# Prenex normal form and empty set

Wikipedia states the following regarding conjunction and disjunction rules for turning a formula into its prenex form:

Conjunction:
$$(\forall x \phi) \land \psi$$ is equivalent to $$\forall x (\phi \land \psi)$$ under the (mild) additional condition that at least one individual exists.

If I understand universal quantification over an empty set, the sentences are not equivalent if no $$x$$ exists because $$(\forall x \phi) \land \psi$$ will be false due to the second conjunct, while $$\forall x (\phi \land \psi)$$ is true, since $$\forall x (\phi \land \psi) \Leftrightarrow \forall x(x \in \{\} \rightarrow \phi \land \psi)$$, and the antecedent is always false.

Disjunction:
$$(\exists x \phi) \lor \psi$$ is equivalent to $$\exists x (\phi \lor \psi)$$ under the (mild) additional condition that at least one individual exists.

Here I get lost, as the same logic applied to the conjunction rule breaks down. Both formulas should be false for the empty set, so I do not understand in what case they would have different truth values.

Help?

In your first paragraph, you write "$$(\forall x\phi)\land \psi$$ will be false due to the second conjunct". This only works if $$\psi$$ is a sentence which is false in the empty structure. For example, taking $$\psi$$ to be $$\top$$ (the sentence "true"), we have that $$(\forall x \phi)\land \top$$ and $$\forall x\, (\phi\land \top)$$ are both true in the empty structure. On the other hand, taking $$\psi$$ to be $$\bot$$ (the sentence "false"), we have that $$(\forall x \phi)\land \bot$$ is false in the empty structure, while $$\forall x\, (\phi\land \bot)$$ is true.

This observation suggests how to fix the reasoning in your second paragraph. Taking $$\psi$$ to be $$\top$$, $$(\exists x \,\phi) \lor \top$$ is true in the empty structure, while $$\exists x\, (\phi\lor \top)$$ is false.

If you don't include $$\top$$ and $$\bot$$ in your syntax for first-order logic, feel free to replace $$\top$$ by $$\forall x\, (x=x)$$ and replace $$\bot$$ by $$\exists x\, \lnot (x= x)$$.

• this is crystal clear. thank you!
– BRE
Commented Jul 21, 2022 at 18:14
• @BRE You're welcome! Glad I could help. And welcome to MSE. If my answer has settled your question, it's good practice to "accept" it by clicking the green check mark on the left. And rather than writing thanks in the comments, you're encouraged to vote up answers that you find helpful. Commented Jul 21, 2022 at 18:18
• thanks! I upvoted it but do not have the reputation required for it to become visible.
– BRE
Commented Jul 23, 2022 at 10:46