At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
you need to create truth tables and list all possible solutions for p and q
I have a truth table that's completely filled out except for the part with the exact equation, but I'm not sure what to do with it now.
p and not q not (p and q) I think you can distribute the not and get not p or not q but I am not sure, I have not done this in a very long time. All I know is you have to create truth tables. This is the basis of learning how to write proofs down the road when you are doing undergraduate work in mathematics. Hopefully someone will answer that knows.
Ok I checked it According to Math proof demystified by Stan Gibilisco These are called De Morgan's Laws not (x and y) iff not x or not y not(x or y) iff not x and not y It still does not address your question though
(p ∧ ~q) and ~(p ∨ q) r stands for result, truth table for p AND NOT q p q r 0 0 0 0 1 0 1 0 1 1 1 0 NOT (p OR q) p q r 0 0 1 0 1 0 1 0 0 1 1 0 So no, they're not equivalent.
you should be looking for when both of these are true. It still goes back to your truth tables
@bluepig148 , your truth statements are not exactly true
Oops, what ones?
Your truth tables should present only one truth-false value per line. What you have presented are the general possible values for three variables.
There are only 2 variables in my truth table, the final column is what the boolean function returns given the two inputs.
Yes, you are right. I should have taken a closer look. They are not the same. I think, in this case, it would have been appropriate to list the actual DeMorgan laws
I started writing a response that simplifies the two expressions and compares them, but then I deleted it to respond to you because I thought my approach was wrong. Lol. I'll do a quick version.
DM's law: a AND b = NOT (NOT a OR NOT b) a OR b = NOT (NOT a AND NOT b) Comparing p AND NOT q and NOT (p OR q) Take a look at the first one and match the patterns. a is p and b is NOT q. Plugging it in, it's the same as: NOT (NOT p OR NOT NOT q) NOT (NOT p OR q) It becomes: NOT (NOT p OR q) vs NOT (p OR q) We can remove the NOT's. NOT p OR q vs p OR q Notice how both sides are being OR'd by q. Then there's NOT p vs p, so they're not the same thing.