# Mathematical Logic: Predicate Calculus

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.

Premise:

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

Conclusion:

$\exists y (My \land Dyy)$

• @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! – Beginner Dec 5 '16 at 22:16
• Well done!! You did make a small mistake in the premise though ... See my answer. – Bram28 Dec 6 '16 at 0:08

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

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

• 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? – Beginner Dec 5 '16 at 22:26
• 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 $y$. – Patrick Stevens Dec 6 '16 at 7:02