Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know that a universal quantifier can be distributed over conjunction and not disjunction, but I'm having a hard time wrap my head around it. Why is this the case? Is there an example of a statement that would demonstrate this principle?

share|improve this question
"Every chess piece is either black or white" is true. "Every chess piece is black or every chess piece is white" is false. –  Matt E Oct 10 '10 at 3:30
facepalm. You're right, thanks! –  chimeracoder Oct 10 '10 at 3:38
add comment

2 Answers

Matt E, in his comment, gives two statements. He shows that a universal quantifier cannot be distributed over disjunction, because if it could, a false statement would follow logically from a true one. The statement "A universal quantifier can be distributed over disjunction" is itself a universal statement, and requires only one counterexample to disprove it.

"A universal quantifier can be distributed over conjunction" is also a universal statement. It cannot be proved by giving examples. There might be some counterexample we didn't think of. So how do we know it's true? -- Good question!

share|improve this answer
add comment

Intuitively, the reason that universal quantification can be distributed over conjunction is that universal quantification can already be viewed as conjunction: $(\forall x)\Phi(x)$ can be viwed as the conjunction of $\Phi(d)$ taken over every element $d$ of the domain. Viewed this way, $(\forall x)(\Phi(x) \land \Psi(x))$ and $(\forall x)\Phi(x) \land (\forall x)\Psi(x)$ are equivalent because they both represent a giant conjunction of every instance of $\Phi(d)$ along with every instance of $\Psi(d)$. This can be made more precise by looking at Tarski's schema for truth in a structure.

For the same reason, existential quantification can be distributed over disjunction, because $(\exists x)\Phi(x)$ can be viewed as the disjunction of every possible substitution instance of $\Phi$.

In very old literature, people actually used $\bigwedge_x$ for $\forall$ and $\bigvee_x$ for $\exists$, for this reason.

Now, continuing this informal viewpoint, the reason that $\forall$ cannot be distributed over disjunction is that a certain distributive rule does not hold: $$ \bigwedge_x\left(\Phi(x) \lor \Psi(x)\right) $$ is not in general the same as $$ (\bigwedge_x \Phi(x)) \lor (\bigwedge_x \Psi(x)) $$ Indeed, this rule already fails when there are just two elements in the domain, because the propositional formula $(P \lor Q) \land (R \lor S)$ is not equivalent to $(P \land R) \lor (Q \land S)$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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