Question #00484

1 Answer
Jan 27, 2016

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.