/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Left Termination of the query pattern palindrome(g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof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, 8 ms] (27) QDP (28) PisEmptyProof [EQUIVALENT, 0 ms] (29) YES ---------------------------------------- (0) Obligation: Clauses: palindrome(L) :- ','(halves(L, X1s, X2s, EvenOdd), ','(eq(EvenOdd, even), eq(X1s, X2s))). palindrome(L) :- ','(halves(L, X1s, X2s, EvenOdd), ','(eq(EvenOdd, odd), last(X1s, X1, X2s))). halves([], [], [], even). halves(.(X, []), .(X, []), [], odd). halves(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) :- ','(last(.(Y, Xs), R, Rests), halves(Rests, Ts, Rs, EvenOdd)). last(.(T, []), T, []). last(.(H, T), X, .(H, M)) :- last(T, X, M). eq(X, X). Query: palindrome(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: palindrome_in_1: (b) halves_in_4: (b,f,f,f) last_in_3: (b,f,f) (b,f,b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: palindrome_in_g(L) -> U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd)) halves_in_gaaa([], [], [], even) -> halves_out_gaaa([], [], [], even) halves_in_gaaa(.(X, []), .(X, []), [], odd) -> halves_out_gaaa(.(X, []), .(X, []), [], odd) halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) last_in_gaa(.(T, []), T, []) -> last_out_gaa(.(T, []), T, []) last_in_gaa(.(H, T), X, .(H, M)) -> U8_gaa(H, T, X, M, last_in_gaa(T, X, M)) U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) -> last_out_gaa(.(H, T), X, .(H, M)) U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd)) U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) -> halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even)) eq_in_gg(X, X) -> eq_out_gg(X, X) U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> U3_g(L, eq_in_gg(X1s, X2s)) U3_g(L, eq_out_gg(X1s, X2s)) -> palindrome_out_g(L) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd)) U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> U5_g(L, last_in_gag(X1s, X1, X2s)) last_in_gag(.(T, []), T, []) -> last_out_gag(.(T, []), T, []) last_in_gag(.(H, T), X, .(H, M)) -> U8_gag(H, T, X, M, last_in_gag(T, X, M)) U8_gag(H, T, X, M, last_out_gag(T, X, M)) -> last_out_gag(.(H, T), X, .(H, M)) U5_g(L, last_out_gag(X1s, X1, X2s)) -> palindrome_out_g(L) The argument filtering Pi contains the following mapping: palindrome_in_g(x1) = palindrome_in_g(x1) U1_g(x1, x2) = U1_g(x2) halves_in_gaaa(x1, x2, x3, x4) = halves_in_gaaa(x1) [] = [] halves_out_gaaa(x1, x2, x3, x4) = halves_out_gaaa(x2, x3, x4) .(x1, x2) = .(x1, x2) U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U6_gaaa(x1, x8) last_in_gaa(x1, x2, x3) = last_in_gaa(x1) last_out_gaa(x1, x2, x3) = last_out_gaa(x2, x3) U8_gaa(x1, x2, x3, x4, x5) = U8_gaa(x1, x5) U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U7_gaaa(x1, x5, x8) U2_g(x1, x2, x3, x4) = U2_g(x2, x3, x4) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg even = even U3_g(x1, x2) = U3_g(x2) palindrome_out_g(x1) = palindrome_out_g U4_g(x1, x2, x3, x4) = U4_g(x2, x3, x4) odd = odd U5_g(x1, x2) = U5_g(x2) last_in_gag(x1, x2, x3) = last_in_gag(x1, x3) last_out_gag(x1, x2, x3) = last_out_gag(x2) U8_gag(x1, x2, x3, x4, x5) = U8_gag(x5) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: palindrome_in_g(L) -> U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd)) halves_in_gaaa([], [], [], even) -> halves_out_gaaa([], [], [], even) halves_in_gaaa(.(X, []), .(X, []), [], odd) -> halves_out_gaaa(.(X, []), .(X, []), [], odd) halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) last_in_gaa(.