I'd like to find out the equivalence existing between these two logical functions, I was trying it just by applying Boole's theorems in just one of them, but still have not arrived to a conclusion yet. $$\overline{ab}h + \overline{c}fgh + \overline{d}fgh + \overline{e}fgh = \overline{(a+b)(cde+ \overline{fg})}h$$
How should Boole's theorems be applied to demonstrate the equivalence between both functions?
Thanks so much in advance.
If you don't mind, I'll use accent marks instead of overline's ...
$$[(a+b)(cde+(fg)')]'h=\text{ (DeMorgan)}$$
$$[(a+b)'+ (cde+(fg)')']h= \text{ (DeMorgan)}$$
$$[a'b'+((cde)'fg)]h = \text{ (DeMorgan)}$$
$$[a'b'+(c'+d'+e')fg]h = \text{ (Distribution)}$$
$$a'b'h+(c'+d'+e')fgh= \text{ (Distribution)}$$
$$a'b'h + c'fgh+ d'fgh+ e'fgh$$
OK, so that is almost, but not quite the same as your left side, which seems to be:
$$(ab)'h +c'fgh+d'fgh+e'fgh$$
that is: that first term is a little different. But: with these overlines it is sometimes hard to tell the difference between $\overline{ab}$ and $\overline{a}\overline{b}$ (which is exactly why I personally avoid overlines)... So I think that is what is going on!