Question number 1: Is this correct?

Question number 2: If yes, would be correct if for the conclusion I use x instead of y? For the context, it seems clear that the right choice was y. However, as x and y are variables I was wondering if it would be the same. Thank you so much for your help!

Translate to logic symbols:

There is a man whom all men despise. Therefore, there is a man who despises himself.


$\exists y \forall x (Mx \land My \land Dxy)$


$\exists y (My \land Dyy)$

  • 1
    $\begingroup$ @bram28 I wrote my first Latex code. I am so happy! However, it is just two lines. Take a look when you have time. Thanks for the tips and encouragement for Latex and logic! $\endgroup$ – Beginner Dec 5 '16 at 22:16
  • $\begingroup$ Well done!! You did make a small mistake in the premise though ... See my answer. $\endgroup$ – Bram28 Dec 6 '16 at 0:08

The premise is not correct. It should be $\exists y (My \land \forall x (Mx \rightarrow Dxy))$

  • $\begingroup$ Thank you for the correction. Is the conclusion right? $\endgroup$ – Beginner Dec 6 '16 at 1:42
  • $\begingroup$ @Beginner yes, conclusion is correct! $\endgroup$ – Bram28 Dec 6 '16 at 2:16
  • 1
    $\begingroup$ It depends what you're quantifying over. I presumed that we were quantifying over men only. $\endgroup$ – Patrick Stevens Dec 6 '16 at 7:03
  • $\begingroup$ @PatrickStevens Yes, you're quite right! It's always a good thing to first get clear on the domain. $\endgroup$ – Bram28 Dec 6 '16 at 12:52

[EDIT: the question changed after I posted this answer. the original premise was $(\exists y)(\forall x)(Dxy)$.]

If you've defined $Dxy$ as "$x$ despises $y$", then your premise and conclusion are correctly translated into symbols.

You're right that $(\exists x)(Dxx)$ is equivalent to $(\exists y)(Dyy)$. That's because there is a rule of the predicate calculus which states that we can rename the "bound variable" $a$ to $b$ in $(\exists a) \phi$ without restriction, as long as $\phi$ doesn't contain any free [unbound] occurrence of the symbol $b$. [A "bound variable" is one which appears inside the bracket of an $\exists$ or $\forall$.]

That is to say, "the name of a variable doesn't matter".

Aside: We do need $b$ not to appear free in $\phi$. Otherwise, from the line $(\exists a)(a < b-1)$ we may relabel $a$ to $b$ and deduce $(\exists b)(b < b-1)$. But this latter formula is obviously not true even though $(\exists a)(a < b-1)$ may be true if $b$ is chosen correctly.

  • $\begingroup$ Thank you for your fast and helpful answer! One new question: In general, the name of the variable does not matter. Does your answer include the current situation when the conclusion is connected to a premise? $\endgroup$ – Beginner Dec 5 '16 at 22:26
  • $\begingroup$ Indeed, my answer includes that situation. The exception would be if $D$ is an abbreviation for something which contains a free occurrence of the symbol $y$. $\endgroup$ – Patrick Stevens Dec 6 '16 at 7:02

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.