Simplify $A!B + A!C + BC$ Could someone help me simplify this logical expression: $A!B + A!C + BC$ ?
I know the identities:
$A + AB = A$
$A + !AB = A + B$
$(A + B)(A + C) = A + BC$
but I'm not sure what the first step in simplification would be.
Any help would be appreciated. Thank you.
 A: Use the third identity to have:
$$(A!B+A!C)+BC\equiv(A!B+A!C+B)(A!B+A!C+C)$$
Then use the second identity to have (and commutativity):
$$(A!B+A!C+B)(A!B+A!C+C)\equiv(B+A+A!C)(C+A+A!B)$$
Finally (almost), use the first identity to have:
$$(A+B+A!C)(A+C+A!B)\equiv(A+B)(A+C)$$
...and I'll let you finish on your own from here.
A: There is no such thing as simplest form.  You can literally apply any of the transforms to your problem and claim to be done.
One form that could be considered the answer is what Asaf Karagila suggests, $(A + B)(A + C)$.  That is called product of sums.  He showed how to derive that.
Another possible answer is $A + BC$.  This is called sum of products. Another possible answer is $A!BC + AB!C + ABC + A!B!C + A!BC$.  That is just an enumeration of satisfying values.
Another possible answer is $A \text{ xor } !B \text{ xor } BC$.  Your relations don't support xor, but it is a possible "simplified answer".
To apply the identities, I suggest you look at what each relation does.  The first removes an entire variable.  The second removes a negation.  The third reduces the number of operations.  Then decide on what you need for your application and apply as necessary.
