I have some homework I can't seem to figure out. The assignment causing problems is devided into two parts; The first is to determine the inverse formula for a given formula (so the S = F'). The second part asks me to simplify the answer to the first question. I've looked trough all materials I have, but I can't figure out what happens here:
(I've used the ' symbol as NOT)
The first formula is as follows:
(a·c·(b + d) + b·c'·d + a'·b)
Which, when i invert it without simplifying, results into:
(a' + c' + b'·d')·(b' + c + d')·(a + b')
So far, so good. However, when I look at the given solution for the second part of the assignment, I can't seem to figure out the logic behind the given steps.
(a' + c' + b'·d')·(b' + c + d')·(a + b')
= (a'·b' + a·c' + b'·c' + a·b'·d' + b'·d')·(b' + c + d') =
= (a'·b' + a·c' + b'·c' + b'·d')·(b' + c + d')
= a'·b' + a·b'·c' + b'·c' + b'·d' + a'·b'·c + b'·c·d' + a'·b'·d' + a·c'·d' + b'·c'·d'+ b'·d' =
= a'·b' + b'·c' + b'·d' + a·c'·d
Can anyone point me in the direction of what steps are taken here? Thanks in advance!
Best Answer
Let me try this-- I have added two additional step in bold to make the steps clear.
$(a' + c' + b'·d')·(b' + c + d')·(a + b')$
$=\mathbf{(a' + c' + b'·d').(a + b')·(b' + c + d')}$ (by coomutativity of $.$)
$= (a'·b' + a·c' + b'·c' + \underline{a·b'·d' + b'·d'})·(b' + c + d')$ (multiplying and rearranging the first two factors; commutativity and associativity of $.$ and $+$, just like addition and multiplication of real number, being used )
$= (a'·b' + a·c' + b'·c' + \underline{b'·d'})·(b' + c + d')$ (using $\mathbf{x+x.y}=x:(1+y)=x.1\mathbf{=x}$ in the underlined)
= a'·b' + a·b'·c' + b'·c' + b'·d' + a'·b'·c + b'·c·d' + a'·b'·d' + a·c'·d' + b'·c'·d'+ b'·d' (multiplying and rearranging just like one step above; also $x.x=x$ has been used in all the 2-product terms)
$=\mathbf{a'.b'+\underline{b'.c'.a+b'.c'}+\underline{b'.d'+b'.d'.(a'+c+c')}+a.c'.d}$ (note that $b'.d'+b'.d'=b'.d'$ as $x+x=x\quad\forall x$; also the rearrangement is for the next step)
$= a'·b' + b'·c' + b'·d' + a·c'·d$, using the rule $x+x.y=x$ with $x=b'.c', y=a$ in the first underline and $x=b'.d', y=(a'+c+c')$ in the second underline.