LOGIC OF COMPUTER IN PAPER AS WELL AS PRACTICAL USE


LOGIC OF COMPUTER IN PAPER AS WELL AS PRACTICAL USE
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We always use logic in every day matters, so do the machines.  But we were taught by experiences we count and the people we trust and these things matters most because it will help in dark times.  But to make the machine understand the logic is very hefty work. But this is certainly not impossible.  In order to understand logic we use integrated circuits.  But to do this we need to design the integrated circuits on paper.  This requires certain rules of algebra.  This I am going talk about the rule of an algebra called the duality principal.

It states that every algebraic expression deducible from the postulates of Boolean algebra remains valid if the operators and identity elements are interchanged.  In a two value Boolean algebra, the identity elements and the elements of the set B are same 1 and 0.  The duality principal has many applications.  If the dual of an algebraic expression is desired, we simply interchange OR and AND operators and replace 1’s by 0’s and 0’s and 1’s.

I am going to show some of the rules below which some are Boolean algebra theorems and some are postulates, there Is some problem because the notation contains some . which is sometimes misunderstood.  Now the theorems and postulates listed are the most basic relationships in Boolean algebra.  The theorems, like the postulates are listed in pairs each relation is dual of the one paired it.  The postulates are basic axioms of the algebraic and need no proof.  The theorems must be proven from the postulates.  The proof of the theorem with one variable is presented below.  At the right is listed the number of postulates which justifies each step of the proof.

Postulate 2
(a) x + 0 = x
(b) x.1 = x
Postulate 5
(a) x+ x’=1
(b) x*x’ = 0
Theorem 1
(a) x + x=x
(b)x*x=x
Theorem 2
(a)x + 1= 1
(b)x*0=0
Theorem 3, involution
(a)(x’)’ =x

Postulate 3 commutative
(a) x + y = y + x
xy = yx
Theorem 4 associative
(a) x + (y + z) = (x + y) + z
(b) x(yz) = (xy)z
Postulate 4 Distributive
(a)x(y+z) = xy +xz
(b) x + yz=(x + y)(x + z)
Theorem 5, DeMorgan
(a)(x+y)’ = x’y’
(b) (xy) = x’ + y’
Theorem 6, Absorption
(a) x + xy = x
(b) x(x+y) = x





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