Proving a=a: Using Natural Deduction to Show Equality in Set Theory

[quantum123]

Today while day dreaming I discovered something interesting. I can prove a=a.
Here's how:

You can prove P=>P using natural deduction rules.(=> Intro)
So you can prove that x is an element of a => x is an element of a
Hence a is subset of a, and vice versa.
By ZFC axiom of extension, a=a

So a=a need not be an axiom, because it can be proven. In this sense, equality is not the fundamental concept. Set membership is.

 

[Stephen Tashi]

Today while day dreaming I discovered something interesting. I can prove a=a.

Unfortunately there are several different defintiions of "=". It appears that what you did was prove that the set consisting of only the element 'a' is equal to the set consisting of only the element 'a'. This doesn't help in other contexts. (For example {a,a} = {a} as sets but not as strings of characters. It also isn't clear whether you are relying on those other contexts in your proof. You'd have to check that the concept of "equvalence relation" isn't used in developing he set theory and propositional logic that you cited.

 

[quantum123]

I am using only set theory as the basis.
I did not mention the set consisting of a, but rather a is an arbitrary set. The axiom of extension in set theory does in fact tell you what = means, unequivocally.

 

[xxxx0xxxx]

Sorry Q, but you've only stated half of the equivalence, the other half does not follow..., unless you us a=a, that is, you have assumed the consequent.