 So the idea is when we want to calculate c, we look at every single rule where c could possibly appear, where it appears on the right-hand side. So that's for anything. For all of these things, when calculating the first of s, the first of a, the first of b, we look at every single rule where that non-terminal appears on the right-hand side. That's what we care about. And then, so this is why we look first at this rule. So we say, OK, it's not the last, so this doesn't apply. But it could be the last because there's an epsilon in the first of big d. And so that's why we add the follow of s to the follow of c. Then we, so that's what, and then rule four says, OK, so we're trying to calculate c, so we add the first of d to the first of c. And the last one says, if there was anything after d, because there's an epsilon in the first of d, then we could add those. But that doesn't, case doesn't apply. But we got to remember, right, we have to look at every place where c is used. So then we have to do the same thing we just did to this rule here. OK. So we say, is it the last most symbol? No. No, there's a d in the way, right? But this says, hey, if every symbol after me, after c, to the end of the rule, has an epsilon in its first set, so is it? Not for the whole rule, no. What do you mean? For the d it does, but the a it doesn't. For c here. So what symbols are after c? d. Just d. Just d. So is there an epsilon in the first set of d? Yeah. Yeah, exactly. So it's an important thing about this rule, right? It doesn't say anything about first being in a here, like in this rule. It just says, hey, if the things that come after you, if there's epsilon's in all the things that come after you, then you can add the follow of, in this case, b to the follow of a. Here this means that if there's a first set in the, if there's an epsilon in the first set, so that everything that comes after c to the end of that rule, right? And there's only d here, so that's the last one. There's an epsilon there. So then we can say, then we can add the follow of a to the follow of c, which is exactly, this is the same rule that we applied up here, right? This is the same rule when we were looking at rule one. This is the same rule we applied, and we said, is that possible? So that's why we do this here as well. And then we just go through and add the rules, and then we also look at this other one to make sure there's not anything we missed. All right, awesome. Thank you. Yeah. Sure. And the first problem, can I, so can I go through derivation for like b and a to get that same string? Or it has to be good through the first? You have to start from s. You always have, your derivations have to start with s. So both of them, so you need two or more ways to define, so I have to get another one. Go through s, right? Exactly, yeah. So that's the idea, right? Is if you can, it's part of proving that it's, exactly, proving it's ambiguous by starting from s and being able to derive the same string with different leftmost derivations. Okay, all right, cool. Yeah. Got it, sure. Any other class stuff, questions? See you post this explanation too? Yeah, I'm recording all this. All right, great. Is there any like pre-order, post-order, in-order tree walk that would somewhat correspond to fall of sets or is that just, does it not really relate? I thought maybe pre-order? In the end, that's what I got. Don't? Thanks. So because it doesn't have to do with your tree necessarily, it has to do with like all possibilities that are going to happen after you, so. Yeah, I'm not gonna make it to there. Yeah, this way, I don't think so. Okay. All right, a little bit of an answer. Any other questions? Okay.