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January 11th, 2017, 06:51 PM   #1
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Test

(Excuse the title name = `Test.` I was testing the not operator and forgot to change it to something relevant.)

Apologies -
So I'm pondering this question on Boolean algebra
$\displaystyle a\overline{b}+a \overline{c}+bc$

It began as a much longer term, I've got it this far.
Now I was looking at it and I thought - well surely that must simplify more.
I tried a K-Map and it comes out, as below:-

$\displaystyle a+bc$

Problem is - I don't know what rule to use, to simplify it past the first term.
There's no common factors but the (a).

Last edited by Kevineamon; January 11th, 2017 at 06:57 PM.
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January 11th, 2017, 07:01 PM   #2
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I would try applying the distributive law to the first two terms (ab' + bc').

That is, $A (B + C) = (AB) + (AC)$, then use DeMorgan's and look for a further simplification.
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January 11th, 2017, 07:25 PM   #3
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Yeh I think that's the problem Joppy. I don't really know - how to apply over DeMorgan's across parentheses.

And I'm a little unsure about using it backwards.

I know to use DeMorgan's when given a problem, then asked how to `take it off.` Like break it up.

But I'm not 100 percent about how to `put it on.`and use it.

I found one video where the guy said - `not` each term - then - `not them all.`

Ok well say I followed that
Is this the term I should get?

$\displaystyle \overline{A}\overline{B}\overline{C}$ - ALL NOT`D ???
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January 11th, 2017, 07:32 PM   #4
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Sorry (silly me ) that would be - ABC.
Sorry trying to think through this

So using DeMorgans - A(B+C) = ABC

Edit (Nope that's wrong - truth tables are different.)

Last edited by Kevineamon; January 11th, 2017 at 07:41 PM.
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January 11th, 2017, 07:47 PM   #5
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I'll post my working below.

$\displaystyle a \overline{b} + a \overline{c} + bc$ ----- Distributive law

$\displaystyle a(\overline{b} + \overline{c}) + bc$

This is where you are confused? I should have been clear in that (as i know it) we are really applying what is a result of the DeMorgan's theorem (in this case), and not actually applying it to the entire expression.

i.e., $\displaystyle \overline{b} + \overline{c}$ is equivalent to $\displaystyle \overline{bc}$. This in itself can be verified easily enough from a truth table, and overtime will become obvious upon inspection.

Continuing we have,

$\displaystyle a(\overline{bc}) + bc$

To perhaps make things more obvious, let $\displaystyle d = bc$.

$\displaystyle a \overline{d} + d$ ------ Law of common identities gives,

$\displaystyle a + bc.$ (d = bc)

Hope this helps.
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Last edited by Joppy; January 11th, 2017 at 07:59 PM. Reason: Flow of working
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January 11th, 2017, 08:11 PM   #6
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Thanks I'll write it out and have a good look through it. This has been bugging me for quite some hours, so thanks a lot.

Still though I suppose the exploration is never wasted. I've begun to like the feeling of when maths gives up its secrets. You definitely have to spend some time with it though.
Fascinating stuff.
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January 11th, 2017, 08:23 PM   #7
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Ha lol I see it now. My personal name for this one is - "The partner goes."
Stroke him out.
I think I nearly got to earlier - but I calculated it as 1 instead, canceling the `ab` term out.
And I went back to the drawing board. Lul...

Very good

Sigh silly me.
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January 11th, 2017, 09:15 PM   #8
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Quote:
Originally Posted by Kevineamon View Post
I've begun to like the feeling of when maths gives up its secrets.
Isn't it satisfying!


Quote:
So using DeMorgans - A(B+C) = ABC
For starters, this expression is already in a minimal form, so there's not a whole lot to be gained from applying DeMorgan's. However, i will try to show below as an example, and as you will see we cycle right back around to where we started.

$\displaystyle A(B + C) = AB + AC$

Negate the entire expression,

$\displaystyle \overline{AB + AC}.$ Let $\displaystyle AB = D, AC = E$, then, $\displaystyle \overline{AB + AC} = \overline{D + E}$.

Use DeMorgan's Law: $\displaystyle \overline{X + Y} = \overline{X} \overline{Y}$

$\displaystyle \overline{D + E} = \overline{D} \overline{E} = (\overline{AB}) (\overline{AC})$

Again using DeMorgan's Law: $\displaystyle \overline{XY} = \overline{X} + \overline{Y}$

Finally we have,

$\displaystyle (\overline{A} + \overline{C})(\overline{A} + \overline{B}).$

Note that the truth tables are different. In fact they are the exact 'opposite' if you will. i.e.,

$\displaystyle \overline{(\overline{A} + \overline{C})(\overline{A} + \overline{B})} = A(B + C).$

DeMorgan's rules are really useful if we want to switch from SOP to POS expressions (not sure if you've covered this yet).

Note that if we didn't use the distributive law at the start the process would have been much simpler, albeit not in a very useful form.

$\displaystyle A(B + C) $

Negating entire expression,

$\displaystyle \overline{A(B + C)} = \overline{A} + \overline{(B + C)} $

$\displaystyle = \overline{\overline{A} + \overline{B} \overline{C}} = (\overline{A})(\overline{\overline{B} \overline{C}}) = A(B + C)$
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January 12th, 2017, 11:31 AM   #9
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Yes I see - Interesting, however as you say - the initial expression is the exact opposite of the product. I would be thinking therefore, we shouldn't use that.
Wouldn't that be like using an identity that isn't equal?

Perhaps then this is why the guy in the video said:-

Not each term - then not them all?

A(B+C)

$\displaystyle \overline{\overline{A}(\overline{B}+\overline{C}})$

$\displaystyle (\overline{\overline{A}\overline{B}}) {\overline{C`}}$

(A+B)C
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January 12th, 2017, 07:20 PM   #10
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It depends what you want. Consider actually building the circuit.

In certain cases you may have limited gates. Maybe only have access to a pile of NAND gates and you need to build some circuit that requires OR gates. Putting it in a different form can allow you to work around this.

EDIT: This is only really a hardware concern though. As far as i know, this would rarely (if ever) be a problem in CS or programmable logic.
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