 Welcome back to another video, and in this one we're going to continue our look at logical equivalents, except slightly differently without using truth tables this time. Here are some facts that we've learned about logical equivalents, and all of these have been established with truth tables, and they're all collected at the end of your section 2.2 in what's called theorem 2.8, important logical equivalencies. Again, all of these were established by truth tables earlier, and you're familiar with these. Here's De Morgan's laws. Here's the equivalent to its original. Here's the negation of a conditional statement and a bunch of other stuff that you've seen. Having proven all of these statements, all of these equivalencies with truth tables already, we can then use them as a platform to prove more logical equivalencies without having to use truth tables. And to see how this works, we're going to go back to something we already know, that the negation of P implies Q is equivalent to P and not Q. That's actually one of these conditional statements, and we proved it with truth tables, as you did in one of the previous activities. Let's just sort of pretend we didn't know this. Let's just cross it completely out. Now I claim that I can use all the remaining equivalencies here, or some of the remaining equivalencies, none of which depend on the thing I crossed out, by the way. Those are all completely independently developed to prove this statement here. And let's do it like this. I'm going to start with the statement on the left-hand side, the negation of P implies Q. Now we have to ask ourselves, what's another way to write some or all of the elements of this statement? I'm looking especially at P implies Q, and down here in conditional statements part two, I'm just going to circle that. Okay, now that says that P implies Q is equivalent to the statement not P or Q. So what I'm going to do here is replace a piece of this left-hand side with something else to which it is logically equivalent. That will keep the entire statement, including with the negation, logically equivalent. So I'm going to replace the stuff in parentheses with what conditional statements part two allows me to say, and that's not P or Q. Again, that is true because of conditional statements part two. Very good. So now what I'm looking at at this point on the right is a negation of a disjunction, a negation of an or statement. There's some other stuff happening in here. I mean the P is negated too, but on the highest most superficial level, it's a negation of an or statement. And I'm thinking back to De Morgan's laws, especially rule number two of De Morgan's laws. So right there, it gives me a way to rewrite the negation of a disjunction. So I'm going to do that. I'm going to rewrite, De Morgan's law is going to let me rewrite this as, says if I negate a disjunction, I negate both pieces of the disjunction and flip this to an and. So I'm going to negate the not P. So it's not not P. And I have not Q and this has to go to an and. And that's true because of De Morgan's laws part two. And there's one last thing I can say here, not not P, that is handled by the double negation property. Another thing that you looked at, another fact that you looked at in a preview activity. It just says that not not P is the same as P. So here I have what I had set out to do. I have P and not Q. And that's exactly what I wanted to prove. I have a string just like in calculus or algebra, you can have a string of mathematical expressions that are all equal, even though they don't look the same, that they're all equal through algebra operations. Here what we're doing is we have a string of logical statements that are all not equal, but logically equivalent through a collection of logical moves that we've established previously through truth tables. But there's no truth tables involved in this, except under the hood. The truth tables take care of the statements that we're using, but we don't actually use truth tables in this proof here. So let's see how well we understand this idea by looking at a new equivalency that I want to prove. I want to prove that the negation of P implies not Q is equivalent to P and Q. So we're not going to be able to just jump straight from here to here using theorem 2.8. We're going to have to go through a series of statements where we gradually work like in an algebra problem, work this statement over until it turns into this statement. So what would be the first line of the proof? And more to the point in this question, what would be the best first step and what rule from theorem 2.8 would justify the best first step in a proof? So read through, I don't have enough room to reprint all these statements on here, so go to your book and read these things through or go to a previous slide and make your selection and come back when you're ready. So probably the very best thing to do here is to use conditional statements 3 and rewrite the negation of this implication. And the reason that I think this is probably the best way to start is just precisely because conditional statements 3 deals with just this situation. I'm going to rewrite that up here. I'm going to swap the P and Q out from theorem 2.8 with A and B just to keep it a little clearer here. Conditional statements part 3 would say that the negation of A implies B is equivalent to A and the negation of B. And again, the reason why I think this rule applies best out of all the rules in theorem 2.8 is because on the most superficial level the statement that you're working with is a negation of a conditional statement. There's some other stuff happening in here, the Q is negated, but on the most superficial topmost level this is the rule that most closely resembles the situation we're in. So we're going to use that rule to transform the statement here. And that's the next concept check here. If you use conditional statements 3 to transform the statement, what goes in the blank? It's going to be P and something. So what's in there? Read over your choices and carefully consider conditional statements part 3 and then come back after you pause the video. So the answer here is going to be B. It helps again to rewrite that conditional statements part 3 with different letters I think. If you're going to negate a conditional statement it's just saying that that will be A and not B. In other words this conditional statement's got a hypothesis and a conclusion and what we're going to do is take the hypothesis and the negation of the conclusion and join them with an and. And the conclusion here is not Q. So I have to negate the not Q. And that's where we get B. So let's finish off the proof here and we're just going to do this in one more step actually. So here's where we were after this first line where we get by conditional statements 3 and applying it to our situation where we had a not Q in the conclusion slot. And then just very simply to get from line 1 to line 2 I'm going to rewrite not not Q as Q by the double negation property. So there's a complete proof that the negation of P implies not Q is actually equivalent to P and Q. And we didn't use any truth tables. All the truth table activity was hidden in the parts of theorem 2.8 that we were using. Thanks for watching.