/sci/ can you clarify the definition of order pairs for me?

>>8604522
Compare the representation of (a, b) with (a, a) and (b, a) and you'll see they are all different.

{{a}, {a, a}} = {{a}, {a}} = {{a}}

>>8604522
>From here how do you reduce this to {{a}}?

Not possible. {{a}} and {{a},{a,a}} are two different sets.

Anyways to know that {{a},{a,b}} is (a,b) you simply need to know:

The first element is a if and only if {a} is an element of the ordered pair {{a},{a,b}}

And the second element is b if and only if {(first element, in this case a),b} is an element of the set.

It is as simple as that. Say you have the set {{1},{1,2}}

The first element is the one that such that {element} is in the set, and if we try {2}, that is not in the set. If we try {1} then that is in the set.

Now we know what the first element is so we construct {1,x} and then set x=2 and find that such an element is in the set, therefore that x is our second element. And x=2 so 2 is our second element.

>>8604532
Thanks here is my follow up question:

We have {{1},{1,2}} representing (a,b)

Since the first element is {1} then we have (a,b)

Now we look for {2} in {{1},{1,2}}

but is {2} an element of {1,2} ? this confuses me.

>>8604549
Correction: Since the first element is {1} then we have (1, b).

>>8604531
So, {{a}, {a,a}} is reduced to {{a}} because {a,a} is just {a} and since we already have the other {a} around we can get rid of it to just get the set containing {a} or {{a}}

>>8604532
I re-read your statement. I think I get it now....

So say you have the ordered pair (2,1)

Then what you do is check to see if {2} is an element of the set {{2}, {2,1}} which it is, so you get (1,b)

Then you check to see if {2,1} is in your set which it is so you then get (2,1)

Is that how it works?

What if a=b?

Say {{2}, {2,2}} then you check if {2} is in your set and if {2,2} is in the set?

>>8604557
Sorry typo again meant to say

>Then what you do is check to see if {2} is an element of the set {{2}, {2,1}} which it is, so you get (2,b)