Add De Morgan's Laws

Well to do that, you have to know that

$\displaystyle P\rightarrow Q\equiv \lnot (P\land \lnot Q)$

And then do a bunch of playing around. I think only mathematicians think it's easier. That being said, now that I worked it out, I might as well write that solution as well...

Or simply

$P\Rightarrow Q\equiv \lnot P\lor Q$

Either way, good that we have this additional solution, thanks for working it out. One last thing, should we not be more precise in the difference between \rightarrow and \Rightarrow? So far we use them interchangeably, maybe we should stick to one notation.

) ah yes, that should make it more easy. I'll simplify the proof with that. And as for the arrows, I don't think students will care much... the Rightarrow html equivalent is really ugly in the wiki font in my memory, but we can make the changes if you think it's important.

Actually, the solution now doesn't use Morgan's Laws at all... It's much simpler, but I'm still unsure if in the case of MATH220, the truth table isn't what is intended. In any case, we have both now :)

Beauty! And Rightarrow or rightarrow may not be that big of a deal after all. We can still change it if people start commenting that it's confusing.

deal!