(T, []), T, []) -> last_out_gaa(.(T, []), T, []) last_in_gaa(.(H, T), X, .(H, M)) -> U8_gaa(H, T, X, M, last_in_gaa(T, X, M)) U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) -> last_out_gaa(.(H, T), X, .(H, M)) U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd)) U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) -> halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even)) eq_in_gg(X, X) -> eq_out_gg(X, X) U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> U3_g(L, eq_in_gg(X1s, X2s)) U3_g(L, eq_out_gg(X1s, X2s)) -> palindrome_out_g(L) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd)) U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> U5_g(L, last_in_gag(X1s, X1, X2s)) last_in_gag(.(T, []), T, []) -> last_out_gag(.(T, []), T, []) last_in_gag(.(H, T), X, .(H, M)) -> U8_gag(H, T, X, M, last_in_gag(T, X, M)) U8_gag(H, T, X, M, last_out_gag(T, X, M)) -> last_out_gag(.(H, T), X, .(H, M)) U5_g(L, last_out_gag(X1s, X1, X2s)) -> palindrome_out_g(L) The argument filtering Pi contains the following mapping: palindrome_in_g(x1) = palindrome_in_g(x1) U1_g(x1, x2) = U1_g(x2) halves_in_gaaa(x1, x2, x3, x4) = halves_in_gaaa(x1) [] = [] halves_out_gaaa(x1, x2, x3, x4) = halves_out_gaaa(x2, x3, x4) .(x1, x2) = .(x1, x2) U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U6_gaaa(x1, x8) last_in_gaa(x1, x2, x3) = last_in_gaa(x1) last_out_gaa(x1, x2, x3) = last_out_gaa(x2, x3) U8_gaa(x1, x2, x3, x4, x5) = U8_gaa(x1, x5) U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U7_gaaa(x1, x5, x8) U2_g(x1, x2, x3, x4) = U2_g(x2, x3, x4) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg even = even U3_g(x1, x2) = U3_g(x2) palindrome_out_g(x1) = palindrome_out_g U4_g(x1, x2, x3, x4) = U4_g(x2, x3, x4) odd = odd U5_g(x1, x2) = U5_g(x2) last_in_gag(x1, x2, x3) = last_in_gag(x1, x3) last_out_gag(x1, x2, x3) = last_out_gag(x2) U8_gag(x1, x2, x3, x4, x5) = U8_gag(x5) ---------------------------------------- (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: PALINDROME_IN_G(L) -> U1_G(L, halves_in_gaaa(L, X1s, X2s, EvenOdd)) PALINDROME_IN_G(L) -> HALVES_IN_GAAA(L, X1s, X2s, EvenOdd) HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> LAST_IN_GAA(.(Y, Xs), R, Rests) LAST_IN_GAA(.(H, T), X, .(H, M)) -> U8_GAA(H, T, X, M, last_in_gaa(T, X, M)) LAST_IN_GAA(.(H, T), X, .(H, M)) -> LAST_IN_GAA(T, X, M) U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> U7_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd)) U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> HALVES_IN_GAAA(Rests, Ts, Rs, EvenOdd) U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U2_G(L, X1s, X2s, eq_in_gg(EvenOdd, even)) U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> EQ_IN_GG(EvenOdd, even) U2_G(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> U3_G(L, eq_in_gg(X1s, X2s)) U2_G(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> EQ_IN_GG(X1s, X2s) U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U4_G(L, X1s, X2s, eq_in_gg(EvenOdd, odd)) U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> EQ_IN_GG(EvenOdd, odd) U4_G(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> U5_G(L, last_in_gag(X1s, X1, X2s)) U4_G(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> LAST_IN_GAG(X1s, X1, X2s) LAST_IN_GAG(.(H, T), X, .(H, M)) -> U8_GAG(H, T, X, M, last_in_gag(T, X, M)) LAST_IN_GAG(.(H, T), X, .(H, M)) -> LAST_IN_GAG(T, X, M) The TRS R consists of the following rules: palindrome_in_g(L) -> U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd)) halves_in_gaaa([], [], [], even) -> halves_out_gaaa([], [], [], even) halves_in_gaaa(.