show that if P and Q are sets then (P - Q) u (Q-P)= (P u Q) - (P n Q)

Question

Set theories 

Answer 1

I shall use the symbol U for union
^ for intersection
~ for complement

 

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

---------------------------------------------

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)

end of proof