# Give the negation of the statement $X→(Y \lor Z)$.

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?

• 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 $(x\ne 2) \rightarrow (x\le1)$
– Joe
Jun 16, 2020 at 19:23
• Recall that $A\to B$ is the same as $¬ A\: V\: B$ Jun 16, 2020 at 19:26

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 $$X \wedge \neg Y \wedge \neg Z$$.
• 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.
$$$$\neg(X \implies (Y \lor Z))\\ \neg(\neg X \lor (Y \lor Z))\\ \neg(\neg X \lor Y \lor Z)\\ X \land \neg Y \land \neg Z$$$$
An implication is false exactly when its antecedent is true and its consequent is false. $$\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)$$