Simplify Boolean expression (ABCD) or (ABC’D’) or (A’B’CD)

Question

I know the answer but I wanna know how to get there.

The answer is (A or C)(A or B’)(A’ or B)(C or D’)(C’ or D)

This has the same truth table, and in this form you can simplify further to

(A or C) (A XOR B’) (C XOR D’)

Karnaugh map is no help.

Answer 1

Karnaugh map is no help.

Not true. Put the terms you have in a K-Map, but instead of trying to group the cells where the expression is true, focus on the ones where it is false:

OK, so this tells you that the expression is false in the case of:

$$A'C'+A'B+AB'+C'D+CD'$$

And that means that the expression is true in the case of:

$$(A'C'+A'B+AB'+C'D+CD')'=(A+C)(A+B')(A'+B)(C+D')(C'+D)$$

Now, if you want to do this purely algebraically, you'll need to do Distribution. This is no fun, and in fact a good bit of work, as it would start with:

$$ABCD+ABC'D'+A'B'CD=$$

$$(A+A+A')(A+A+B')(A+A+C)(A+A+D)(A+A+A')(A+B+B')(A+B+B')...$$

Do you see how this works? It's like a generalized FOIL principle, but instead of getting $2\cdot 2 = 4$ terms you get $4\cdot4\cdot4=64$ terms!

When you have all $64$ terms though, you will find that many terms will have a variable and its negation, and can therefore be erased, e.g. $A+A+A'$ and $A+B+B'$ can be removed.

Of the remaining ones, you can remove duplicate variables, e.g. $A+A+B'$ becomes just $A+B'$.

And finally, you can use Absorption to get rid of the more complicated terms, e.g $A+B'$ will absorb $A+C'+B'$, i.e. the $A+C'+B'$ can be removed.

In the end, you should be left with exactly the $5$ terms from your answer.