Exercise 1859Logic, sets and proofs › Quantifiers and negationL1Added 3 Nov 2026

When distribution fails

#distribution#counterexample#quantifier

Statement

Show that $\forall x,\ (P(x)\lor Q(x))$ is not equivalent to $(\forall x,\ P(x))\lor(\forall x,\ Q(x))$ .

Hint

Find a domain and predicates where the first is true but the second is false.

Show solutionHide solution

Counterexample. Domain $\{1,2\}$ . $P(x)$ : " $x=1$ ". $Q(x)$ : " $x=2$ ".

First. $\forall x,\ P(x)\lor Q(x)$ : at $x=1$ , $P$ true; at $x=2$ , $Q$ true. So true.

Second. $\forall x,\ P(x)$ is false ( $x=2$ fails). $\forall x,\ Q(x)$ is false ( $x=1$ fails). So the disjunction is false.

Conclusion. Distribution of $\forall$ over $\lor$ fails in general.

Dual. $\exists x,\ (P(x)\land Q(x))$ does not distribute as $\exists x,\ P(x)\land\exists x,\ Q(x)$ .