I am struggling to convert the sum of maxterms:
((¬b ∧ ¬d) ∨ ((b ∧ (¬c ∧ d)) ∨ (¬a ∧ (b ∧ d))))
into a product of minterms.
I use Morgan and get this:
((¬b ∨ ¬d) ∧ ((b ∨ (¬c ∨ d)) ∧ (¬a ∨ (b ∨ d))))
which doesn't have an equivalent truth table.
(btw, this is a very handy site that is helping me
http://web.stanford.edu/class/cs103/tools/truth-table-tool/ )
I know drawing Karnaugh's make it easier to visualize, but I wanted to learn how to do the conversion algebrically, just using Morgan and distribution laws. Any good soul to enlight me on this puzzle?
thank you really much and sorry for the dumb question!
Martin
Best Answer
Start as follows:
$$\begin{align*} \overline{b}\:\overline{d} + b\: d\: \overline{c} + b\: d\: \overline{a}\tag{1}\\\\ \overline{\overline{\overline{b}\:\overline{d} + b\: d\: \overline{c} + b\: d\: \overline{a}}}\tag{2}\\\\ \overline{\overline{\overline{b}\:\overline{d}} \cdot \overline{b\: d\: \overline{c}} \cdot \overline{b\: d\: \overline{a}}}\tag{3}\\\\ \overline{(b+d) \cdot (\overline{b}+ \overline{d}+ c) \cdot (\overline{b}+ \overline{d}+ a)}\tag{4}\\\\ \overline{(b\:\overline{b}+ b\:\overline{d}+ b\:c+d\:\overline{b}+ d\:\overline{d}+ d\:c) \cdot (\overline{b}+ \overline{d}+ a)}\tag{5}\\\\ \overline{(b\:\overline{d}+ b\:c+d\:\overline{b}+ d\:c) \cdot (\overline{b}+ \overline{d}+ a)}\tag{6}\\\\ \overline{\overline{b}\:d+ \overline{b}\:d\:c+b\:\overline{d}+ \overline{d}\:b\:c+ a\:b\:\overline{d}+ a\:b\:c+a\:d\:\overline{b}+ a\:d\:c}\tag{7}\\\\ \overline{\overline{b}\:d+b\:\overline{d}+ c(\overline{b}\:d+ \overline{d}\:b)+ a(b\:\overline{d}+d\:\overline{b}+ b\:c+ d\:c)}\tag{8}\\\\ \overline{\overline{b}\:d+b\:\overline{d}+ a\:c(b+ d)}\tag{9} \end{align*}$$
NOTE
The last, going from (8) to (9) above, can be handled several ways. Rather than belabor things with algebra, I'll just explain in words.
It's pretty easy to see that if \$\overline{b}\:d+b\:\overline{d}\$ is true or false, then \$c(\overline{b}\:d+ \overline{d}\:b)\$ is irrelevant. If \$c\$ is false, then the earlier term \$\overline{b}\:d+b\:\overline{d}\$ still over-rides. And if \$c\$ is true, all it does is permit the additional factor \$\overline{b}\:d+b\:\overline{d}\$ to express itself. But it already has done so. So \$c(\overline{b}\:d+ \overline{d}\:b)\$ is entirely redundant. It can be removed. I could write this algebraically, adding opposite conditions and then simplifying, but I think these words are sufficent to get the point across.
A similar argument also says that the same terms, \$\overline{b}\:d+b\:\overline{d}\$, that are inside of \$ a(b\:\overline{d}+d\:\overline{b}+ b\:c+ d\:c) \$ are equally pointless and can be removed, reducing it down nicely.
At this point, there are two completely equivalent statements in (9) above. You can choose either one of them:
$$\begin{align*} \overline{\overline{b}\:d+b\:\overline{d}+ a\:b\:c}&=\overline{\overline{b}\:d+b\:\overline{d}+ a\:c\: d}\tag{10} \end{align*}$$
The reason should be obvious. Examine \$\overline{b}\:d+b\:\overline{d}+ a\:c(b+ d)\$. If \$b\$ is false and \$d\$ is true, then the first term picks that up. If \$b\$ is true and \$d\$ is false, then the second term picks that up. So the only way that matters for the third term is when either \$b\$ or \$d\$ is true. But none of this matters unless both of the first two terms are false. And the only remaining possibility is that if one of \$b\$ or \$d\$ is true, then the other must also be true. So you can neglect either one of them; your choice. Hence, (10) must be a true statement.
So I'll pick the left side now:
$$\begin{align*} \overline{\overline{b}\:d+b\:\overline{d}+ a\:b\:c}\tag{11}\\\\ (b+\overline{d})\cdot (\overline{b}+d)\cdot (\overline{a}+\overline{b}+\overline{c})\tag{12} \end{align*}$$
But you could just as well have picked the right side.
Algebra works.
Ask questions if you have any about the above process. I did a few things that you may consider "subtle" but are easily explained.