Simplify Boolean Algebra: F = (AB’C + A(BC)’+ (ABC)’)’

[pirtu]

Can someone help how to simplyfy this boolean algebra? F = (AB’C + A(BC)’ + (ABC)’)’ answers should be ABC but I can`t make it, my calculations gives always wrong answer.

 

[mathman]

Can someone help how to simplyfy this boolean algebra? F = (AB’C + A(BC)’ + (ABC)’)’ answers should be ABC but I can`t make it, my calculations gives always wrong answer.

Show your work - it could help analyzing what is going on.

 

[pirtu]

Here is my "last try" so far:
F = (AB’C + A(BC)’ + (ABC)’)’
F = A` B``C` + A`B``C`` + A``B``C``
F = A`BC` + A`BC + ABC
F = A´BC` + BC (A´ + A) >>>>: A` + A = 1
F = BC + A`BC´
F = B (C + A´C´)

 

[dodo]

Hi, pirtu,
yes, the expression does simplify to ABC. (And, also, something that can always be checked before you engage in attempts at proofs is that the truth table of your expression coincides with the truth table for ABC. Which it does.)

Rather than me giving the answer, why don't you show what you have done so far? Then someone (or me) can tell you where you veered off.

EDIT: someone was faster than me. :)

EDIT2:
pirtu:
On your first step, you tried to apply DeMorgan's Laws, but forgot to change the ANDs to ORs and vice versa.
And that would yield something too long... maybe it is easier if you apply DeMorgan's only to the inner terms that contain parenthesis, as a first step; the resulting expression should be easier to manage.

 

[pirtu]

tried something different but still no, can´t understand.
F = (AB’C + A(BC)’ + (ABC)’)’
F´ = AB´C + AB´C´ + A´B´C´
F` = AB´ (C + C´) + A´B´C´
F´ = AB´ + A´B´C´
F´ = B´ (A + A` C`)
F` = B` (A + C`)
F = B (A´ + C)

 

[dodo]

Watch your first step: (BC)' is not B'C'... it is (B' + C'). Remember to interchange ANDs and ORs when to reduce negations.

Also, if you are working with the inside terms only, preserve the outer negation. If you have an expression that looks like (blah blah blah)', and you want to transform the "blah blah blah" inside, then the outer ()' needs to stay there.

Here is a general hint: if you started with something like (blah blah blah)' and want to arrive at ABC, notice that your destination ABC can be transformed (using DeMorgan's Laws) as ---> ABC = (A' + B' + C')'. So, if you can make the inner "blah blah blah" look like A' + B' + C', you know that you're in the right direction -- no need to worry too much with the outer ()'.

EDIT:
Here is a summary of DeMorgan's Laws, if you're having trouble with them:
(A + B + C + D + E + ... + Z)' = A'B'C'D'E'...Z'
(ABCDE...Z)' = A' + B' + C' + D' + E' + ... + Z'
The negation of a sum becomes a PRODUCT of negations, and the negation of a product becomes a SUM of negations.

 

[pirtu]

I´m confused. Tried to change "sums", here is my result then:
F = (AB’C + A(BC)’ + (ABC)’)’
F` = AB`C + A (B` + C´) + (A´ + B´ + C´)
F´ = AB´C + AB´ + AC´ + (A´ + B´ + C´)
F´ = AB´ (C + 1) + AC´ + (A´ + B´ + C´)
F´ = AB´C + AC´ + (A´ + B´ + C´)
F´ = A (B`C + C´) + (A´ + B´ + C´)
F` = A (C´ B´) + (A´ + B´ + C´)
F´ = AC´ + AB´ + (A´ + B´ + C´)

 

[dodo]

Ahhh, sorry, I didn't notice the first time that you changed F = ... into F' = ..., so the outer negation is there alright.

Your first step is now fine. Then you distributed, and took AB' as a common factor... but (C + 1) is not equal to C. It is equal to 1. ("Anything OR True" is always True.) Check the step from line #4 to line #5, and then go ahead... you're doing OK.

 

[pirtu]

and new try, why this does not open to me?!? :)
F = (AB’C + A(BC)’ + (ABC)’)’
F` = AB`C + A (B` + C´) + (A´ + B´ + C´)
F´ = AB´C + AB´ + AC´ + (A´ + B´ + C´)
F´ = AB´ (C + 1) + AC´ + (A´ + B´ + C´) >>>>: (C + 1) = 1
F´ = AB´ + AC´ + (A´ + B´ + C´)
F´ = A (B´ + C´) + (A´ + B´ + C´)

Thank you Dodo for your help so far, I could not have made it this far without your help!

 

[dodo]

By the next-to-last line, where you have AB' + AC' + A' + B' + C' (the last set of parenthesis is superfluous, as the "sum" (OR) is associative... all "sums" are at the same "level", so to speak), you are almost there.

You would like AB' + AC' to go away, so that you are left with A' + B' + C'... therefore your strategy should be to get rid of the AB' + AC' somehow. Notice that B' is repeated, and that you can take it as a common factor... and the same with the repeated C'.

 

[pirtu]

Am I going right ridection if I change it like this:
F´ = (A + B´) * (A + C´) + A´ + B´ + C´
F´ = F = A + (B´C´) + A´ + B´ + C´
I don't know how to make B´and C´ as a common factor?

 

[dodo]

I'm not sure how did you get to F' = (A + B')(A + C') + A' + B' + C' ... but what I meant was something much simpler.

In post #9, next-to-last line, you had F' = AB' + AC' + A' + B' + C'. Changing the order a little, you have F' = A' + (AB' + B') + (AC' + C'). See it now? (The parentheses are superfluous, but I just intend to show what can be combined with what.)

By the way, your last move in post #9 could be fine, it is no error; I just wanted you to see the common factors first. You could very well go from that last line, F' = A (B' + C') + A' + B' + C', write it as F' = A (B' + C') + A' + (B' + C'), and recognize that you have a (B' + C') repeated (that is, the two common factors I mentioned before, but in one sweep of hand).

 

[pirtu]

Thank you very much! Now I got it:
F' = A' + (AB' + B') + (AC' + C') >>>>: (AB' + B') = B´ ja (AC' + C') = C´

 

[dodo]

You're welcome!... when you write AB' + B', the second B' is actually B' * 1, so you can take B' as a common factor and write (A + 1) B'... and you already know that A + 1 = 1.

If, instead, you had A(B' + C') + A' + B' + C', the second B' + C' is also (B' + C')*1, and (B' + C') can be taken as common factor, so you can write (A + 1)(B' + C') + A', and again A+1 = 1. Any of the two roads are equivalent.

One last piece of advise, pirtu: exercises are not just practicing stuff. Most of us learn our mathematics while doing the exercises, not the first time the teacher said something. So don't worry if you don't know everything *before* doing an exercise, because that's normal. Exercises are there to teach you something, not to practice things that you already know completely.