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

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>>8343285

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

Then I will answer you.

>>8343289

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

How many shits did I take?

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>>8343301

Define "shit"

2+2=11

>>8343285


o o
o o

>>8343319

Prove that there are 4 circles.

>>8343285
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

>>8343344
>Prove that there are 4 circles.

I never said how many circles there were.

You did.

>>8343301
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?

>>8343403
Oooooooooooooh fucking rekt sooooooon

∆∆ + ∆∆ = ∆∆∆∆

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>>8343285
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.

>>8343561
Define define

>>8343285
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

>>8343301
billions and billions of shit molecules.

>>8343569
you first

>>8343561
Even better:

Let F[x,y,z] be the free group generated by x,y, and z and let G=F[x,y,z]/<xyxz^{-1}>.
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.

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>>8343403
rekd

>>8343678
you are a monster

Also, prove that the free group exists :)

>>8343857
Prove math exists

>>8343301
Hopefully none, taking shit out of the toilet is missing the point of toilets

>>8344922
fucking kek

>>8343678
even better? this is pedantic, ridiculous, and off-point. anon got it right before, this is dumb

>>8343289
Gorilla poster fighting another shitposter?

>>8343315
2+2=22 pleb

>>8343403
YEET

>>8343403
*applause*

>>8343285
>nazi frog
>>>/pol/