(X, []), .(X, []), [], odd) -> halves_out_gaaa(.(X, []), .(X, []), [], odd) halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) last_in_gaa(.(T, []), T, []) -> last_out_gaa(.(T, []), T, []) last_in_gaa(.(H, T), X, .(H, M)) -> U8_gaa(H, T, X, M, last_in_gaa(T, X, M)) U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) -> last_out_gaa(.(H, T), X, .(H, M)) U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd)) U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) -> halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even)) eq_in_gg(X, X) -> eq_out_gg(X, X) U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> U3_g(L, eq_in_gg(X1s, X2s)) U3_g(L, eq_out_gg(X1s, X2s)) -> palindrome_out_g(L) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd)) U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> U5_g(L, last_in_gag(X1s, X1, X2s)) last_in_gag(.(T, []), T, []) -> last_out_gag(.(T, []), T, []) last_in_gag(.(H, T), X, .(H, M)) -> U8_gag(H, T, X, M, last_in_gag(T, X, M)) U8_gag(H, T, X, M, last_out_gag(T, X, M)) -> last_out_gag(.(H, T), X, .(H, M)) U5_g(L, last_out_gag(X1s, X1, X2s)) -> palindrome_out_g(L) The argument filtering Pi contains the following mapping: palindrome_in_g(x1) = palindrome_in_g(x1) U1_g(x1, x2) = U1_g(x2) halves_in_gaaa(x1, x2, x3, x4) = halves_in_gaaa(x1) [] = [] halves_out_gaaa(x1, x2, x3, x4) = halves_out_gaaa(x2, x3, x4) .(x1, x2) = .(x1, x2) U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U6_gaaa(x1, x8) last_in_gaa(x1, x2, x3) = last_in_gaa(x1) last_out_gaa(x1, x2, x3) = last_out_gaa(x2, x3) U8_gaa(x1, x2, x3, x4, x5) = U8_gaa(x1, x5) U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U7_gaaa(x1, x5, x8) U2_g(x1, x2, x3, x4) = U2_g(x2, x3, x4) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg even = even U3_g(x1, x2) = U3_g(x2) palindrome_out_g(x1) = palindrome_out_g U4_g(x1, x2, x3, x4) = U4_g(x2, x3, x4) odd = odd U5_g(x1, x2) = U5_g(x2) last_in_gag(x1, x2, x3) = last_in_gag(x1, x3) last_out_gag(x1, x2, x3) = last_out_gag(x2) U8_gag(x1, x2, x3, x4, x5) = U8_gag(x5) PALINDROME_IN_G(x1) = PALINDROME_IN_G(x1) U1_G(x1, x2) = U1_G(x2) HALVES_IN_GAAA(x1, x2, x3, x4) = HALVES_IN_GAAA(x1) U6_GAAA(x1, x2, x3, x4, x5, x6, x7, x8) = U6_GAAA(x1, x8) LAST_IN_GAA(x1, x2, x3) = LAST_IN_GAA(x1) U8_GAA(x1, x2, x3, x4, x5) = U8_GAA(x1, x5) U7_GAAA(x1, x2, x3, x4, x5, x6, x7, x8) = U7_GAAA(x1, x5, x8) U2_G(x1, x2, x3, x4) = U2_G(x2, x3, x4) EQ_IN_GG(x1, x2) = EQ_IN_GG(x1, x2) U3_G(x1, x2) = U3_G(x2) U4_G(x1, x2, x3, x4) = U4_G(x2, x3, x4) U5_G(x1, x2) = U5_G(x2) LAST_IN_GAG(x1, x2, x3) = LAST_IN_GAG(x1, x3) U8_GAG(x1, x2, x3, x4, x5) = U8_GAG(x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: PALINDROME_IN_G(L) -> U1_G(L, halves_in_gaaa(L, X1s, X2s, EvenOdd)) PALINDROME_IN_G(L) -> HALVES_IN_GAAA(L, X1s, X2s, EvenOdd) HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> LAST_IN_GAA(.(Y, Xs), R, Rests) LAST_IN_GAA(.(H, T), X, .(H, M)) -> U8_GAA(H, T, X, M, last_in_gaa(T, X, M)) LAST_IN_GAA(.(H, T), X, .