**Xyxy Boolean Algebra**

Xyxy Boolean Algebra,

Xyxy Boolean Algebra,

Due to the properties of Stone space for finite algebras, this result means that every finite Boolean algebra is isomorphic to the Boolean algebra of subsets of some finite set S. Example: The following structures are most popular Boolean algebras: 1. Twovalue (or binary) Boolean algebra 2 ({01} 01) B = , ,+,⋅,, is the smallest, but the most important, model of general Boolean algebra. It has only two elements, 0 and 1. The binary operations + and ⋅ are defined by max{ } x y x y + = , and For convenient reference, here is a list of the main laws in Boolean algebra: ¢¢ = x x double complement law xx x + = x x x ́ = idempotent laws ( ) x y x y + ¢= ¢ ́ ¢ ( ) x y x y ́ ¢= ¢+ ¢ de Morgan's laws x +=11 x ́ = 0 0 annihilation.laws x x y x + ́ = ( ) xx y x ́ + = ( ) absorption laws ¢ = 0 1 ¢ = 1 0 complement laws These laws can all be proved from the axioms, just as we have done for the idempotent laws, although the proofs can be tricky unless you are given some hints. (De Morgan's Trends and Future Directions Hazarika, Shyamanta M. logical commitments is discussed. To complete the chapter, we highlight points that we consider most important for future The set complement of a regular open set is regular closed. It is well known that the regular open sets form a Boolean algebra under the operations x y x y + = ∪ : int(cl( )), x y x y ⋅ = ∩ : , and − = x X x : int( \), see (Halmos, 1963). Algebras over regular sets have been defined.by McKinsey and Tarski (1944).is taken from the classical logic based on the twovalued realization. From the point of the Boolean algebra, this principle is adequate or correct only in twovalued case. The reason is simple: the Boolean function has the analyzed Boolean algebra. In order to illustrate the main idea, we will use the Boolean function of two independent variables x and y, from the famous Boolean paper from 1848 [2]: ()() ()()()() ()()() ,1,1 1,010,11 0,011 xyxy xyxy xy φ=φ +φ −+φ− +φ−−. (1) Actually, this On an MValgebra A we may define the constant 1:= ¬ 0 and the operations: x y x y ⊗ =¬ ⊕¬ : x y x y := ⊗ ¬ x y x y x ∨ = ⊕ : ( ) , We have that (, , ,,) A ∨ ∧01 is a bounded distributive.lattice with least element 0, and greatest element 1. We shall summarize a few basic There is also a parallel between the trivial Boolean algebra 2 01 01 ≡ ∨ ∧ ({,}, , ,,), and standard MValgebra ([,], , ,) 01 0 ⊕ ¬: Any equation holding in 2, will hold in any Boolean algebra. • Categorical Equivalence: Ans. (a) Solution: Given expression Y AB AC BC = + + For standard SOP form Y ABCC ACBB BCAA = + + ++ + () () () = + + + ABC ABC ABC ABC ABC ABC + + Y ABC ABC ABC ABC = + + + This is the standard SOP expression for the Boolean expression. 65. Ans. (d) Solution: Y ABC ABC ABC ABC = + + + = + + + ABCC ACBB () () = +AB AC Complement of Y Y ABAC = + =()()ABAC = + + ( )( ) A BA C = + + ( )( ) A BA C 66. Ans. (b) Solution:.f x y xy = +⋅ =( )( ) x y x y ⋅ ⋅ = + + ( )( ) x y x y y s: E Schriider~Bernstein Theorem. If there is a oneone correspondence f between X and a subset of Y and a oneone correspondence g between Y and a subset of X, then there is a oneone correspondence between X and Y. (In terms of cardinal numbers m and n, if min and aim, then m = :1.) Proof. For every subset Z QX, let ¢(Z) = X ~ g[Y ~ f[Z]] (Fig. Dl). (Recall that, for any function h, h[C] = {h(u): 146 C}.) X Y X Y as. feFig. Dl Fig. DZ Now, 2. gz, > f[Z,] g; {[22] > Y ~ f[z,] g Y ~ 112.Actually, a number of possible sets of postulates may be used to define the algebra. The particular treatment of boolean algebra given here is derived from that of E. V. Huntington.and M. H. Stone. The author would also like to acknowledge the influence of 1. S. Reed and S. H. Caldwell on this development of the concepts of the algebra. 68 TABLE 3.10 BOOLEAN ALGEBRA RULES BOOLEAN ALGEBRA AND GATE. TABLE 3.6 X Y X Y 0 0 1 1 0 1 0 0 1 1 1 0 0 0 1 1 TABLE 3.7 Alternatively, a POS expression off can be obtained from the dual principle (defined next) applied to an SOP expression of f: Dual Principle—The dual of an expression + xy xy is equal to ( )( ) x yx y + + where AND and OR operators are interchanged; ANDs are converted to ORs and ORs are converted to ANDs, but the variable names remain the same in their complemented or uncomplemented form. In general, the dual of a Boolean algebraic rule, such as x(y.+ z) = xy + xz, results in x XX cancels, and combining XY with X + Z + XY makes the ANDOR form (XYX + XYZ + XYXY) available for any further simplification. Examination of the ANDOR form shows that, in this case, algebraic simplification is relatively simple. The first term is rewritten in the form XXY = 0, simplifying the term to 0 Y. The theorems of Boolean algebra state that conjunctively combining any variable with 0 results in a logic value of 0. Therefore the first term simplifies to 0. The last term in the