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:
- $(P \land Q) \equiv \lnot(P \to \lnot Q)$
- $\lnot(P \to \lnot Q) \equiv \lnot P \to Q$ (Double Negation)
- $\lnot P \to Q \equiv P \lor Q$
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.
$\endgroup$ 21 Answer
$\begingroup$Simply consider the truth table to see that $P\vee Q\not\equiv P\wedge Q$:
$$ \begin{array}{|c|c|c|c|}\hline P&Q&P\vee Q&P\wedge Q \\\hline T&T&T&T\\\hline F&T&T&F\\\hline T&F&T&F\\\hline F&F&F&F\\\hline \end{array} $$
$\endgroup$ 4
