By Anderson J. M.

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For each set of conditions for which the function is to be 1, a corresponding term is included in the disjunctive normal form selected, as indicated in the proof of Theorem 2. The sum of these terms gives the function, although not necessarily in simplest form. The following example indicates this method. Some short cuts for simplification will be pointed out in Chapter 5, but for the present any simplifying will be performed in the usual way after the function is obtained in disjunctive normal form.

Proof. The reasons for the steps in the proofs have been omitted, but should be supplied by the reader. (a) x c y is equivalent to xy' = 0 and y c z is equivalent to yz' = 0. Hence xz' = xz'(y + y') = xyz' + xy'z' = 0 + 0 = 0. But xz' = 0 is equivalent to x c z, which was to be shown. (b) From xcy and x c z we have xy' = 0 and xz' = 0. Hence xy' + xz' = 0 and x(y' + z') = 0. But by Theorem 9, y' + z' = (yz)' and thus x(yz)' = 0 or, equivalently, x c yz. (c) From x c y we have xy' = 0 and hence x(y + z)' = x(y'z') _ (xy')z' = 0.

In the following chapter on symbolic logic, we will see that it is often helpful to think of logical problems as problems in an algebra of sets. That this is possible is due to the representation theorem. REFERENCES ANDREE, R. , Selections from Modern Abstract Algebra, Holt, 1958. BIRKHOFF and 1IAcLANE, A Survey of Modern Algebra, MacMillan, 1948. HUNTINGTON, E. , Postulates for the Algebra of Logic, Trans. Am. Math. Soc. 5 (1904). JOHNSON, R. , First Course in Abstract Algebra, Prentice-Hall, 1953.

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