>> Okay. So these five rules
are going to be referred to as rules of replacement. These are rules where you can
exchange one thing for another. The last eight we had
learned, were referred to as rules of implication. Those are kind of
multistep processes, where you're creating
something new. These rules, you can
just exchange one thing for the other. So let's get working
on the first one. This is referred to
as, de Morgan's law. These four dots just mean "are
exchangeable with each other." So this left side, anytime you
see this left side anywhere in an argument, it doesn't
have to be the main operator. It can be buried inside of
parenthesis or set braces. You can just exchange it, with
what you see on the right side. And this goes in
either directions, that'll be for the
next ten rules you see. So here's one set, you
can exchange these two with each other, or you could
exchange the bottom pair with each other as well. So those go in either
direction as well. So anytime you see negation
parenthesis P wedge Q, you can swap that out with negative P dot
negative Q. Likewise, anytime you see negation
parenthesis P.Q, close parenthesis,
you can swap that out with negation P wedge negation
Q. I think this rule is kind of intuitive, once you think
about it with statement. So, let me walk you
through it with a statement. The first one, negation
parenthesis P wedge Q. If I go into a grocery store, it's
Thanksgiving day, and I forgot to shop, and I'm,
trying to get turkey and stuffing from the store. And the person at the
cashier counter says "Oh! I'm sorry. We neither have turkey
nor stuffing." That person is telling me,
that I'm going to walk out of that grocery store empty
handed, if I go in there. Because I'm not going to a
find turkey, I'm not going to find stuffing, and that's
what I'm trying to get. And so what that means
is, I don't have turkey, and I don't have stuffing. So you're sending the
negation through to the P, and you're sending it
through to the Q, but you have to change this wedge
operator to a dot operator. So I neither have
turkey nor stuffing, means I don't have turkey,
and I don't have stuffing. Neither a P nor Q, means not
P, and not Q. The bottom pair, now let's say going to
different grocery store, I'm looking for turkey
and stuffing again. But this time, the person
says I don't have both turkey and stuffing. But I can't remember which
one it is I didn't have, that tells me I might walk
out of there with something. I just, I'm not going to walk
out of there with both things. So if they said I don't have
both turkey and stuffing, either they don't have turkey
or they don't have stuffing. They don't, one of them
is missing for sure. They just can't remember
which one was missing. But I'm going to walk out
with at best, one thing. So that's de Morgan's rule, be
sure on this bottom one that, you send your negation through
to the P, you send the negation through to the Q. And then,
the dot operator needs to turn to a wedge operator,
thare's de Morgan's. Right, let's look at the
next one, commutativity. I think it's pretty
straightforward . Anytime two things are
connected with a wedge operator, or a dot operator you
can just switch places. So the top set says P
wedge Q, you just switch. And now, that's a replaceable. With Q wedge P, they're
just switchable. Then the bottom set says, P dot
Q, and that's just replaceable with Q dot P. Commutativity,
pretty straightforward. Next one, associativity. This one just means that,
any time you have two wedges or two dots, the parenthesis
in there becomes irrelevant. The parenthesis location
becomes irrelevant. So you can shift
the parenthesis set, in this case the parenthesis
is covering the P wedge Q, and the second case, it's
covering the Q wedge R. So you can go, parenthesis
P wedge Q, closed parenthesis
wedge R, is replaceable with P wedge parenthesis Q
wedge R, closed parenthesis. And the same thing goes
for a dot operator. Parenthesis P dot Q,
closed parenthesis dot R, is replaceable with P
dot parenthesis Q dot R, closed parenthesis. So all they're doing is shifting
the parenthesis place over, from this spot on the
left, you're just moving it over to the right one more spot. And again, these four dots
mean that you could do this in either direction, or
anytime you see either side, you can swap them
with each other. Next slide, we have
distribution. P wedge, open parenthesis
Q dot R, close parenthesis, goes to parenthesis P
wedge Q, close parenthesis, dot P wedge R, close
parenthesis. And what's happening, is you're
