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By Garrett Birkhoff

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We first assume a A (b v c) = (a A b) v (a A c), and apply the preceding lemma to the three elements a, x (a A b) v c, y (a v c) A (b V c). These elements satisfy x ~ y by the distributive inequality. Also, a v x = a v (a A b) v c = a v c, a v y =y v a = (a v = [(a v c) A (b V c)] v a c) A [(b av c v c) V a] by L5 by IA. Likewise, a A x = x A a [(a A b) v c] A a = (a A b) v (c A a) = (a A b) v (a A c) a A y =a a A (b v c) by L5 by hypothesis. A [(a v c) A (b V c)] = [a A (a V c)] A (b V c) =a A (b v c).

Kline, Trans. AMS 18 (1917), 301-25; E. V. , 38 (1935), 1-19. t V. Glivenko, Amer J. Math. 58 (1936), 799-828, and 59 (1937), 941-56. See also E. Pitcher and M. F. Smiley, Trans. AMS ti: (1942), 95-114; and. M. F. Smiley and R. Transue, Bull. AMS 49 {1943), 280-7. §10 BOOLEAN ALGEBRAS 43 which is "between" any two of the three elements a, b, c. For the study of these properties, the reader is referred to the periodical literature. t Exercises for §§8-9: I. If one defines (a, z, b)fl by condition (i) of Lemma l, which of conditions (Bl)-(BS) holds every poset?

The prime weals of a given lattice L are the kernels of the lattice epimorphisms 0: L __,. 2. Unlike the situation in rings, maximal ideals need not in general be primeneither is the converse true. But we note THEOREM 7 (STONE [I]). and only if it i8 maximal. s prime if PRooF. If Pis a prime ideal of A, then for any a¢ P we have a' E P since a A a' = 0 E P; hence any ideal J > P contains some a ¢ P and a', and so 1 = a v a'. Conversely, let M be maximal. Suppose x A y EM and x ¢ M; then x v M > M must contain 1.

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