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.