LOGIC OF COMPUTER IN PAPER AS WELL AS PRACTICAL USE
LOGIC OF COMPUTER IN PAPER AS WELL AS
PRACTICAL USE
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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