I'm confused on your removed redundant terms. at that point I have (x+w+x)(x+w+y+z+z)(z'+x)(x'+y+z+z). So I'm removing the x in the first term, z in the second, and z in the third to get (z+w)(x+w+y+z)(z'+x)(z'+y+z) right? What did I miss there?

Made a mistake there: $(w+x)(w+x+y+z)=(w+x)$, I did the reverse. The other reduction I made is that $$(z+y+z')=(y+(z+z'))=y+1=1$$

Is it possible you could supply the answer by finding the inverse first then the sum of products for that inverse? I know its the same as you have above but I'm having more trouble that way. :(

All right, I'll write it up that way. Apparently, I was having trouble that way too :P. Are you supposed to do it algebraically, or can you just do it with a truth table?

Algebraically. With a truth table it might be huge since there are four variables.

It is surprisingly effective for this stuff. Remember that each 1 on the truth table corresponds to a term in the SOP form

Thanks for helping out further.

Also note that $w'x'y'z' + w'x' = w'x'(y'z'+1)=w'x'(1)=w'x'$

huh, formatting doesn't work here apparently. Anyway, I guess that's everything.

Honestly though, whenever you need a SOP (or POS for that matter), the truth table saves a bunch of time if you know how to use it

So you weren't able to get SOP using the F'?

I was, check the latest edit

I got it to the same point that I had gotten the last one

Do you know minterm and maxterm notation, like $$m_1 + m_2+m_5$$ or something?

For that problem? No. But in general, yes I know how to get it.

Are there any laws for x'*x' or x*x'? Similar to how x'+x = 1?

yeah. x'*x = 0, since x and (not x) is never true.

How are you getting (w+x)(w+x+y+z) = (w+x)? I'm not seeing it...

the general rule would be a(a+b) = (a+b)

remember that we can distribute addition over multiplication, so that

oh wait

hmm I guess I'm rustier at this than I thought

anyway, the truth table method never fails

:P

Can you expand the binomials individually? so, [x'w'+z][x'+y'z']? and then [x'

...then [x'+z']?

or are you using some easier method>

?

the easiest method for binomials is to just take every combination

Okay, did the truth table, the answer is going to be:

F'=m_0+m_1+m_2+m_3+m_6+m_7

I uploaded the truth table into my answer

back to binomials though

Do you have a link that explains the method you're talking about for binomial expansion?

I'll try to look for one

Thanks.

But to try to explain briefly, if we have (a+b)(c+d)(e+f)

just look at every combination of one element from each sum, and that will be the product

so we'd have ace+acf+ade+adf+bce+...

and it works for things that aren't binomials too, and it's easier than foiling every time

and thinking "okay, now firsts..."

I can't find a link

I guess you could do two at a time

and go from there

Could you apply it to the F' that we have?

F' = [x

w

maybe then I'll see it better.

I'll take a try at that later. Anyway, does the way that I got the minterms from the table make sense?

yes, absolutely.

Also, do you see how that's way less work if you're looking for the minterms anyway?

So what I was saying before, when you have a(a+b), since a implies a(a+b) we should have a(a+b)=a

If we were dealing with Max terms on that table (all rows with 0 on the F column) what does that give?

Using algebra, we could say a(a+b)=aa+ab=a+ab=a(1+b)=a

so I was right the first time

man this stuff gets confusing

As in what would be the maxterm expression for F?

What is the significance of the max term expression for F?

The heading for that column is F right? not F'

the heading for the column is F'

and if min terms gives F' then what does max terms give?

there's a "not" in front of the expression as it began

sorry have to go. Good luck, hope this has helped

Oh i see it. :P

thanks a lot for your help!