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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?

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    $\exists xP(x)\equiv \neg \forall x \neg P(x)$ – 2017-01-27
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    The third one isn't right, as $P\iff Q$ isn't a tautology, whereas $P\land Q \implies Q \land P$ is. You must've mistaken $\land $ and $\implies$ ;) The fourth one should be $\exists x, P$ if and only if $\neg \forall x, \neg P$ – 2017-01-27
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    @Reveillark: It was a typo what a meant was $\exists\, v : P = \lnot \left( \forall\, v:P \right)$ which is equivalent to what you said right? – 2017-01-27
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    @Max : right I switched the $\wedge $ and the $\implies $ around. As for the fourth on it was a typo which I edited, is it correct now? – 2017-01-27
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    @TheStrikeofRunningWater, use $A \to B \iff \neg A \vee B$ in 1),2),3).. – 2017-01-27
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    Seems fine to me. Though be careful that these are only true in classical logic (e.g. not in intuitionistc logic) – 2017-01-27
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    Use truth tables to convince you that the first three are correct. – 2017-01-28
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    When I was in uni, we used the following definition: $A\Rightarrow B := B\vee\lnot A$. Also $\Leftrightarrow$ is defined as $A\Leftrightarrow B := (A\Rightarrow B)\wedge (B\Rightarrow A)$ This makes things very easy. – 2017-01-28

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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

  • $¬[\forall x\ P(x)] = \exists x\ ¬P(x)$

Equivalence of Implication

  • $A \to B = ¬A \lor B$

Demorgan's Law

  • $¬(A \lor B)=¬A \land ¬B$

Interestingly, (4) cannot be proved with truth tables.