(H, M)) -> LAST_IN_GAA(T, X, M) U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> U7_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd)) U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> HALVES_IN_GAAA(Rests, Ts, Rs, EvenOdd) U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U2_G(L, X1s, X2s, eq_in_gg(EvenOdd, even)) U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> EQ_IN_GG(EvenOdd, even) U2_G(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> U3_G(L, eq_in_gg(X1s, X2s)) U2_G(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> EQ_IN_GG(X1s, X2s) U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U4_G(L, X1s, X2s, eq_in_gg(EvenOdd, odd)) U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> EQ_IN_GG(EvenOdd, odd) U4_G(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> U5_G(L, last_in_gag(X1s, X1, X2s)) U4_G(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> LAST_IN_GAG(X1s, X1, X2s) LAST_IN_GAG(.(H, T), X, .(H, M)) -> U8_GAG(H, T, X, M, last_in_gag(T, X, M)) LAST_IN_GAG(.(H, T), X, .(H, M)) -> LAST_IN_GAG(T, X, M) The TRS R consists of the following rules: palindrome_in_g(L) -> U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd)) halves_in_gaaa([], [], [], even) -> halves_out_gaaa([], [], [], even) halves_in_gaaa(.(X, []), .(X, []), [], odd) -> halves_out_gaaa(.(X, []), .(X, []), [], odd) halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) last_in_gaa(.(T, []), T, []) -> last_out_gaa(.(T, []), T, []) last_in_gaa(.(H, T), X, .(H, M)) -> U8_gaa(H, T, X, M, last_in_gaa(T, X, M)) U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) -> last_out_gaa(.(H, T), X, .(H, M)) U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd)) U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) -> halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even)) eq_in_gg(X, X) -> eq_out_gg(X, X) U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> U3_g(L, eq_in_gg(X1s, X2s)) U3_g(L, eq_out_gg(X1s, X2s)) -> palindrome_out_g(L) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd)) U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> U5_g(L, last_in_gag(X1s, X1, X2s)) last_in_gag(.(T, []), T, []) -> last_out_gag(.(T, []), T, []) last_in_gag(.(H, T), X, .(H, M)) -> U8_gag(H, T, X, M, last_in_gag(T, X, M)) U8_gag(H, T, X, M, last_out_gag(T, X, M)) -> last_out_gag(.(H, T), X, .(H, M)) U5_g(L, last_out_gag(X1s, X1, X2s)) -> palindrome_out_g(L) The argument filtering Pi contains the following mapping: palindrome_in_g(x1) = palindrome_in_g(x1) U1_g(x1, x2) = U1_g(x2) halves_in_gaaa(x1, x2, x3, x4) = halves_in_gaaa(x1) [] = [] halves_out_gaaa(x1, x2, x3, x4) = halves_out_gaaa(x2, x3, x4) .(x1, x2) = .(x1, x2) U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U6_gaaa(x1, x8) last_in_gaa(x1, x2, x3) = last_in_gaa(x1) last_out_gaa(x1, x2, x3) = last_out_gaa(x2, x3) U8_gaa(x1, x2, x3, x4, x5) = U8_gaa(x1, x5) U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U7_gaaa(x1, x5, x8) U2_g(x1, x2, x3, x4) = U2_g(x2, x3, x4) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg even = even U3_g(x1, x2) = U3_g(x2) palindrome_out_g(x1) = palindrome_out_g U4_g(x1, x2, x3, x4) = U4_g(x2, x3, x4) odd = odd U5_g(x1, x2) = U5_g(x2) last_in_gag(x1, x2, x3) = last_in_gag(x1, x3) last_out_gag(x1, x2, x3) = last_out_gag(x2) U8_gag(x1, x2, x3, x4, x5) = U8_gag(x5) PALINDROME_IN_G(x1) = PALINDROME_IN_G(x1) U1_G(x1, x2) = U1_G(x2) HALVES_IN_GAAA(x1, x2, x3, x4) = HALVES_IN_GAAA(x1) U6_GAAA(x1, x2, x3, x4, x5, x6, x7, x8) = U6_GAAA(x1, x8) LAST_IN_GAA(x1, x2, x3) = LAST_IN_GAA(x1) U8_GAA(x1, x2, x3, x4, x5) = U8_GAA(x1, x5) U7_GAAA(x1, x2, x3, x4, x5, x6, x7, x8) = U7_GAAA(x1, x5, x8) U2_G(x1, x2, x3, x4) = U2_G(x2, x3, x4) EQ_IN_GG(x1, x2) = EQ_IN_GG(x1, x2) U3_G(x1, x2) = U3_G(x2) U4_G(x1, x2, x3, x4) = U4_G(x2, x3, x4) U5_G(x1, x2) = U5_G(x2) LAST_IN_GAG(x1, x2, x3) = LAST_IN_GAG(x1, x3) U8_GAG(x1, x2, x3, x4, x5) = U8_GAG(x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 14 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: LAST_IN_GAG(.