Prove that

Prove that
2 + 2 = 4
brainlets need not apply.

$I I = 2$
$I I I I = 4$
$2+2 = I I + I I = I I I I = 4$

Two men have gay sex. Each one has 2 balls but when they fuck there's 4 balls total. Atheists BTFO.

Set theory already did my guy.

I count two fingers on my one hand. I count two fingers on my other hand. I count all fingers erected on both hands and count the humber 4.

Thus two fingers on one hand, counted with two fingers on another hand equals four fingers in total.

PRESCHOOL KNOWLEDGE WHATUKNOBOUTDATHAHA

First we need to define addition:
In general, for each [math]m\in\mathbb{N}[/math] there exists a unique function [math]A_m[/math] such that, for all [math]n\in\mathbb{N},~A_m(0)=m~\text{and}~A_m(S(n))=S(A_m(n)),~S[/math] denoting the successor function. Then [math]+:=\{((n,\,m),\,k):k=A_n(m)=k,~n,m\in\mathbb{N}\}[/math]. Addition now being defined, we can move on the the problem at hand: 2+2=[math]A_2(2)=A_2(S(S(0)))=S(S(A_2(0)))=S(S(2))=4[/math].

It's trivial

engineer detected lol

4 = 4 • 1= 1 + 1 + 1 + 1
2 = 2 • 1= 1 + 1
4 = (1 + 1) + (1 + 1)
4 = (2) + (2)
4 = 2 + 2

2+2 = s(2) + 1 = s(s(2)) + 0 = s(s(2)) = s(3) = 4

I want to make an addition to that ass, if you can contemplate what I'm implying

2+2=1+1+2=1+1+1+1=4

what axioms are you willing to accept?

No axioms are allowed.

f(x)=x+x=2x
Let x = 2
f(2)=2(2)
2*2=4=2+2

>/math]denotingthesuccessorfunction
...rather than set cardinality?

>I count two fingers
I count one finger extended upward to you.

because of how we define = + 2 and 4

Definition 1
[math]\forall n \in \mathbb{N} \exists ! s(n) \in \mathbb{N} | s(n) = n+1 [/math]
>For all natural number n exists only one natural number called successor of n s(n) such as [math]s(n) = n + 1[/math].
Definition 2
[math]s(1)= 2 = 1+1[/math] , [math]s(2)= 3 = 2+1[/math] , [math] s(3)= 4 = 3+1[/math]
Theorem [math]2+2= 4.[/math]
Proof
[math]4 = 3+1 = 2+1+1 = 2+2 \to 2+2=4 \blacksquare[/math]

[s4s] visiting

>■

wuts cumming out of her butt?

I don't know what construction of [math]\mathbb{N}[/math] you're using, but with the von Neumann construction, the cardinality of [math]n[/math] is [math]n[/math].