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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!

1 Answers 1

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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.

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    In such cases, you are welcome to ask what is the difficulty in which step. Just for an alternate easy way to Boolean algebra, may I know if you have read basics operations of set theory? The rules are exactly the same in both cases: $(U,\phi, \cup, \cap,')$ in set theory is exactly same as $(1,0,+,.)$ in Boolean algebra. So, if have read one, the other is obviously done (this is true for basic operations though).2011-11-01