I shall use the symbol U for union
^ for intersection
~ for complement
<h3 /><h3>----------------------------------------------------------------------------------------------------</h3>
The property of set subtraction states that for any sets X and Y, X-Y = X ^ Y~
That is in words, X throw away Y is the same as the intersection of X and the complement of Y
or X and anything NOT in Y
At the SET level, the proof looks like this:
(P-Q) U (Q-P) = « given
(P^Q~) U (Q ^ P~) = « property of set subtraction
(P U Q) ^ (P U P~) ^ (Q~ U Q) ^ (Q~ U P~) = « distributive property
(P u Q) ^ (Q~ U P~) = « Union of any set with it's complement is the universal
set which gets dropped by any intersection of a non-universal set
(P u Q ) ^ (Q ^ P)~ = « De Morgan's Law
(P u Q) - (Q ^ P) = « property of set subtraction
(P u Q) - (P ^ Q) « commutative property
<h3>----------------------------------------------------------------------------------------------------</h3>At the ELEMENT level, the proof looks like this:
» Proves left side is subset of right side
If x is in (P-Q) U (Q-P) then x is in (P-Q) OR x is in (Q-P)
If x is in P-Q then x is in P but not in Q.
Then X is in P U Q.
Since x is not in Q, x is NOT in P ^ Q
Hence x is in P U Q but NOT in P^Q, which
means it is an element if (P U Q) - (P ^ Q)
Therefore x is an element of the set defined by the right side, which means (P-Q) U (Q-P) is a subset of (P U Q) - (P ^ Q)
« proves right side is subset of the left side
If x is an element of (P U Q) - (P ^ Q) then
x is in (P U Q) but NOT in (P^Q)
Since x is in P U Q, x is in P OR x is in Q.
Since x is NOT in (P^Q) then x is NOT in BOTH P and Q.
Hence x is in either P or Q but not both.
If x is in P, then x is not in Q, so x is in P-Q.
Therefore x is in (P-Q) U (Q-P).if x is in Q, then x is not in P, so x is in Q-P.Therefore x is in (Q-P) U (P-Q).There x is an element of set defined by the left side, which menas (P u Q)-(P ^Q) is a subset of (P-Q) U (Q-P).
#CarryOnLearning
#LetsEnjoyTheSummer
<h3><u>»<em>X</em><em> </em></u><em><u>x</u></em><em><u> </u></em><em><u>K</u></em><em><u> </u></em><em><u>i</u></em><em><u> </u></em><em><u>m</u></em><em><u> </u></em><em><u>0</u></em><em><u> </u></em><em><u>2</u></em><em><u> </u></em><em><u>x</u></em><em><u> </u></em><em><u>X</u></em></h3>