Who else struggling with the foundations of arithmetic here?

who else struggling with the foundations of arithmetic here?

We don't

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

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

multiplication was a stupid idea anyway

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.

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.

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

Fucking this.

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

:^)

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

Expanding

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

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.

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.

Its an axiom, you dont need to prove it

are you 12?

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 ???

correction onyl neutral element for addition, not multiplication

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

eq_refl

proof by contradiction.

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

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

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.