just taking your P wedge, and sending it to your Q, and
then sending it to your R, and your dot stays in place. So look, P wedge Q, that's
the left side of the dot, and you keep your dot in place. And then the other one goes
to P wedge, with your R, there is the right
side of your dot. Same thing happens
if you flip it. So you have P dot, parenthesis
Q wedge R, close parenthesis, is exchangeable with
parenthesis P dot Q, close parenthesis
wedge, open parenthesis, P dot R, close parenthesis. You just take your P dot, and
you carry it through to your Q, and then you carry
it through to your R, and you leave that
wedge in place. So the left side says P dot
Q, there's your P dot carried through to your Q. And then
the right side says P dot R, there's your P dot
carried through to your R, and you just leave
your wedge in place. So it's distribution. Next slide, double negation. So anytime you have
P, you can add two, or remove negation symbols, so long as you do
it in sets of two. So if you think about this
statement "my computer's on." If I say my computer is not
on, that would mean, it's off. But then if I say, it is not the
case that my computer is not on. It is not the case that
my computer is not on, so you have two negations here. What is this do to my computer,
it puts it back on again. So this means the same
thing as my computer is on. So, so long as you add or
remove negations in sets of two, it doesn't change the
meaning of the statement from a logical standpoint,
logic can't tell the difference. That's double negation, and
that's last one for today. So let's go ahead, and start
practicing with these new rules. Alright, first one. Q wedge parenthesis L
wedge C, negation C, conclusion L wedge Q. Let's
get our rules out here. We have square horseshoe
circle, square circle, square horseshoe circle. Not circle therefore, not
square, square wedge circle, not square therefore, circle. Square horseshoe circle,
circle horseshoe triangle, square horseshoe triangle,
square square wedge circle, square dot circle, square. Let's see all these rules so
far we have MP, MT, DS, HS, this is addition,
simplification, we have conch, square circle, square
dot circle. And then we have
constructive dilemma, square horseshoe circle, and
triangle horseshoe diamond, square wedge triangle,
circle wedge diamond. Then the last four we had,
negations squared dot circle, the bottom version was
square wedge circle. Then you carried your
negation through, and changed your operator. Negation square wedge
negation circle, negation squared dot
negation circle, that was DM. Then we had com square dot
circle, square wedge circle, those were just flippable. So circle dot square,
circle wedge square. Then we had distribution, we had square dot
circle wedge triangle, square wedge circle
dot triangle, that was swappable
with on the top. Square dot circle wedge
square dot triangle, and on the bottom it was
square wedge circle dot, square wedge triangle. Let's see, we need
two more in here. There was associativity, ASSOC. That was square wedge
circle wedge triangle, square dot circle dot triangle. And you could just shift
the parenthesis over. So you go, square wedge
circle wedge triangle, square dot circle dot triangle. And then the last one
was double negation, and it went square is the same
as negation negation square. Okay, with all those
rules, let's go back, and look at this here. So that looks like a setup for
a DS, except for that's a Q on the left side, and this
is a negation C. So you need that to be a C, it's not
that, that won't work. I see two wedges here, DS
is the only thing that works with a wedge, excepts like
de Morgan's would work, but we don't have
the parenthesis, there's no negations going
on, com, you could com it. Yeah okay, let's try and com it. So we'll go L wedge C,
keep those in parenthesis, wedge Q. That's going to
be 1 com, alright, line 4, anything else you
can do with that. You can't distribute
it, because they need to be a dot and a wedge. You can do associativity. So shift to the parenthesis
over -- you know what? We need to shift it over
so we get the C by itself, because then you
can have line 2. If we have associativity
here you would have, L wedge C wedge Q, like that. The L is by its self, but
we want that C exposed, and if you just do associativity
to line 1 right off the bud, that would make the C exposed. So I'm going to backup on
this one, and try this again. So we'll go line 3, we're going
to do associativity right away, we'll Q wedge L wedge C.
