Answer:
Proved
Explanation:
Given
[tex]X = (a.\bar b)+(\bar a.b)[/tex]
[tex](a + b)\ .[/tex] [tex]\frac{}{a.b}[/tex]
Required
Find out why they represent the same
To do this, we simplify either [tex](a + b)\ .[/tex] [tex]\frac{}{a.b}[/tex] or [tex]X = (a.\bar b)+(\bar a.b)[/tex]
In this question, I will simplify [tex](a + b)\ .[/tex] [tex]\frac{}{a.b}[/tex]
Apply de morgan's law
[tex](a + b)\ .[/tex] [tex]\frac{}{a.b}[/tex] [tex]=[/tex] [tex](a + b) . (\bar a + \bar b)[/tex]
Apply distribution property
[tex](a + b)\ .[/tex] [tex]\frac{}{a.b}[/tex] [tex]=[/tex] [tex]a.\bar a + a.\bar b + \bar a . b + b . \bar b[/tex]
To simplify, we apply the following rules:
[tex]a.\bar a = 0[/tex] --- Inverse law
[tex]b.\bar b = 0[/tex] --- Inverse law
So, the expression becomes
[tex](a + b)\ .[/tex] [tex]\frac{}{a.b}[/tex] [tex]=[/tex] [tex]0 + a.\bar b + \bar a.b + 0[/tex]
[tex](a + b)\ .[/tex] [tex]\frac{}{a.b}[/tex] [tex]=[/tex] [tex]a.\bar b + \bar a.b[/tex]
Rewrite as:
[tex](a + b)\ .[/tex] [tex]\frac{}{a.b}[/tex] [tex]=[/tex] [tex](a.\bar b)+(\bar a.b)[/tex]
From the given parameters:
[tex]X = (a.\bar b)+(\bar a.b)[/tex]
This implies that:
[tex](a + b)\ .[/tex] [tex]\frac{}{a.b}[/tex] when simplified is X or [tex](a.\bar b)+(\bar a.b)[/tex]
