# $\models P \wedge Q \iff \neg ( \neg P \vee \neg Q )$ is a valid argument

I have to show that $$\models (P \wedge Q) \iff \neg ( \neg P \vee \neg Q )$$ is a valid argument.

However, I have no idea how to interpret a $$\models$$ symbol WITHOUT a LHS. I have always seen it like this $$P, P \implies Q \models Q$$, which I read as: given the premises on the LHS hold, then RHS is logically entailed. Which I would then just rephrase as a formula like $$P \wedge (P \implies Q) \implies Q$$, and then use a truth table to show it is true for all $$P,Q$$.

Now, my intuition is, since there is nothing in the LHS, then RHS is just entailed for everything. And so the formula will be $$⊤ \implies [ \ (P \wedge Q) \iff \neg ( \neg P \vee \neg Q ) \ ]$$ which, simplifies back to $$(P \wedge Q) \iff \neg ( \neg P \vee \neg Q )$$ (using disjunction elimination and so on) and now I can make a truth table for this.

Is my approach correct? Or did I get completely lost.

• Yes, this is correct. With nothing on the LHS, the claim is that the statement on the RHS is true in every case (which is correct). And this is equivalent to having the single statement $\top$ on the LHS (much as, with $P$ and $P\mathbin\rightarrow Q$ on the LHS, this is equivalent to having the single statement $P\wedge(P\mathbin\rightarrow Q)$ on the LHS). And yes, you can verify this with a truth table. Commented Nov 28, 2023 at 22:39
• See Tautology. Commented Nov 29, 2023 at 6:46