Church–rosser Theorem Puns

Rosser's theorem with improved bounds gives an inequality of

ln(n) +ln(ln(n)) -1 < P(n)/n < ln(n) +ln(ln(n)) for n>= 6

I assumed there lies a function f(n), where f(n) always outputs a value of -1 < f(n) < 0 and will eventually start converging for larger N.

It can be written in the form of:

P(n)/n = ln(n) + ln(ln(n)) + f(n) (for n>=6)

By taking f(n) = P(n)/(n) - { ln(n) + ln(ln(n))}

Computing different values. I did find that

e^f(n) converges at around 0.385

i.e for n = 10^(12),

e^f(n) ~ 0.3852.....

But the convergence is rather slow and keeps slowing down.

I also got this inequality

0.374 < e^(Pn/n)/(nln(n)) < 0.833

(for n >= 6)

Hence,

The inequality can be written as

0.374 < e^f(n)< 0.833

I don't know how why I got the result. Any help would be appreciated. Also, I don't know whether I should keep trying to work on this project. Thank you.

I'm trying to understand the CR theorem; ie if you have a lambda expression M that beta reduces to N1 and N2 then there is a common reduct N3.

How does this apply to an expression like M = (λa.b)(λx.xxx)(λx.xxx)?

It seems like if you take the leftmost redex you get (λa.b)(λx.xxx)(λx.xxx) -> b(λx.xxx) = N1

But if you take the middle redex you get: (λa.b)(λx.xxx)(λx.xxx) -> (λa.b)(λx.xxx)(λx.xxx)(λx.xxx) = N2

And in this case it's not clear to me that there is a common N3 such that N1 -> N3 <- N2 ?

Thanks!

Just incase anyone missed it because it premiered on the same dat as Battle in the Valley - Tom Lawlor literally ate Fred Rossers hair on the last episode of strong

So Filthy.. Great angle though!