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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.

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    $\begingroup$ 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. $\endgroup$ – Mario Carneiro Feb 18 '15 at 1:43
  • $\begingroup$ 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. $\endgroup$ – Horo BEAMS Feb 18 '15 at 1:58
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Simply consider the truth table to see that $P\vee Q\not\equiv P\wedge Q$:

$$ \begin{array}{|c|c|c|c|}\hline P&Q&P\vee Q&P\wedge Q \\\hline T&T&T&T\\\hline F&T&T&F\\\hline T&F&T&F\\\hline F&F&F&F\\\hline \end{array} $$

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  • $\begingroup$ I thank you for the truth table, but I was aware of that - I was trying to see where I had gone wrong in my line of reasoning with logical equivalences, listed in the steps above. $\endgroup$ – Horo BEAMS Feb 18 '15 at 2:07
  • $\begingroup$ Sorry about being pedantic then. $\endgroup$ – Laars Helenius Feb 18 '15 at 2:19
  • $\begingroup$ Oh! No, I didn't mean to offend. My bad, I was just saying. $\endgroup$ – Horo BEAMS Feb 18 '15 at 2:27
  • $\begingroup$ No worries. I'm not upset. I was trying to be genuine. Sometimes it is hard to communicate effectively, right. $\endgroup$ – Laars Helenius Feb 18 '15 at 2:28

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