00:01
So in this problem, we're asked to prove this congruency that's called the distributive law for conjunction over disjunction.
00:11
And the way to prove this is to set up first, just set up a truth table.
00:18
And so we start with the left -hand side.
00:20
So we start with q and r.
00:23
And they can either be true or false.
00:26
So they can both be true.
00:28
Q could be true with r -faults.
00:31
I want these a little bit closer because i got several columns to put on here.
00:42
The other way around, false and true.
00:45
Or they could both be false.
00:48
Okay, so now, starting with the left -hand side, q, or, r.
00:54
The v there is an or.
00:59
This is an or.
01:00
So that makes this one an and, okay? a little hat there.
01:06
The little mountain thing is an and.
01:09
Okay, so if this is an or, as long as one of these two is true, this is going to be true.
01:14
So this is true, this is true, and this is false.
01:19
Okay, now let's set up p.
01:23
We'll make p.
01:24
P has to either be true or false, right? while q or r, that combination is true or false.
01:32
So i could set up true here and false here.
01:36
And then when they're false, i could set up true, and i'll set up another fault.
01:41
Here to make it false.
01:43
So let me do this one again, just for completeness of my table.
01:47
Okay, so then that makes p and q or r.
01:53
This is an and, which means it's going to be in both.
02:03
No, it just has to be in one of them.
02:11
It has to be in both.
02:15
So this is true, and then this is false, since p's false, and then this one's false, and this one's false.
02:28
Now let's look at the right hand side.
02:31
P and q.
02:34
Okay, so p is here.
02:38
Q is here...