Given formulas $P$, $Q$ and a variable $x$, define the operators $\wedge,\vee$ and the quantifier $\exists$ in terms of the operators $\lnot,\implies$ and the quantifier $\forall$.
$1)$ $$\qquad P\wedge Q \quad = \quad \lnot \left( P \implies \lnot Q \right)$$
$2)$ $$ \qquad P\vee Q \quad = \quad \lnot P \implies Q$$
$3)$ $$ \qquad P\iff Q \quad = \quad P\implies Q \;\wedge \;Q\implies P$$
$4)$ $$\exists\, x : P(x) \quad = \quad \lnot \forall\, x : \lnot P(x) $$
Are these correct? What is the best way to prove this without the use of truth tables?
The first, second & fourth are fine; however the third is in need of brackets.
It should instead be written
P⟺Q=(P⟹Q)∧(Q⟹P)$\tag{3}\label{3}$
While $\ref{3}$ may be correct under some order of operations defined in your textbook; it is usually better to use brackets, because there is no agreed upon order of operations in logic.
The best way to prove each of these are correct without the use of truth tables is to simply use properties of each operator & Demorgan's Laws.
Negation of For all
Equivalence of Implication
Demorgan's Law
Interestingly, (4) cannot be proved with truth tables.