Now your C is exposed, and so you've got
line 1a sock ilne 4. Now, we'll do the com,
and so it'll be C wedge, Q wedge L. That's going
to be a line 3 com. And now we're set up for
that disjunctive syllogism. So you go line 5, Q wedge L is
going to be all you have left. Let me write that out. So it's C wedge Q wedge,
Wedge L, and negation C, square circle negation
square, therefore Q wedge L. And that is, can you give
us Q wedge L by, 2 and 4 DS. We're almost there, now we
just got to switch these two. They're opposites of the
conclusion, L wedge Q by 5 com, and that's going to gives us
our conclusion, and we made it. Now in this one, you
noticed I got to line 4, and then realized I was heading
in a direction that didn't work. You can leave those lines in
place, and keep building on it. That's fine, I just
wanted to erase it, so you had a clean
template to work with. But just know, so long as you're
following steps that work, you're okay, you
haven't blown it. The only thing that'll get you
in trouble, is if you do a step that doesn't, that
can't be done. So let's say you try and do
DS, but you're really working on a horseshoe operator, but DS
only works on a wedge operator, then you've made a mistake. But in this case, if I just
stayed with those lines that turned out to be taking me
in a direction that didn't work, then you just you kind
of left them there, but kept working around,
that would've been fine too. You'll find that
happens quite a bit, you can take a few different, some of these problems you
can take a few different paths to get there. Some paths are longer
than others. This one shortens
things up a bit. Had I stayed with the path
I initially started on, it just would have been a longer
proof, and that's totally fine. You can take longer
proof if you want. Alright. Let's go ahead,
and work on the next one. Oh! this looks like a lot of simplifications are
going to be happening here. So, I'm just going start by
simplifying right off the bat. So we're going to pull
the H out by one simp. And then we will, let's see 4, let's just keep simplifying
out on line one. So I'm going to do an
A sock, to get that H, to get the T by itself. So go H dot C dot T,
that's going to be 1 A sock. And now we're going to simplify
out, the H dot C, by 4 simp, this probably is going
to be doing more steps than are needed, but it's okay. Oh well, for this one we're
going to definitely have to do some simplifications here. I'm just going to go ahead and simplify everything
out of line 1. Probably, it won't
need to use them all, but just to be sure we
get them all knocked out, it seems like an easy step to
take, so let's do that first. So you have, H by 1 simp,
and then we can com that. So we'll go, C dot T
dot H, that's 1 com, and then 5 we'll pull out
the C dot T, by 4 simp, and then will pull
out the C, by 5 simp. And then, we'll com line
5, go T dot C, by 5com, and then we'll pull
out the T by 7 simp. Okay. Now let's look at line
2, what can we do with that? Shoot, there's nothing, you
could double negate that, but you can't work
inside and outside it. If we did a double
negation on this, there's two ways
that would look. So you could go negation,
negation, negation, negation F dot T. You can't
do your note double negation, when it's separated with
this parenthesis line. So you're either adding
two negations to this side, or you'd add two
negations to the end inside. And I don't think either way
would work, it's not going to do anything for us. But what else we could
do, is de Morgan this. So if we take line 2, you can
send that negation through. So it would be negation,
negation F. Then you get to change that operator to
wedge negation T, by 2 DM. Let me draw that out
on the side for us. We have negations square dot
circle, goes to negation square, wedge negation circle,
and you have to add that first one parenthesis. So this is what we're
looking at. Now inside the square
there's suck time, you have a negation F,
and the circle is your T, and so the final answers
is going to be negation, negation F wedge, negation T. And that's what we have
done here on line 9. So there's a de Morgan's. Now, okay, I see where
we're going here. Take line 10, we're going
to flip this around line 9, we're going to com it, 9 com
negation T, wedge negation, negation F. And now look
through, specifically look at lines 8 and 10, and
we're close to something. We're not quite there,
but we're close. So if we have negation T,
wedge negation, negation F, there's a rule that works
with the wedge operator. We already did a de Morgan,
so it's not that one. There's another rule that works
with a single wedge operator. And that rule is
going to be the DS. So we have square wedge circle,
you need a negation square, and that will allow
you to say circle. Now, the negation square, is
going to be that negated T. So we need two negations
in front of a T. Line 8, we don't have it there, but
we just learned the new rule of double negation, that gives
us two negations in front of that T, that's
going to be 8 DN. Now line 12, we're set for
that disjunctive syllogism, which is going to
give us negation, negation F, and that's 12. I'm sorry, that's 11 and 10. So 10, 11 DS. And our final step, line 13. We have our two negations F, we
just double negation that out, and so we have F by 12 DN. And there is that proof. Yeah! We made it. So I think here, we probably
took some extra steps. Line 1, that's line 3,
that's simplification of 1, we didn't need that
simplified out. And then, when we simplified
out the C at line 6, I don't think we ever
used line 6 again, no. So we didn't line,
need line 3 or line 6, those extra simplifications. If you don't want to do
those, you don't have to. For me I just, when I see some
easy steps, I tend to take them. It takes more steps
than is necessary. But at least, I have everything
laid out, and I can go back and play with it, if I need it. So I just kind of simplify
everything out right away. But those were wasted steps, you totally didn't
need to take those. Alright, there's that one,
let's try another one. Let's see line 4, 5. I'm assuming it's going to take
at least 6 steps probably more. Alright let's see here
-- uh, hmmm, yikes. Okay. This looks like,
maybe it's setting up for a distribution, you have
that wedge and the dot there, and this conclusion
looks like maybe, you would de Morgan your
way into that conclusion. Let me rewrite this conclusion. So we know, we might be looking for something a little
bit different here. We might be looking
for negation R wedge, negation Q. If we find this. So really if we can get
a negation R by itself, then we can add the negation Q,
and then de Morgan out of it. So if we see a negation R,
I think that's the goal. That's the thing we're looking
for in this whole proof, it's a negation R.
