Prove 2+2=4 without asking me to have faith

Prove 2+2=4 without asking me to have faith.

Define "2"
Define "4"
Define "+"
Define "="

Then I will answer you.

Let's say I took a shit. Then I took another shit. Then I took two more shits.

How many shits did I take?

Define "shit"

2+2=11

o o
o o

Prove that there are 4 circles.

DEFINITIONS:
a=a for all a
2 = ss0
4 = ssss0
a+0=a for all a
a+sb=s(a+b) for all a and b

2+2=4 iff
ss0+ss0=ssss0 iff
s(ss0+s0) = ssss0 iff
ss(ss0+0) = ssss0 iff
ssss0 = ssss0 iff
true

qed, chum

>Prove that there are 4 circles.

I never said how many circles there were.

You did.

is 1 shit = 1 turd?
and what qualifies as a turd? is there a minimum measurement? because I would argue a dingleberry doesn't qualify as a turd but it's still a piece of shit. And what happens if the shit hits the bowl and snaps in half? does it still count as one shit, or does it become two shits?

Oooooooooooooh fucking rekt sooooooon

∆∆ + ∆∆ = ∆∆∆∆

Define 0 to be the empty set { }.

Define the successor function S(a) = a U {a} for all a.

Define 1 to be the successor of 0, which is equal to 0 U {0} = {{ }}.

Define 2 to be the successor of 1. This is 1 U {1} = {{ }, {{ }}}.

Define 3 to be the successor of 2, which is 2 U {2} = {{ }, {{ }}, {{ }, {{ }}}}.

Define 4 to be the successor of 3, which is 3 U {3} = {{ }, {{ }}, {{ }, {{ }}}, {{ }, {{ }}, {{ }, {{ }}}}}.

Define the addition operator as follows: a + 0 = a and a + S(b) = S(a + b).


We shall now prove that 2 + 2 = 4.

We know that 2 + 0 = 2, by the definition of the addition operator.
We also know that 2 + 1 = 2 + S(0) = S(2 + 0).
And we also know that 2 + 2 = 2 + S(1) = S(2 + 1).

Okay, so 2 + 0 = 2. That means that S(2 + 0) = S(2) = 3 = 2 + 1.
Therefore, S(2 + 1) = S(3) = 4. Therefore, 2 + 2 = 4.

Define define

Okay so take a pencil, and then another, and then another, and then another.
We gave that the arbitrary number of four
You'll notice we can split that into one pencil and another, and one pencil and another
we gave those groups the arbitrary name two

axioms

billions and billions of shit molecules.

you first

Even better:

Let F[x,y,z] be the free group generated by x,y, and z and let G=F[x,y,z]/.
Let [math] \pi: F[x,y,z]\to G [/math] be the quotient map.
Define the symbols '2', '+' and '4' as follows:

[math] 2=\pi(x) [/math]
[math] +=\pi(y) [/math]
[math] 4=\pi(z) [/math]

Then the equation 2+2=4 holds in the group G.

rekd

you are a monster

Also, prove that the free group exists :)

Prove math exists

Hopefully none, taking shit out of the toilet is missing the point of toilets

fucking kek

even better? this is pedantic, ridiculous, and off-point. user got it right before, this is dumb

Gorilla poster fighting another shitposter?

2+2=22 pleb

YEET

*applause*

>nazi frog