2
$\begingroup$

I have a very simple question. Is the following proposition true or false from a logic point of view ? $$\forall x \in \emptyset, 2 + 2 = 5$$

I think that the previous proposition can be written as $$\forall x, x \in \emptyset \Rightarrow 2 + 2 = 5$$ which is always true, but I'm not sure...

I did not follow pure logic courses and I actually have a more complex problem which is in the same spirit as this one, so your answers to this question (and explainations) will help me a lot !

Thank you for your help.

$\endgroup$
  • 6
    $\begingroup$ Yes, that's right. $\endgroup$ – David C. Ullrich Jul 5 at 16:57
  • $\begingroup$ Thank you ! So, if I take for example this proposition : $\forall x \in X, \exists y \in Y, \exists z \in Z, P(x, y, z)$, can I write it as : $\forall x, \exists y, \exists z, [x \in X \wedge y \in Y \wedge z \in Z] \Rightarrow P(x, y, z)$ ? $\endgroup$ – deeppinkwater Jul 5 at 17:01
  • 3
    $\begingroup$ No, that doesn't follow at all. $\forall_{x\in X}P(x)$ is an abbreviation for $\forall x(x\in X\implies P(x))$, but there's no $\implies$ in $\exists_{x\in X} P(x)$; that's $\exists x(x\in X\land P(x))$. $\endgroup$ – David C. Ullrich Jul 5 at 17:08
  • $\begingroup$ So I can write the previous proposition as : $\forall x, x \in X \Rightarrow [\exists y, y \in Y \wedge [\exists z, z \in Z \wedge P(x, y, z)]]$ ? $\endgroup$ – deeppinkwater Jul 5 at 17:14
  • $\begingroup$ Yes............ $\endgroup$ – David C. Ullrich Jul 5 at 17:16

Your Answer

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

Browse other questions tagged or ask your own question.