**X Xy Boolean Algebra 3**

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Concise text begins with overview of elementary mathematical concepts and outlines theory of Boolean algebras; defines operators for elimination, division, and expansion; covers syllogistic reasoning, solution of Boolean equations, In this CaSe, (a) 2. e8# is equationally compact iff 3 = 3 Pix.xy ". where 3' e!'ls and B1,, B, are complete Boolean algebras, (b) 3' e!!! is weakly injective in #1 iff 31 ey,” x.xy,”, where 3' e!!! max and B1,B, are complete Boolean algebras, (c) 3 e!" is topologically compact iff we have 3 =3'x.x3' for some sets II,,I, and some algebras 2', e! is: (d) Jeff is injective in m iff 3/ e3!,”x.xy,”, where 31, em any isomorphism between nonsingleton subalgebras of the algebras.2, extended to the For instance, it is clear that the main condition of idempotency is satisfied since the equalities xx = x/\x = x hold always. There is a theorem converse to Theorem 24 in some sense. Namely, each Boolean ring 2' can be made into a partially ordered set with the order relation defined by the following condition: x < y whenever x . y = x. Endowed with this order, 3£ is a Boolean algebra whose zero and unity coincide with those of the ring. The proof of this is not complicated. Many facts will Examples of such functions are x f x', xy*, xyz' + x'yz + xy'z in one, two, and three variables, respectively. The following definition gives a name to such functions. DEFINITION. A Boolean function.is said to be in disjunctive normal form in n variables ,TI, xz, . . , xn, for n > 0, if the function is a sum of terms of the type Afci)^^) • • • f*(xn), where /,(z,) is x± or x[ for each i = 1, 2, . . . , n, and no two terms are identical. In addition, 0 and 1 are said to be in disjunctive normal form in n variables In the second example of Section III it is shown that there is in general no embedding f of an orthocomplemented set P in a Boolean algebra which satisfies (iii)* f(x vy)=/(x)v/(y) whenever x commutes with y. Consider also the apparently weaker (iii)" f(x v y) =/(x) v/(y) whenever x 1 y. LEMMA 1.5. Let f be a homomorphism satisfying (iii)" of P in an ortho modular set Q. Then f also satisfies.(iii)*. Proof. Suppose x commutes with y. Then x = xy' v xy and y= xy v x'y so f(x v y) =f(xy' v xy v xy v Which one of the following statements is not correct? (a) X XY X + = (b) XX +Y XY ( ) = (c) XY XY X + = (d) ZX ZXY ZX ZY + = + Which of the following Boolean algebra rules is correct? (a) A A ⋅ =1 (b) A + AB = A + B (c) A AB A B + = + (d) A + (A + B) = B What are the ultimate purposes of minimizing logic expressions? 1. To get a small size expression. 2. To reduce the number of variables in the given expression. 3. To implement the function of the logic expression with least hardware. 4.Continuing in a similar manner, (xyrz) 1 (xyz) reduces to just xz, and (xyzr) 1 (xyz) reduces to just xy. x y ? zr) 1 (y((.? y z)z 1 0 is (x 1 yr 1 z)z ? (x 1 y 1 zr)? (y(( 1 z)z? 1 The duality principle states that if a Boolean expression is true, then its dual is also true. The duality principle does not say that a Boolean expression is equivalent to its dual. We will see in Section 2.7.3 that the inverse of a Boolean expression can be obtained easily by first taking the dual of that expression Now we want to give an abstract characterization of pseudotree algebras. For this purpose Let R be a disjunctive ramification set of nonzero elements and let X, Y be finite subsets of R. Then II X < XD Y iff one Obviously any of (1)–(3) implies that II X < XD. Y. Now suppose that II X = XXY and (1) and (2) do not hold. Note that if X = 0 then XD Y =.1; hence X % (). From the falsity of (2) it then follows that II Xe X and Y # (). Then disjunctiveness yields (3). L Proposition 2.14. Let R C At Rewriting /using Boolean algebra rules, we have /= (yz)' . (wx)' + w + x + y + xy + yz + yw = (yz)' . (wx)' + (w + yw) + (x + xy) + (y + yz) = (yz)' . (wx)' + w + x + y IMHKi (yz)' . (wx)' is not rewritten as (y'+ z') . (w' + x') = y'w' + x'y' + z'w' + z'x', as the modified form requires more gates and more inverters than the original form. 2.50(b) Example 2.24 Simplify the following Boolean expressions using Boolean algebra: (i) (x + y + xy)(x + z) (ii) x[y + z(xy + xz)'] (iii) xy +z + (x' + y) z! '..(i) (x + y + xy)(x