Simplify the following expression using Boolean Algebra into sum-of-products (SOP) expressions

. refers to AND + refers to OR

a'.b'.c' + a.b'.c' + a.b.c'

This is what I have so far.

a'.b'.c' + a.b'.c' + a.b.c'

= (b'.c').(a'+a) + a.b.c'

= b'.c' + a.b.c'

Anyway to simplify it further?


You can not use associativity when operators are mixed:

$$a'.b'.c' + a.b'.c' + a.b.c' \neq (b'.c').(a'+ a) + a.b.c'\tag{1}$$

It is true that $a'.b'.c' = b'.c'.a'$, by commutativity of ".", but it is not legitimate/not valid to impose parentheses to group $(a' + a)$ as you did.

You can use the distributive laws, and you might want to use Demorgan's Laws, as well.

Example: for using the distributive law, and the fact that $b' + b = 1$ $$a.b'.c' + a.b.c' = a.c'.b' + a.c'.b = a.c'(b'+b) = a.c'$$

Now we've simplified our expression to

$$a'.b'.c' + a.c'\tag{2}$$

There's a common term of $c'$ in each of these products, so we can simplify further. See what you come up with, and I'll work with you to clarify/confirm, etc. if you have any more questions.

$$a'.b'.c' + a.c'=(a'.b' + a).c' $$ $$= [(a'+a).(b'+a)].c' $$ $$= (b'+a).c' = b'.c' + a.c' $$ $$= a.c' + b'.c'$$

  • $\begingroup$ I see. Then how do I approach the question above? $\endgroup$ – Lawrence Wong Feb 3 '13 at 16:17
  • $\begingroup$ Do you know how the distributive laws work? Do you know DeMorgan's? $\endgroup$ – Namaste Feb 3 '13 at 16:19
  • $\begingroup$ distributive rule x.(y+z) = (x.y) + (x.z) I know DeMorgan rule. $\endgroup$ – Lawrence Wong Feb 3 '13 at 16:21
  • $\begingroup$ a.b′.c′+a.b.c′=a.(b′.c′+b.c′) is not equal to a.b′.c′+a.b.c′=a.c'(b′+b) ? $\endgroup$ – Lawrence Wong Feb 3 '13 at 16:25
  • $\begingroup$ I am really lost. Do you mind guiding me through this question? $\endgroup$ – Lawrence Wong Feb 3 '13 at 16:32

Please follow @amWhy's approach. Once you get it, you may try to understand this:

$$a'.b'.c' + a.b'.c' + a.b.c'$$ $$=a'.b'.c' + a.b'.c'\mathbf{+ a.b'.c'}+ a.b.c'$$ $$=b'.c'(a'+a)+a.c'(b+b')$$ $$=b'c'+a.c'$$


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