Question

Question #00484

Answer 1

Let me offer you a suggestion of the proof.

As you would like to prove a "`<=>`" relation, you can prove it with proving first "`rArr`" and then "`lArr`".

**1) Prove ** "`rArr`":

Let `A uu B = A nn B`.

We need to prove that `A = B`. To do that, we need to prove that for every `x in A`, it follows `x in B` (i.e. `A sube B`), and for every `x in B`, `x in A` (i.e. `B sube A`).

Let me also remind you of the formal definitions of `uu` and `nn`:

`A uu B = { x | x in A vv x in B }`
`A nn B = { x | x in A ^^ x in B }`

Let `x in A`.

As `A sube A uu B`, so all elements from `A` are included in `A uu B`, we can safely assume that `x in A uu B`.

However, we know that `A uu B = A nn B`, so `=> x in A nn B`.

But since `x in A nn B` and all elements of `A nn B` are included in `B`, `x in B` must immediately hold.

Thus, `x in A => x in B`. `=> A sube B`

Let `x in B`.

The reasoning is exactly the same, really. Let me formulate this with mathematical terms exclusively though.

`x in B stackrel(B sube A uu B)(rArr) x in A uu B`

`stackrel(A uu B = A nn B)(rArr) x in A nn B`

`stackrel (A nn B sube B)(rArr) x in A`

Thus, `x in B => x in A`, thus `B sube A`.

So we have proven that

`(A uu B = A nn B) rArr (A = B)`

**2) Prove ** "`lArr`":

Let's prove the other direction then.

Let `A = B`. We need to prove that `A uu B = A nn B` needs to be true now.

However since `A = B`, we see that

`A uu B = A uu A = A = A nn B = A nn B`.

In case of doubt, we can also check the formal definitions:

`x in A uu B stackrel(A = B)(<=>) x in A uu A <=> x in A <=> x in A nn A stackrel(A = B)(<=>) x in A nn B`.

Thus, `A uu B = A nn B`.

Here, we have proven that

`(A uu B = A nn B) lArr (A = B)`

3)

Since both

`(A uu B = A nn B) rArr (A = B)`

and

`(A uu B = A nn B) lArr (A = B)`

are true, we know that

`(A uu B = A nn B) <=> (A = B)`

is true as well.

q.e.d.