How is this proposition false? I have a proposition:
((x v y) <=> (~x => ~y))
When I solve this, I end up getting True but the answer is False. Here's how I solved it:
when we have a <=> b, we can write it as ~a.~b + a.b and a => b becomes ~a+b
So the above equation becomes:
=> ~(x + y).~(~x => ~y) + (x + y).(~x => ~y)
=> ~(x+y).~(x+~y) + (x+y).(x+~y)
Using De Morgan's law (~(a+b) = ~a.~b and ~(a.b) = ~a+~b)
=> (~x.~y).(~x.y) + (x+y).(x+~y)
=> (~x.~x + ~x.y + ~y.~x + ~y.y) + (x.x + x.~y + y.x + y.~y)
=> (~x + ~x(y+~y) + 0) + (x + x(~y + y) + 0)
=> (~x + ~x) + (x + x)
=> ~x + x which is always True
But, according to another solution, <=> is only True when both sides result in the same solution, So, if we go by this:
=> LHS = x+y and RHS is (x + ~y), which can never be equal and hence, False.
So what am I doing wrong?
EDIT:
PS: This is where I got the question from.
 A: There is at least one mistake when you go from:
$$\neg (x+y).\neg (x+ \neg y) + (x+y).(x+ \neg y)$$ to
$$(\neg x. \neg y).(\neg x.y) + (x.y).(x+ \neg y)$$ using De Morgan laws. You switched $x+y$ into $x . y$.
Also I don't understand how you go from this line to the next one:
$$(\neg x.\neg x + \neg x.y + \neg y.\neg x + \neg y.y) + (x.x + x.\neg y + y.x + y.\neg y).$$
It seems that you apply a kind of distributivity of $.$ over itself???
A: After line "Using De Morgan's law"
You should have  $$(\neg x.\neg y).(\neg x.y) + (x+y).(x+\neg y)$$
$$0+x+x .\neg y+y.x +y.\neg y$$
$$x+x.(y+\neg y) = x$$
So you end up with $x$: initial equivalence is True together with $x$
A: 
When I solve this, I end up getting True but the answer is False

The answer is not False. You can check it yourself by setting x=True, y=True, in which case you get
(T+T)<=>(F=>F)

which simplifies to
T <=> T

and further to just T.
The answer is also not True, and you can check this by setting x=False, y=True and you get
(T+T)<=>(T=>F)

which simplifies to
T <=> F

and further to just F.


=> LHS = x+y and RHS is (x + ~y), which can never be equal and hence, False.

This is incorrect. So long as x is true, x+y and x+~y can be equal (i.e., both true.
