# Question #00484

##### 1 Answer

Let me offer you a suggestion of the proof.

As you would like to prove a "

**1) Prove ** "

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 ** "

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.