Boolean Algebra Simplification

[Rossy]

Homework Statement

[/B]
How to simplife it? I would like to see a process of simplification.

Y = A'B'C'D + A'B'CD + A'BC'D + A'BCD' + A'BCD + AB'C'D + AB'CD + ABC'D'

The ' denotes a bar over the previous letter.My second question is:

Y = BC + AB'D = (uploaded picture)

Y = A'D + B'D + A'BC + ABC'D' = ??

Homework Equations

Simplification Rules

The Attempt at a Solution

I tried it but can't get to the correct result.

The correct result should be (made through Karnaugh map):

Y = A'D + B'D + A'BC + ABC'D'Thank you for any advices and I'm sorry for my english.

 

[Rossy]

uploaded pic

 

[NascentOxygen]

How to simplife it? I would like to see a process of simplification.

Y = A'B'C'D + A'B'CD + A'BC'D + A'BCD' + A'BCD + AB'C'D + AB'CD + ABC'D'

Hi Rossy.

Break it down into manageable parts, e.g., look at "taking out a common factor"

A'B'C'D + A'B'CD = A'B'D(C' + C) = ?

Can you complete this simplification?

 

[Rossy]

Hi NascantOxygen,

it's A'B'D

Could you look at the image I uploaded, it's my attempt.. as you can see I can't get to the right solution.

 

[NascentOxygen]

The most obvious mistake is that you have ignored one term in the given expression, viz.,

+ A'BC'D +

 

[NascentOxygen]

In Boolean algebra you may encounter an expression such as (C + C' D) and wish to simplify it. One way is to AND the C term with something having the value TRUE, and I'll choose (1 + D) which has the value 1 because anything ORed with TRUE is TRUE.

C(1 + D) + C' D

= C.1 + CD + C'D

= C + D(C + C')

= C + D

 

[Rossy]

So if I take this expression A'BC'D + ABC'D' I can simplyfy it like this BC'(A + D)(A' + D') but how do I continue?

 

[NascentOxygen]

So if I take this expression A'BC'D + ABC'D' I can simplyfy it like this BC'(A + D)(A' + D') but how do I continue?

You can't go further with that. You have extracted the factor BC' and that's it.

 

[Rossy]

So you can't get from this expression

Y = A'B'C'D + A'B'CD + A'BC'D + A'BCD' + A'BCD + AB'C'D + AB'CD + ABC'D'

to this expression by Boolean Algebra

Y = A'D + B'D + A'BC + ABC'D'

 

[NascentOxygen]

So you can't get from this expression

Y = A'B'C'D + A'B'CD + A'BC'D + A'BCD' + A'BCD + AB'C'D + AB'CD + ABC'D'

to this expression by Boolean Algebra

Y = A'D + B'D + A'BC + ABC'D'

Yes, it can be simplified to that.

 

[Rossy]

Can't get further...

 

[NascentOxygen]

Take two of these terms (shown in green) and simplify them using the help shown in post #6

A'BC + B'D + A'BC'D + ABC'D'

 

[Rossy]

Is it right?

 

[NascentOxygen]

That is right.

 

[Rossy]

Thank you for your time and patience with me! :) The problem is solved.