why is ∃x [P(x) or Q(x)] ≡ ∃x P(x) or ∃x Q(x) ?

Question

why is ∃x [P(x)  or Q(x)]  ≡ ∃x P(x) or ∃x Q(x) ?

anonymous · Answer

Looks like that backwards EX is multiplied by the functions inside the brackets and it might be a constant

anonymous · Answer

xD
do u see any contradiction ?

anonymous · Answer

the thing that bother me is the x on the LHS is singular while the x's on the RHS might not be the same object

anonymous · Answer

lets prove it using this , 
let A $\in $ [P(x) or Q(x)]  
then A $\in$  p(x) or A$\in$ Q(x)
xD
haha lolz it doesnt need a proof

anonymous · Answer

its ok :)
ill give u an example
let
 p(x)=2x
and 
Q(x) =x^2
A=2
then A might be in P(x) or Q(x) 
:)

anonymous · Answer

for group example
P={1,2,3,4}
Q={5,6,7,8}
and A=5
then A$in$ [Q or p]
5 in Q or 5 in P
it satisfy
5 in Q
its ok if its not in P

anonymous · Answer

If there's an $x$ that makes $P(x)$ or $Q(x)$ a true statement, then there's an $x$ that makes $P(x)$ true or there's an $x$ that makes $Q(x)$ true.