Yes.

The question marks in the types were obviously to not give you the types because otherwise

the problem would be trivial.

Right?

Yeah, yeah.

Here those question marks in there, right?

Yeah, those are because otherwise I would have had to put a polytype here which would

have given the answer away.

It's a placeholder.

I mean you can treat it as a type in which case you have to like unify that variable

type with one of the other types which is like perfectly fine, like ends up being the

same answer either way.

Okay.

Question one.

Yeah.

Two and three?

Yeah.

This one says classes, this one says it.

Oh, that's just a copy paste error which thankfully does not impact the question whatsoever.

It's really just a duplicate answer.

I don't know.

Yeah.

I think I wanted to move it up and then copy it up instead apparently.

I will fix that.

Okay.

The second one.

Yes.

I maybe should have warned you that you may want to use a pen and a paper.

It might have helped in this particular instance.

Okay.

So the way that I would do this without pen and paper is stare at this thing and notice,

okay, so this is one thing applied to another thing applied to another thing.

So we have three terms.

So what matters is that we are applying the second and the first one to the second and

the third.

So we apply the first one.

Therefore, we're interested in the first two arguments, m and n.

So we get rid of m and n and we replace m and n by those two in here.

So the first one goes in here.

The second one goes in here.

Now I would actually use a pen and paper.

Or I mean you already have a laptop open.

Just use an editor.

So we need to get rid of m and n and substitute by the stuff that we otherwise get.

And then what we get is lambda s, lambda o.

By the way, there is a reason why I called those s and o.

Now n is the second one.

So lambda s, lambda o, s, o.

Then the first one, lambda s, lambda o.

Are we s, s, o?