Question:
${Let S = \{x_1, x_2, x_3, ..., x_n\}}$
Show that
${\backsim\Bigl(\forall x {\in} S,P(x) \vee Q(x)\Bigr) \Leftrightarrow \exists x {\in} S, \Bigl( \backsim P(x) \wedge \backsim Q(x)\Bigr)}$
This is what I have done so far ...
${\backsim\Bigl(\forall x {\in} S,P(x) \vee Q(x)\Bigr) \Leftrightarrow \exists x {\in} S, \Bigl( \backsim P(x) \wedge \backsim Q(x)\Bigr)}$
${\exists x {\in} S, \backsim P(x) \wedge \backsim Q(x)\Leftrightarrow \exists x {\in} S, \Bigl( \backsim P(x) \wedge \backsim Q(x)\Bigr)}$
And then I have no I idea how to continue from here.
Any hint will be appreciated.
Without nailing down which axioms of logic can be used one has to be satisfied with an intuitive verification.
Pls, consider the following figures.
In $S$, the red elements are those that have property $P$, the blue elements are those that have property $Q$, and the purple elements have both properties. The upper figure depicts the situation when $$\forall x {\in} S,\color{red}{P(x)} \vee \color{blue}{Q(x)}.$$
The lower figure depicts the negation: There are elements that neither of the two properties have (white). This can be described by
$$\sim \left[\forall x {\in} S,\color{red}{P(x)} \vee \color{blue}{Q(x)}\right].$$
The lower figure clearly shows that there are then elements that neither the two properties have (white). In other word these existing white elements that don't have property $Q$ and they don't have property $P$. This can be described as
$$\exists x {\in} S, \left[ \backsim \color{red}{P(x)} \wedge \backsim \color{blue}{Q(x)}\right].$$
The argumentation can be replayed in the other direction.