Question

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Answer 1

Answer:

Proved

Explanation:

Given

$X = (a.\bar b)+(\bar a.b)$

$(a + b)\ .$  $\frac{}{a.b}$

Required

Find out why they represent the same

To do this, we simplify either $(a + b)\ .$  $\frac{}{a.b}$ or $X = (a.\bar b)+(\bar a.b)$

In this question, I will simplify $(a + b)\ .$  $\frac{}{a.b}$

Apply de morgan's law

$(a + b)\ .$  $\frac{}{a.b}$ $=$ $(a + b) . (\bar a + \bar b)$

Apply distribution property

$(a + b)\ .$  $\frac{}{a.b}$ $=$ $a.\bar a + a.\bar b + \bar a . b + b . \bar b$

To simplify, we apply the following rules:

$a.\bar a = 0$ --- Inverse law

$b.\bar b = 0$ --- Inverse law

So, the expression becomes

$(a + b)\ .$  $\frac{}{a.b}$ $=$ $0 + a.\bar b + \bar a.b + 0$

$(a + b)\ .$  $\frac{}{a.b}$  $=$ $a.\bar b + \bar a.b$

Rewrite as:

$(a + b)\ .$  $\frac{}{a.b}$  $=$ $(a.\bar b)+(\bar a.b)$

From the given parameters:

$X = (a.\bar b)+(\bar a.b)$

This implies that:

$(a + b)\ .$  $\frac{}{a.b}$ when simplified is X or $(a.\bar b)+(\bar a.b)$