In my study of the Prenex Normal Form I encountered the following property about conjunctions with universal quantifiers: $$\forall x(p\land q)\iff\forall x\ p\land\forall x\ q$$ This property seems to make sense if I apply Universal Instantiation. I however encountered the following property about disjunctions with universal quantifiers that made me wonder why it doesn't use a biconditional connective like in the first formula? It goes as follows: $$\forall x\ p\lor\forall x\ q\implies\forall x(p\lor q)$$ I would have guessed that, using Universal Instantiation just as in the first formula, it would have also a biconditional connective instead of a mere conditional one. I found the same exact question with conjunctions of existential quantifiers $$\exists x(p\land q)\implies\exists x\ p\land\exists x\ q$$ and disjunctions with existential quantifiers $$\exists x(p\lor q)\iff\exists x\ p\lor\exists x\ q$$ Shouldn't all be biconditionals? Your help, as always, will be greatly appreciated :)

  • $\begingroup$ Please use MathJax so that we can read your mathematics in line. $\endgroup$
    – Rob Arthan
    Sep 15 '20 at 14:56
  • $\begingroup$ No; not all hold in both ways. Example: "Every number is (either Even or Odd)" is True while Either (every number is Even) or (every number is Odd)" is False. $\endgroup$ Sep 15 '20 at 15:13

Every number is either odd or even, but "at least one of the following holds: (1) all numbers are odd; (2) all numbers are even" is certainly false.

$\exists x(p\land q)\Rightarrow \exists xp\land\exists xq$ is basically the contraposition of $\forall xp\lor\forall xq\Rightarrow\forall x(p\lor q)$. That is why only one direction holds in both cases.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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