help w/ simplifying boolean expressions

[RichieZ]

I can't figure out how they went from setup 1 to 2 (6 terms to 3 terms)

click here to see it

 

[Beattie]

there is essentially... 2(ABCD) + !A + !C + !A!B!C!D
so
one ABCD cancels one !A!B!C!D leaving !A + !C + ABCD

where ! is the same as the bar.

It's been a long time since I have done this, so I am not sure if it's right

 

[RichieZ]

2(ABCD) + !A + !C + !A!B!C!D

isn't this only 5 terms?

there are six terms:
!A + !C + ABCD + AB!C!D + !A!B!C!D + !A!BCD

 

[dighn]

i believe they used a k-map simplification. it can be done algebraicly but i dont think you'd wnt to...
basically it involves padding each term so every term has 4 variables in them. you end up with 16 terms of 4 variables then you start the factoring and stuff. kmap does it graphically

 

[RichieZ]

Originally posted by: dighn
i believe they used a k-map simplification. it can be done algebraicly but i dont think you'd wnt to...
basically it involves padding each term so every term has 4 variables in them. you end up with 16 terms of 4 variables then you start the factoring and stuff. kmap does it graphically

We learn Karnaugh maps next week, this is supposed to be done algebraicly

 

[dighn]

Originally posted by: RichieZZZZ

Originally posted by: dighn
i believe they used a k-map simplification. it can be done algebraicly but i dont think you'd wnt to...
basically it involves padding each term so every term has 4 variables in them. you end up with 16 terms of 4 variables then you start the factoring and stuff. kmap does it graphically

We learn Karnaugh maps next week, this is supposed to be done algebraicly

in that case, expand !A and !C to be full min terms by multiplying !A by (B+!B)(C+!C)(D+!D) and !C by (A+!A)(B+!B)(D+!D). even on paper that would look complicated and if i do it here using the ! notiation ti would be unreadable 😛 (i tried in notepad and it's a nightmare)

 

[TuxDave]

I believe it's because the C' will encompass all terms with C' in it, like: ABC'D' or A'B'C'D', the same goes with the A' encompassing all other terms. So all other terms with C' goes away. You can see this by this example:

C' + ABC'D' = (1)*C' + ABC'D' = C'(1+ABD') = C'

 

[dighn]

Originally posted by: TuxDave
I believe it's because the C' will encompass all terms with C' in it, l

oh yeah... oops

 

[RichieZ]

Originally posted by: TuxDave
I believe it's because the C' will encompass all terms with C' in it, like: ABC'D' or A'B'C'D', the same goes with the A' encompassing all other terms. So all other terms with C' goes away. You can see this by this example:

C' + ABC'D' = (1)*C' + ABC'D' = C'(1+ABD') = C'

THANK YOU, damn I hate it when its this simple but I coudln't see it. I get the rest of it, I was just having trouble with that step

 

[Yomicron]

by DeMorgan's Theorem, (AC)' = A' + C'

 

[Beattie]

Originally posted by: RichieZZZZ
2(ABCD) + !A + !C + !A!B!C!D

isn't this only 5 terms?

there are six terms:
!A + !C + ABCD + AB!C!D + !A!B!C!D + !A!BCD

oops, I guess I screwed up.