nil balance_out_gaaaa(x1, x2, x3, x4, x5) = balance_out_gaaaa(x1) tree(x1, x2, x3) = tree(x1, x2, x3) U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_gaaaa(x1, x2, x3, x18) U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_gaaaa(x1, x2, x3, x18) balance_out_ga(x1, x2) = balance_out_ga(x1) BALANCE_IN_GA(x1, x2) = BALANCE_IN_GA(x1) U1_GA(x1, x2, x3) = U1_GA(x1, x3) BALANCE_IN_GAAAA(x1, x2, x3, x4, x5) = BALANCE_IN_GAAAA(x1) U2_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_GAAAA(x1, x2, x3, x18) U3_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_GAAAA(x1, x2, x3, x18) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: BALANCE_IN_GA(T, TB) -> U1_GA(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) BALANCE_IN_GA(T, TB) -> BALANCE_IN_GAAAA(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, [])) BALANCE_IN_GAAAA(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) BALANCE_IN_GAAAA(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> BALANCE_IN_GAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)) U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> BALANCE_IN_GAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)) The TRS R consists of the following rules: balance_in_ga(T, TB) -> U1_ga(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) balance_in_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) U1_ga(T, TB, balance_out_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) -> balance_out_ga(T, TB) The argument filtering Pi contains the following mapping: balance_in_ga(x1, x2) = balance_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x1, x3) balance_in_gaaaa(x1, x2, x3, x4, x5) = balance_in_gaaaa(x1) nil = nil balance_out_gaaaa(x1, x2, x3, x4, x5) = balance_out_gaaaa(x1) tree(x1, x2, x3) = tree(x1, x2, x3) U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_gaaaa(x1, x2, x3, x18) U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_gaaaa(x1, x2, x3, x18) balance_out_ga(x1, x2) = balance_out_ga(x1) BALANCE_IN_GA(x1, x2) = BALANCE_IN_GA(x1) U1_GA(x1, x2, x3) = U1_GA(x1, x3) BALANCE_IN_GAAAA(x1, x2, x3, x4, x5) = BALANCE_IN_GAAAA(x1) U2_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_GAAAA(x1, x2, x3, x18) U3_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_GAAAA(x1, x2, x3, x18) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes. ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> BALANCE_IN_GAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)) BALANCE_IN_GAAAA(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) BALANCE_IN_GAAAA(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> BALANCE_IN_GAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)) The TRS R consists of the following rules: balance_in_ga(T, TB) -> U1_ga(T, TB, balance_in_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) balance_in_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) U1_ga(T, TB, balance_out_gaaaa(T, -(I, []), -(.(','(TB, -(I, [])), X), X), -(Rest, []), -(Rest, []))) -> balance_out_ga(T, TB) The argument filtering Pi contains the following mapping: balance_in_ga(x1, x2) = balance_in_ga(x1) U1_ga(x1, x2, x3) = U1_ga(x1, x3) balance_in_gaaaa(x1, x2, x3, x4, x5) = balance_in_gaaaa(x1) nil = nil balance_out_gaaaa(x1, x2, x3, x4, x5) = balance_out_gaaaa(x1) tree(x1, x2, x3) = tree(x1, x2, x3) U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_gaaaa(x1, x2, x3, x18) U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_gaaaa(x1, x2, x3, x18) balance_out_ga(x1, x2) = balance_out_ga(x1) BALANCE_IN_GAAAA(x1, x2, x3, x4, x5) = BALANCE_IN_GAAAA(x1) U2_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_GAAAA(x1, x2, x3, x18) We have to consider all (P,R,Pi)-chains ---------------------------------------- (7) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (8) Obligation: Pi DP problem: The TRS P consists of the following rules: U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> BALANCE_IN_GAAAA(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT)) BALANCE_IN_GAAAA(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> U2_GAAAA(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) BALANCE_IN_GAAAA(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> BALANCE_IN_GAAAA(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1)) The TRS R consists of the following rules: balance_in_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) -> balance_out_gaaaa(nil, -(X, X), -(A, B), -(A, B), -(.(','(nil, -(C, C)), T), T)) balance_in_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) -> U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) U2_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(L, -(IH, .(V, IT1)), -(H, T), -(HR1, TR1), -(NH, NT1))) -> U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_in_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) U3_gaaaa(L, V, R, IH, IT, LB, VB, RB, A, D, H, X, T, HR, TR, NH, NT, balance_out_gaaaa(R, -(IT1, IT), -(HR1, TR1), -(HR, TR), -(NT1, NT))) -> balance_out_gaaaa(tree(L, V, R), -(IH, IT), -(.(','(tree(LB, VB, RB), -(A, D)), H), .(','(LB, -(A, .(VB, X))), .(','(RB, -(X, D)), T))), -(HR, TR), -(NH, NT)) The argument filtering Pi contains the following mapping: balance_in_gaaaa(x1, x2, x3, x4, x5) = balance_in_gaaaa(x1) nil = nil balance_out_gaaaa(x1, x2, x3, x4, x5) = balance_out_gaaaa(x1) tree(x1, x2, x3) = tree(x1, x2, x3) U2_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_gaaaa(x1, x2, x3, x18) U3_gaaaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U3_gaaaa(x1, x2, x3, x18) BALANCE_IN_GAAAA(x1, x2, x3, x4, x5) = BALANCE_IN_GAAAA(x1) U2_GAAAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15, x16, x17, x18) = U2_GAAAA(x1, x2, x3, x18) We have to consider all (P,R,Pi)-chains ---------------------------------------- (9) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: U2_GAAAA(L, V, R, balance_out_gaaaa(L)) -> BALANCE_IN_GAAAA(R) BALANCE_IN_GAAAA(tree(L, V, R)) -> U2_GAAAA(L, V, R, balance_in_gaaaa(L)) BALANCE_IN_GAAAA(tree(L, V, R)) -> BALANCE_IN_GAAAA(L) The TRS R consists of the following rules: balance_in_gaaaa(nil) -> balance_out_gaaaa(nil) balance_in_gaaaa(tree(L, V, R)) -> U2_gaaaa(L, V, R, balance_in_gaaaa(L)) U2_gaaaa(L, V, R, balance_out_gaaaa(L)) -> U3_gaaaa(L, V, R, balance_in_gaaaa(R)) U3_gaaaa(L, V, R, balance_out_gaaaa(R)) -> balance_out_gaaaa(tree(L, V, R)) The set Q consists of the following terms: balance_in_gaaaa(x0) U2_gaaaa(x0, x1, x2, x3) U3_gaaaa(x0, x1, x2, x3) We have to consider all (P,Q,R)-chains. ---------------------------------------- (11) 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: *BALANCE_IN_GAAAA(tree(L, V, R)) -> U2_GAAAA(L, V, R, balance_in_gaaaa(L)) The graph contains the following edges 1 > 1, 1 > 2, 1 > 3 *BALANCE_IN_GAAAA(tree(L, V, R)) -> BALANCE_IN_GAAAA(L) The graph contains the following edges 1 > 1 *U2_GAAAA(L, V, R, balance_out_gaaaa(L)) -> BALANCE_IN_GAAAA(R) The graph contains the following edges 3 >= 1 ---------------------------------------- (12) YES