(H, T), X, .(H, M)) -> LAST_IN_GAG(T, X, M) The TRS R consists of the following rules: palindrome_in_g(L) -> U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd)) halves_in_gaaa([], [], [], even) -> halves_out_gaaa([], [], [], even) halves_in_gaaa(.(X, []), .(X, []), [], odd) -> halves_out_gaaa(.(X, []), .(X, []), [], odd) halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) last_in_gaa(.(T, []), T, []) -> last_out_gaa(.(T, []), T, []) last_in_gaa(.(H, T), X, .(H, M)) -> U8_gaa(H, T, X, M, last_in_gaa(T, X, M)) U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) -> last_out_gaa(.(H, T), X, .(H, M)) U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd)) U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) -> halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even)) eq_in_gg(X, X) -> eq_out_gg(X, X) U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> U3_g(L, eq_in_gg(X1s, X2s)) U3_g(L, eq_out_gg(X1s, X2s)) -> palindrome_out_g(L) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd)) U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> U5_g(L, last_in_gag(X1s, X1, X2s)) last_in_gag(.(T, []), T, []) -> last_out_gag(.(T, []), T, []) last_in_gag(.(H, T), X, .(H, M)) -> U8_gag(H, T, X, M, last_in_gag(T, X, M)) U8_gag(H, T, X, M, last_out_gag(T, X, M)) -> last_out_gag(.(H, T), X, .(H, M)) U5_g(L, last_out_gag(X1s, X1, X2s)) -> palindrome_out_g(L) The argument filtering Pi contains the following mapping: palindrome_in_g(x1) = palindrome_in_g(x1) U1_g(x1, x2) = U1_g(x2) halves_in_gaaa(x1, x2, x3, x4) = halves_in_gaaa(x1) [] = [] halves_out_gaaa(x1, x2, x3, x4) = halves_out_gaaa(x2, x3, x4) .(x1, x2) = .(x1, x2) U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U6_gaaa(x1, x8) last_in_gaa(x1, x2, x3) = last_in_gaa(x1) last_out_gaa(x1, x2, x3) = last_out_gaa(x2, x3) U8_gaa(x1, x2, x3, x4, x5) = U8_gaa(x1, x5) U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U7_gaaa(x1, x5, x8) U2_g(x1, x2, x3, x4) = U2_g(x2, x3, x4) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg even = even U3_g(x1, x2) = U3_g(x2) palindrome_out_g(x1) = palindrome_out_g U4_g(x1, x2, x3, x4) = U4_g(x2, x3, x4) odd = odd U5_g(x1, x2) = U5_g(x2) last_in_gag(x1, x2, x3) = last_in_gag(x1, x3) last_out_gag(x1, x2, x3) = last_out_gag(x2) U8_gag(x1, x2, x3, x4, x5) = U8_gag(x5) LAST_IN_GAG(x1, x2, x3) = LAST_IN_GAG(x1, 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: LAST_IN_GAG(.(H, T), X, .(H, M)) -> LAST_IN_GAG(T, X, M) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) LAST_IN_GAG(x1, x2, x3) = LAST_IN_GAG(x1, 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: LAST_IN_GAG(.(H, T), .(H, M)) -> LAST_IN_GAG(T, M) 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: *LAST_IN_GAG(.(H, T), .(H, M)) -> LAST_IN_GAG(T, M) 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: LAST_IN_GAA(.(H, T), X, .(H, M)) -> LAST_IN_GAA(T, X, M) The TRS R consists of the following rules: palindrome_in_g(L) -> U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd)) halves_in_gaaa([], [], [], even) -> halves_out_gaaa([], [], [], even) halves_in_gaaa(.(X, []), .(X, []), [], odd) -> halves_out_gaaa(.(X, []), .(X, []), [], odd) halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) last_in_gaa(.(T, []), T, []) -> last_out_gaa(.(T, []), T, []) last_in_gaa(.(H, T), X, .(H, M)) -> U8_gaa(H, T, X, M, last_in_gaa(T, X, M)) U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) -> last_out_gaa(.(H, T), X, .