Question

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Answer 1

Proved

Step-by-step 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)`