# Distribution of Universal Quantifiers

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?

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"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

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 $(P \land R) \lor (Q \land S)$.

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