(H, M)) U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd)) U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) -> halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even)) eq_in_gg(X, X) -> eq_out_gg(X, X) U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> U3_g(L, eq_in_gg(X1s, X2s)) U3_g(L, eq_out_gg(X1s, X2s)) -> palindrome_out_g(L) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd)) U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> U5_g(L, last_in_gag(X1s, X1, X2s)) last_in_gag(.(T, []), T, []) -> last_out_gag(.(T, []), T, []) last_in_gag(.(H, T), X, .(H, M)) -> U8_gag(H, T, X, M, last_in_gag(T, X, M)) U8_gag(H, T, X, M, last_out_gag(T, X, M)) -> last_out_gag(.(H, T), X, .(H, M)) U5_g(L, last_out_gag(X1s, X1, X2s)) -> palindrome_out_g(L) The argument filtering Pi contains the following mapping: palindrome_in_g(x1) = palindrome_in_g(x1) U1_g(x1, x2) = U1_g(x2) halves_in_gaaa(x1, x2, x3, x4) = halves_in_gaaa(x1) [] = [] halves_out_gaaa(x1, x2, x3, x4) = halves_out_gaaa(x2, x3, x4) .(x1, x2) = .(x1, x2) U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U6_gaaa(x1, x8) last_in_gaa(x1, x2, x3) = last_in_gaa(x1) last_out_gaa(x1, x2, x3) = last_out_gaa(x2, x3) U8_gaa(x1, x2, x3, x4, x5) = U8_gaa(x1, x5) U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U7_gaaa(x1, x5, x8) U2_g(x1, x2, x3, x4) = U2_g(x2, x3, x4) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg even = even U3_g(x1, x2) = U3_g(x2) palindrome_out_g(x1) = palindrome_out_g U4_g(x1, x2, x3, x4) = U4_g(x2, x3, x4) odd = odd U5_g(x1, x2) = U5_g(x2) last_in_gag(x1, x2, x3) = last_in_gag(x1, x3) last_out_gag(x1, x2, x3) = last_out_gag(x2) U8_gag(x1, x2, x3, x4, x5) = U8_gag(x5) LAST_IN_GAA(x1, x2, x3) = LAST_IN_GAA(x1) 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: LAST_IN_GAA(.(H, T), X, .(H, M)) -> LAST_IN_GAA(T, X, M) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) LAST_IN_GAA(x1, x2, x3) = LAST_IN_GAA(x1) 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: LAST_IN_GAA(.(H, T)) -> LAST_IN_GAA(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: *LAST_IN_GAA(.(H, T)) -> LAST_IN_GAA(T) The graph contains the following edges 1 > 1 ---------------------------------------- (20) YES ---------------------------------------- (21) Obligation: Pi DP problem: The TRS P consists of the following rules: U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> HALVES_IN_GAAA(Rests, Ts, Rs, EvenOdd) HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) The TRS R consists of the following rules: palindrome_in_g(L) -> U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd)) halves_in_gaaa([], [], [], even) -> halves_out_gaaa([], [], [], even) halves_in_gaaa(.(X, []), .(X, []), [], odd) -> halves_out_gaaa(.(X, []), .(X, []), [], odd) halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) last_in_gaa(.(T, []), T, []) -> last_out_gaa(.(T, []), T, []) last_in_gaa(.(H, T), X, .(H, M)) -> U8_gaa(H, T, X, M, last_in_gaa(T, X, M)) U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) -> last_out_gaa(.(H, T), X, .(H, M)) U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd)) U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) -> halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even)) eq_in_gg(X, X) -> eq_out_gg(X, X) U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) -> U3_g(L, eq_in_gg(X1s, X2s)) U3_g(L, eq_out_gg(X1s, X2s)) -> palindrome_out_g(L) U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) -> U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd)) U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) -> U5_g(L, last_in_gag(X1s, X1, X2s)) last_in_gag(.