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The instruction is to provide a statement using laws of logic that negates the given statement. The dilemma I'm having now is to whether show the negation sign (¬) in my answer or simplify the expression.

So, what I did is the following:

$X→(Y \lor Z)$

$¬[X→(Y \lor Z)] \space \space \space \space \space \space \space\space\space\space\space\space\space\space\space\space\space\space$ (assuming that this is what 'negation' means)

$¬X→¬(Y \lor Z)$

$¬X→(¬Y ∧ ¬Z) \space \space \space \space\space\space\space\space\space\space\space\space\space\space$ (De Morgan's Law)

$¬(¬Y ∧ ¬Z) →¬(¬X) \space\space\space\space\space$ (Contraposition)

$(Y \lor Z) → X \space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$ (De Morgan's + Double Negation)

Is my solution correct? Is this what negation means?

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    $\begingroup$ You cannot distribute the negation across the implication like you did. Example: $(x=2) \rightarrow (x>1)$, but incorrectly distributing like you did would give $(x\ne 2) \rightarrow (x\le1)$ $\endgroup$
    – Joe
    Jun 16, 2020 at 19:23
  • $\begingroup$ Recall that $A\to B$ is the same as $ ¬ A\: V\: B$ $\endgroup$
    – Miguel
    Jun 16, 2020 at 19:26

3 Answers 3

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Your third line is incorrect. The negation $\neg (X\rightarrow (Y \lor Z) )$ is equivalent to $X \wedge \neg(Y \lor Z) $ which is precisely $X \wedge \neg Y \wedge \neg Z$.

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  • $\begingroup$ I would add the explanation that what you wrote is true because $A \rightarrow B$ is equivalent to $\neg (A \wedge \neg B)$, in case the OP doesn’t know that. $\endgroup$
    – Joe
    Jun 16, 2020 at 19:28
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\begin{equation} \neg(X \implies (Y \lor Z))\\ \neg(\neg X \lor (Y \lor Z))\\ \neg(\neg X \lor Y \lor Z)\\ X \land \neg Y \land \neg Z \end{equation}

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An implication is false exactly when its antecedent is true and its consequent is false. $$\lnot(P\to Q)~~\iff~~ P\land \lnot Q$$

A disjunction is false exactly when neither disjunct is true. $$\lnot(P\vee Q)~~\iff~~(\lnot P\land\lnot Q)$$

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