Let M be the set of all memes.

Is this well-defined? How can you tell whether something is an element of M?

f(x) is a meme making fun of x for all x in M

Does such an f even exist? Why? Obviously it exists for some x in M but for all?

Thus there exists a normie meme n

What’s a normie meme? Why does its existance follow?

and a unique function F for all natural number k

This again requires f to be well-defined.

The set M is also equipped with a dankness norm.

Prove it has that norm and please also prove it fulfills all properties of a norm.

with property that ||F(k)|| ≤ ||F(k+1)|| for all k in N.

[proof required]. Idea for a counterexample: A meme making fun of a meme in such a terrible way it cannot possibly be “danker”. Though this would require f^-1(terrible meme making fun of meme) to not be empty.

We define a function f’:P(M)->M such that f’(A) is a meme simultaneously making fun of all memes in the set A.

We extend F transfinitely by defining F(l) where l is a limit ordinal to be f’(A) where A is the union of F(a) for a < l.

It seems possible that there might also be finite closed rings of memes that all make fun of another. In this case there would be no normiest meme, and dankness would not be well-ordered.

I would rather take my mathematical memes with adequate proof that functions are well defined, unique and existing 🤢

Oddly enough, if you take the derivative of f, you get the constant function 0.

Many interpret this to mean that “all memes are derivative”, which is true, but not the cause.

The real cause is that the value of any given meme is equal to the value of any other given meme. Doesn’t matter which one you look at, they all fail to make you laugh.

Let M and f be as in the hypotheses of the meme.

Since f is a meme, then f∈M. That means f can be applied to itself. It follows that (f, f(f))∈f. We have then:

f ∈ {f, {f, f(f)}} = (f, f(f)) ∈ f

Thus, f∈f, violating the Axiom of Regularity. We conclude the meme is mathematical bullshit and I will not have it.