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I'm negating this proposition: "If you study you will not fail."

I'm using proposition P: "You study" and proposition Q: "You will fail."

The original statement can be written as "$P → ¬Q.$"

My instructor has the negation of this statement like this:

$¬(P → ¬Q) = ¬(¬P \lor ¬Q) = P\land Q $

Why does $¬(P → ¬Q) = ¬(¬P \lor ¬Q)$ ?

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    $\begingroup$ Re-wording: You will study and you will fail... Not too nice to say that to someone ^^ $\endgroup$
    – AlexR
    Nov 12, 2014 at 21:56
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    $\begingroup$ @AlexR No, that's not said. It is denied that "you will study and your will fail" is true. $\endgroup$ Nov 12, 2014 at 22:03
  • $\begingroup$ @DougSpoonwood The negation is $P\wedge Q$ wich is spoken "$P$ and $Q$"... I don't quite understand your objection? $\endgroup$
    – AlexR
    Nov 12, 2014 at 22:05

3 Answers 3

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Because $A\rightarrow B \equiv \lnot A \lor B\tag{1}$

Think of this as stating: An implication $A\rightarrow B$ is true whenever

  • $A$ is false: $\;(\lnot A)$

    OR: $\;\lor$

  • $B$ is true: $\;(B)$

Hence we have $\quad \lnot A \lor B$.

In your case, we have $\;A = P\;$ and $\;B = \lnot Q$,

So using $(1)$ on your proposition: $$\lnot(P \rightarrow \lnot Q) \equiv \lnot (\lnot P \lor \lnot Q) $$ By DeMorgan's, we get $$\lnot \lnot P \land \lnot \lnot Q \equiv (P \land Q)$$

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To verify both statements are equivalent, notice that the statements can only be false when both $P,Q$ are false.

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I find it more convenient to use: $$[A\implies B]\equiv \neg[A \land \neg B]$$

Then your statement could be restated as: It cannot be that you both study and fail. $$ [P\implies \neg Q]\equiv \neg[P\land Q]$$

The negation is that you both study and fail.
$$P\land Q$$

Or equivalently, by De Morgan's Law:

$$\neg[\neg P \lor \neg Q]$$

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