Lambda Calc, Peano Axioms, & Recreating Maths

6 posts were split to a new topic: Ancient computers

That can't quite be right. The successor of n will be n+1 no matter how many times you call it. It's a good thought, though, but remember that a number is a function of functions.

Think about this:

Jens-Church-starter script pic

Yeah but HOW?

I mean, try plugging things into there and see what results you get. Like, what's 3(0)? What's 3(3)? What's 3(successor)? Etc.

Right. Thanks for the tip. I actually posted a reply to a post right as you moved it to a new topic. I was talking about the "Exploring how computers work" YT series. Sorry 'bout that.

I found that:
Jens-Church-starter 64

Jens-Church-starter 27

Jens-Church-starter script pic

I'm not quite sure what I'm doing, but I came up with
Jens-Church-starter plusMaybe

Though, it acts more like messed-up exponents.

Given F(x) = x+1, n plus q will be F(n) wrapped in F() q times, right?

2 + 3 = F(F(F(2))) = F(F(3)) = F(4) = 5

untitled script pic - 2023-12-13T194552.212

I think I may have misworded my claim.

Yes. It won't be F(n) repeatedly; it'll be, as you say, F(F(F(...(N)))). We don't have CASCADE in this game, so you (that is to say, @bluebaritone21) have to find a way to do that given the available tools.

Indeed. You seem to have solved the exercise after next. :~) You want a bigger hint?

Yes, please. I can't seem to figure this out, no matter how much random junk I try. I've tried adding the successor around the second item, but that didn't work. I tried hard-coding it, but that's cheating. I have no idea what I'm doing.

Okay. What does this do?


Don't just TRY it -- you won't learn anything from that. Actually plug SUCCESSOR into the input to THREE and see what function it reports.

I got this:
Jens-Church-starter (f)
Does successor go wherever f is? Like:
bubble

Or expanded out:
huge
(I see what you mean about ugly programs.)

This. So, tell me in words, what does that function do when you call it with a number. Let's say f(7)?

It plugs 7 into successor, then plugs THAT into successor, and finally, plugs that into Successor, aka S(S(S(7))) = 10

OH!

this(question mark)

would this work?

(Just tested it. YES! IT DOES!)

Thank You!!!!


I've almost got times:

almost times
(EDIT: No I didn't)

Cool! So, remember what 3(f) means, and we want to do 3*5=15, so how do you make 15(f) out of 3(f)?

3+3+3+3+3 (3+3 5 times)? So:

3+(3+(3+(3+(3))))

Hmm....

Remember, 3 is a function, so it's more like 3(3(3(3(3(something))))).

Thanks for the tip.

I tried
nottimes,
but it didn't work.

This doesn't either.
why

See if you can do it without using PLUS.

Thank you.