(T, []), T, []) -> last_out_gag(.(T, []), T, []) last_in_gag(.(H, T), X, .(H, M)) -> U8_gag(H, T, X, M, last_in_gag(T, X, M)) U8_gag(H, T, X, M, last_out_gag(T, X, M)) -> last_out_gag(.(H, T), X, .(H, M)) U5_g(L, last_out_gag(X1s, X1, X2s)) -> palindrome_out_g(L) The argument filtering Pi contains the following mapping: palindrome_in_g(x1) = palindrome_in_g(x1) U1_g(x1, x2) = U1_g(x2) halves_in_gaaa(x1, x2, x3, x4) = halves_in_gaaa(x1) [] = [] halves_out_gaaa(x1, x2, x3, x4) = halves_out_gaaa(x2, x3, x4) .(x1, x2) = .(x1, x2) U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U6_gaaa(x1, x8) last_in_gaa(x1, x2, x3) = last_in_gaa(x1) last_out_gaa(x1, x2, x3) = last_out_gaa(x2, x3) U8_gaa(x1, x2, x3, x4, x5) = U8_gaa(x1, x5) U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8) = U7_gaaa(x1, x5, x8) U2_g(x1, x2, x3, x4) = U2_g(x2, x3, x4) eq_in_gg(x1, x2) = eq_in_gg(x1, x2) eq_out_gg(x1, x2) = eq_out_gg even = even U3_g(x1, x2) = U3_g(x2) palindrome_out_g(x1) = palindrome_out_g U4_g(x1, x2, x3, x4) = U4_g(x2, x3, x4) odd = odd U5_g(x1, x2) = U5_g(x2) last_in_gag(x1, x2, x3) = last_in_gag(x1, x3) last_out_gag(x1, x2, x3) = last_out_gag(x2) U8_gag(x1, x2, x3, x4, x5) = U8_gag(x5) HALVES_IN_GAAA(x1, x2, x3, x4) = HALVES_IN_GAAA(x1) U6_GAAA(x1, x2, x3, x4, x5, x6, x7, x8) = U6_GAAA(x1, x8) 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: U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) -> HALVES_IN_GAAA(Rests, Ts, Rs, EvenOdd) HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) -> U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests)) The TRS R consists of the following rules: last_in_gaa(.(T, []), T, []) -> last_out_gaa(.(T, []), T, []) last_in_gaa(.(H, T), X, .(H, M)) -> U8_gaa(H, T, X, M, last_in_gaa(T, X, M)) U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) -> last_out_gaa(.(H, T), X, .(H, M)) The argument filtering Pi contains the following mapping: [] = [] .(x1, x2) = .(x1, x2) last_in_gaa(x1, x2, x3) = last_in_gaa(x1) last_out_gaa(x1, x2, x3) = last_out_gaa(x2, x3) U8_gaa(x1, x2, x3, x4, x5) = U8_gaa(x1, x5) HALVES_IN_GAAA(x1, x2, x3, x4) = HALVES_IN_GAAA(x1) U6_GAAA(x1, x2, x3, x4, x5, x6, x7, x8) = U6_GAAA(x1, x8) 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: U6_GAAA(T, last_out_gaa(R, Rests)) -> HALVES_IN_GAAA(Rests) HALVES_IN_GAAA(.(T, .(Y, Xs))) -> U6_GAAA(T, last_in_gaa(.(Y, Xs))) The TRS R consists of the following rules: last_in_gaa(.(T, [])) -> last_out_gaa(T, []) last_in_gaa(.(H, T)) -> U8_gaa(H, last_in_gaa(T)) U8_gaa(H, last_out_gaa(X, M)) -> last_out_gaa(X, .(H, M)) The set Q consists of the following terms: last_in_gaa(x0) U8_gaa(x0, x1) 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: U6_GAAA(T, last_out_gaa(R, Rests)) -> HALVES_IN_GAAA(Rests) HALVES_IN_GAAA(.(T, .(Y, Xs))) -> U6_GAAA(T, last_in_gaa(.(Y, Xs))) Strictly oriented rules of the TRS R: last_in_gaa(.(T, [])) -> last_out_gaa(T, []) last_in_gaa(.(H, T)) -> U8_gaa(H, last_in_gaa(T)) U8_gaa(H, last_out_gaa(X, M)) -> last_out_gaa(X, .(H, M)) Used ordering: Knuth-Bendix order [KBO] with precedence:last_in_gaa_1 > U8_gaa_2 > HALVES_IN_GAAA_1 > U6_GAAA_2 > last_out_gaa_2 > [] > ._2 and weight map: []=1 last_in_gaa_1=1 HALVES_IN_GAAA_1=1 ._2=0 last_out_gaa_2=1 U8_gaa_2=0 U6_GAAA_2=0 The variable weight is 1 ---------------------------------------- (27) Obligation: Q DP problem: P is empty. R is empty. The set Q consists of the following terms: last_in_gaa(x0) U8_gaa(x0, x1) 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