who else struggling with the foundations of arithmetic here?

>>8036701
We don't

i'm still bugged with the fact that two negs make a positive

>>8036706

>i'm still bugged with the fact that two nigs make a poz

>>8036706
multiplication was a stupid idea anyway

>>8036706
a - a = 0 = -0 = -(a - a) = (-a - - a)
Therefore -a is the additive inverse of --a.
But the additive inverse is unique. (because a - a = 0 = a - b implies -a = -b -> a = b)
Therefore --a == a.

>>8036701
It's the axiom of identity. The idea is you can't do math (or logic for that matter) without this axiom. You are however free to make up your own system if you so desire.

>>8036701

construct the natural numbers with the succession function and you have 1=1 by the axiom of extensionality.

>>8037016
Fucking this.

Literally all you need is the sucessor function, the iteration theorem, and zero.
Boom. Arithmetic.

>>8036715
:^)

>>8036731
How this -(a - a) became this (-a - - a)?

>>8038839
Expanding

>>8036706
x-x=0 so take: a - (b - c)

Adding zero twice:
a - (b - c) = a - (b - c) + b - b + c - c

by commutation:

= a - (b - c) + b - c - b + c

letting b - c=d

= a - d + d - b + c

=a - b + c

Thus:

a - (b - c) = a - b + c

>>8036701
The Poincare Conjecture implies that 1 is about 1. It does look like itself. As such, it's probably useful to use such identities simply.

>>8036701
We do not.
The concept of identity is fundamental and MUST be true, otherwise literally everything you are doing is a waste of time.

Technically you can prove that it is a universal truth, but it is quite redundant.

>>8036701
Its an axiom, you dont need to prove it

>>8038839
are you 12?

File: picdump-16-04-29-003.jpg (44KB, 405x720px) Image search: [Google]

44KB, 405x720px

eng here.

ok, lets say a = a isnt explaining itself as proof.

but 0 = 0 is, right ?
i mean, isnt equality of neutral element (add., mult.) enough proof ???

>>8040414
correction onyl neutral element for addition, not multiplication

>>8036701
It's an axiom m8. You can't prove it.

>>8036701
eq_refl

proof by contradiction.

if a = b where b != a then a != a
therefore a = a

>>8041358
but how can you prove that b != a if you can't even be sure what a actually equates to?

>>8042004
We define 1 as being 1, that's all there is to it.
If you take that certainty out everything else just shatters.

In any order logic, you have an element in your rho-type structure which satisfies under tarski.