Principle of duality Contdition (math help)

Principle of duality condition:

In mathematics, duality have numerous meanings, and even though it is "a very profound and important concept in (modern) Mathematics" and "an important general theme that has manifestations in almost all areas of Mathematics", there is no single universally agreed definition that unifies all concepts of duality.

If the dual of a is B, the dual is B. As is involutional sometimes contain fixed points, dual a sometimes a itself.

(Source: Wikipedia)

Here, we will learn about the principle of duality.

Examples of principle of duality condition:

Let's look at some example problems in the principle of duality condition.

(Demorgan's law):

Can then be "a" and "b" Boolean algebra,

(a + b)' = construct "

Proof:

We have to prove the complement of a + b = construct '.

The definition of complement, is enough to show it

(a + b) + design ' = 1

(a + b).(construct "") = 0

(a + b) + design ' = (a + b) + construct "(Axiom 3 X)"

= b + a + construct (associative)

= b + (a + a').(a + b "") (Axiom 4y)

b + 1 = (a + b "") (axiom-5)

= b + (a + b "") (axiom 2y)

b + b =' + a (associative by +)

= 1 + (Axiom-5)

a = + 1 (axiom 3 X)

= 1 (Theorem 2 X)

(a + b) + design ' = 1... (1)

(a + b) 'a = ((a + b)' construct) b' associativity

(a '(a + b)) = b = (a'a + construct) b' (axioms 3 x, 4 X)

= (0 + BA') b' (axiom-5)

= (BA') b'

BB =' a' (axiom 3 X)

0.a =' (axiom 5)

= 0

(a + b) construct ' = 0.. (2)

(1) And (2), the complement of a + b is construct ' is

(a + b)' = construct "

More examples of principle of duality condition:

Boolean algebra, for all x, y 'in' B

(XY)' = X' + y'

Proof:

The definition of the complement of an element is sufficient to prove it

AB + (a ' + b') = 1

And (a ' + b') = 0

AB + (a ' + b')

= (AB + a') + b' (associativity +)

= (a + a') (b + a') + b' (axiom 4y)

= 1 (b + a') + b' (axiom-5)

= b + a' + b' (2y) or 1 a = a

b + b =' + a' (axiom 3 X)

= 1 + a' (axiom-5)

= a' + 1 (axiom 3 X)

= 1 (Theorem 2 X)

AB + (a ' + b') = 1...(1)

= (A ' + b')

b = a (a ' + b') (axiom 3y)

b = (aa ' + from ') (axiom 4 X)

b = (0 + AB') (axiom-5)

= Bab' (axiom 2y)

bb ='a (axiom 3 X)

= 0.a

a = (3 X and 2y theorem)

= 0

AB (a ' + b') = 0...(2)

(1) And (2), we have

(AB)' = a' + b'

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