I wanted to show the maxterm expansion of the following Boolean Function below $f(A,B,C,D) = (A + B + D)(A' + C)(C + D)$
What I know:
From what I know, the domain is f(A, B, C, and D).
No terms are in standard form.
I can make use of Zero property $AA' = 0$ to manipulate each term.
The first term is easy to place in standard form. By Zero Property
$(A + B + D) + CC'$
Using the distributive law for "OR" over "AND" $A + BC = (A + B)(A + C)$, the first term can be written into standard form as $(A + B + C + D')(A + B + C' + D)$. where $A = (A + B + D')$ and $BC = CC'$
The problem is, how do I do this to the second and third terms?
You can likewise expand the second and third terms; you just need to do it in two of these transformations. For example, for the second term, you have:
$A'+C= A' + C + BB'= (A' + C +B)(A' + C +B')$
And each of those terms can be expanded. In fact, now that we recognize the pattern, we can do that in one step, for example:
$A'+C+B'= (A'+C+B'+D)(A'+C+B'+D')$
the general pattern is called Adjacency by the way:
Adjacency
$P = (P+Q)(P+Q')$
Finally, once you get used to this pattern, I think you'll be able to do several Adjacency steps at once. For example, for the third term, we'll get:
$C+D=(A+B+C+D)(A'+B+C+D)(A+B'+C+D)(A'+B'+C+D)$