$P \lor Q$ is Logically Equivalent to $P \land Q$?

For some reason I somehow came up with the logical equivalence of $(P \land Q) \equiv (P \lor Q)$ and I was hoping someone could point out the error in my reasoning, as I can't seem to find out where I went wrong. I utilized various logical equivalences and this came out to be the answer. Below was my process:

1. $(P \land Q) \equiv \lnot(P \to \lnot Q)$
2. $\lnot(P \to \lnot Q) \equiv \lnot P \to Q$ (Double Negation)
3. $\lnot P \to Q \equiv P \lor Q$

I've been using Discrete Mathematics and Its Applications by Kenneth Rosen and according to the table of logical equivalences (pgs 24-25 if anyone has the 6th edition), everything seems to check out, but I already know $P \land Q$ and $P \lor Q$ aren't logically equivalent. But why is this happening? Please pardon my ignorance, I'm just terribly new at Discrete Structures.

• Step 2 is wrong, and I don't know what you meant to do either. The negation of an implication is a conjunction: $\neg(P\to Q)\equiv(P\land \neg Q)$, but you knew that already. – Mario Carneiro Feb 18 '15 at 1:43
• OH! I see what you mean. It's just that these problems tend to throw me off so much. Thanks for the advice, I see where I went wrong. – Horo BEAMS Feb 18 '15 at 1:58

Simply consider the truth table to see that $P\vee Q\not\equiv P\wedge Q$:
$$\begin{array}{|c|c|c|c|}\hline P&Q&P\vee Q&P\wedge Q \\\hline T&T&T&T\\\hline F&T&T&F\\\hline T&F&T&F\\\hline F&F&F&F\\\hline \end{array}$$