The Algebra of Logic (Paperback)

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1911 edition. Excerpt: ...U V are not equivalent.2 1 Likewise if we make u = o, J/ = 1, we obtain the equalities (a = o), (=i), which assert still more than the given inclusion. 'According to the remark in the preceding note, it is clear that we have JJ (0 = bu) = (a = b = o), (a + - = ) = (a = b = I), since the equalities affected by the sign T may be likewise verified by the values u = o, # = I and z = o, v = I. If we wish to know within what limits the indeterminates a and v are variable, it is sufficient to solve with respect to them the equations (ab) = (a = bu), (a ) = (a + v = b), or ab'=a'bu-ab'--ati, ab'=ab'--b'v--a'bv', or a'w-)-abu = o, a'b'v + a'bv = o, 35. The Expression of a Double Inclusion by Means of an Indeterminate.--Theorem. The double inclusion is equivalent to the equality x = au + bu together with the condition (b a), u being a term absolutely indeterminate. Demonstration.--Let us develop the equality in question, x(a'u + b'u) + x' (au + bu) = o, (ax + ax')u + (b' x + bx')u = o. Eliminating u from it, a'b'x + abx = o. This equality is equivalent to the double inclusion abxa + b. But, by hypothesis, we have (b a) = (ab = b) = (a + b--a). The double inclusion is therefore reduced to b Cx a. So, whatever the value of u, the equality under consideration involves the double inclusion. Conversely, the double inclusion involves the equality, whatever the value of x may be, for it is equivalent to a' x + bx = o, and then the equality is simplified and reduced to / axu + b'xu =0. 1 from which (by a formula to be demonstrated later on) we derive the solutions u = ab-f-w (a + b'), v = db--w a--b), or simply u = ab-f-wb', v = a b--wa, w being absolutely indeterminate. We would arrive at these solutions simply by asking: By what term must we multiply b in...