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.

  • 5
    $\begingroup$ 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

Simply consider the truth table to see that $P\vee Q\not\equiv P\wedge Q$:

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.