Thread: [a]=[b] <--> a~b

Thm: [a]=[b] <--> a~b

I have proved:
[a]=[b] --> a~b

but cannot figure out how to prove:
[a]=[b] <-- a~b

I don't know if I can just write the steps backwards since I don't see how I would know: (4) --> (5)


This fact was stated without proof in a youtube playlist on equivalence relations:

Originally Posted by lamp23

Thm: [a]=[b] <--> a~b
I have proved:
[a]=[b] --> a~b
but cannot figure out how to prove:
[a]=[b] <-- a~b

Suppose that then if we know that .
Because we know that or

Can you finish?

I think (3) and (4) need an existencial quantifier for the variable x. Otherwise, seems fine.

Originally Posted by ModusPonens

I think (3) and (4) need an existencial quantifier for the variable x. Otherwise, seems fine.

Actually no that is not necessary.
If we start with then because we know that or .

Originally Posted by Plato

Suppose that then if we know that .
Because we know that or

Can you finish?

Yes, thanks a lot. I was missing using the given of "Suppose "

Yes. But isn't that a different proof? I mean if we want to make the proof in the OP correct, isn't there a need of a quantifier in order for a (an element, ultimately, a set) to be in relation with another element (which x is not, because it's a free variable)? I'm unsure.

yes, something along the lines of: let x be in [a] = [b]. we know such an x exists because both [a] and [b] are non-empty (because ~ is an equivalence, so ~ is reflexive. so at the very least, a is in [a], and b is in [b] (a and b may actually be the same, but perhaps not)).

but it's not necessary to introduce x at all. as Plato points out, we have:

a in [a] = [b], so a in [b], so b~a, and by symmetry, a~b. doing so eliminates the need to use transitivity to show [a] = [b] → a~b (we we *do* need transitivity to show the reverse implication, however).