If we find that, then we can add negation Q, and
then we can de Morgan to this over here, the actual
conclusion. Now over on line1, it's
setting up for a distribution. So line one we have, square
wedge circle dot triangle. Is the same as square
wedge circle, and square wedge triangle. Let's put those parentheses up. And this one is an S wedge I,
dot negation J. So S wedge I, and S wedge negation J,
let's write that out. S wedge I dot S dot negation
J, this is one distribution. Now let's see --
so, so, oh look. So line 2, the the left side
of line 2, is the left side of this second part of
line 4s can, conjunct. And then the right,
left side of line 3, is the right side
of that conjunct. See, you have that
part, and that part, these are two horseshoes
and that's a wedge. There's a rule we know,
that works with this. That's the rule we
learned a while back. First off, let me simplify this,
and then see if we can't figure out what that rule is. So we need to get this
edges wedge negation J out. In order to do that, the Rule of Simplication pulls
from the left side. So we have to do a new
rule to get that out. I'm going to flip it, you
think of that new rule. S wedge negation J, dot S wedge
I, and that's going to be 4 com. And then we're going to
go S wedge negation J, by 5 simp, line 7. Okay. So we have S
horseshoe negation R, negation J horseshoe,
negation Q. If we put a dot between those two things,
this rule should start to look familiar to you. It's an old rule. Does that look familiar yet? So you have the left side here, the left side here allows
you pull out the right side, or the other right side. That's the rule of
constructive dilemma. So we need to put
these two together. We need to go S horseshoe
negation R, and negation J horseshoe
negation Q. We're doing the rule of conjunction there on lines 2
and 3, and then line 8 we can do that final line of the
constructive dilemma there. So you going to go negation
R, wedge negation Q, and that's going to be
using lines 6 and 7, and you use the rule of CD. Now, if we go back
to our conclusion, I initially thought we are
trying to get negation R by itself, and then we'd add
a negation Q. But it turns out that wasn't the right hunch, because we already have
negation R wedged negation Q, and so we're there without
this negation R by itself. So that was a faulty guess, it
didn't work but we're there. So let's see line 9, but we're
here and we know that to get from here to our
actual conclusion, was the rule of de Morgan's. So you'll go negation R
and Q, that's eight DM. And now, we finally made
it to our conclusion. So for these, I think
it's kind of useful to have your provisional guesses
or hypotheses about, you know, how you're going to
solve this through. Be willing to walk
away from them, just like that one I thought
you know, at the end of the day, we just needed to get
a negation R by itself. It turned out that never popped
up, but we had on Line 8, the negation R watch negation Q
which was the de Morgan's line. So we're close, but
not quite there, and then as you go through, just
keep revisiting your conclusion and seeing, you know,
am I any closer to it? What steps would it take to
get me to the conclusion? Alright. Well, that's
it for today's lesson, keep working on these. We have five more to go, and
then we'll be done with these. So, get working, memorize
these rules, I know it's a lot but you'll get through it. Take care, have a good
one, see you next time. Double negation, this going to
keep simplifying this line 6. I'm going to simplify out
the H by 5 simp, oops! Now, I think I probably
took some extra steps with these ones. Let's see if I can find the
specific steps we needed to do. Did we need 5? 5 com, we needed it. Do we need the T? We needed the T out of there. Do we need that H simplified? We didn't need to simplify the
H. So line 3 was an extra line, we didn't need to bother
with that simplification. We could had just jumped
straight to the com, and then simplified out
until we got the C. Actually, we didn't need to simplify out. Do we need the C
either and Line 6? We didn't need to
simplify out that C. So I guess we didn't
need line 5 either. Let's see if we could've
done it without that. So we comed it, and then --
I don't know we need that.