# A question about logical equivalence and bound variables within Predicate Logic

Question 1: Are these two translated statements logically equivalent?

Statement A: ¬∃x(Nx ∧ Axc) (It is not the case, that for some x, x is a novelist and x admires Carol) Statement B: ∀x¬(Nx → Axc) (For all x, if x isn't novelist, then x does not admire Carol)

Translated from 'No novelists admire Carol'.

Question 2: If quantifiers (∀x, ¬∀x, ∃x, ¬∃x) bind variables, how would I know if the following statement is true or not? 'Everything in that store is either overpriced or poorly made', or symbolically, '∀x(Sx → (Ox ∨ Px))'.

If the 'x' in 'Sx', for example, is bound and not free, and I am not allowed to ask [with an intention of falsifying that statement] something like: 'Do they sell pistachios in here?', I have no way of finding out whether the pistachios are indeed overpriced and poorly made, and if the initial statement itself is true or not.

The fact that variables that occur after a quantifier are bound, and not free, is taken from Velleman's book

Thanks

Qn 1. No. Here's the correct equivalence:

¬∃x(Nx ∧ Axc)

is equivalent to

∀x¬(Nx ∧ Axc)

for pushing a negation pass a quantifier flips the quantifier to its dual. And that's equivalent to

∀x(Nx → ¬Axc)

by the classical equivalence of ¬(A ∧ B) with (¬A v ¬B) with (A → ¬ B).

Qn. 2 You find out whether ∀x(Sx → (Ox ∨ Px)) by finding out whether everything in that store is either overpriced or poorly made. That might be a practical problem, but it isn't a logical one!

• Q2: Am I getting this right: If I were to find something in that store that is both, not overpriced and poorly made, then it would be a logical refutation by modus tollens, but if I were to 'find' a product that wasn't to be found in that store, that would falsify that statement by empirical means? Or is this the case when such (empirical) falsification doesn't count, because that statement is explicitly formulated, such that, 'x is in that store' and 'x is overpriced' and 'x is poorly made', must be taken together? If that makes any sense at all... – Kas Sep 18 '13 at 15:44
• Q1: Ok, I see. I thought universal quantifiers always used conditionals. Will have to brush up on my conversions. – Kas Sep 18 '13 at 15:46
• These very elementary questions indeed suggest you need to read an elementary logic book on translating in and out of quantificational logic. Try Paul Teller's Modern Logic Primer (freely available if you google for his website). – Peter Smith Sep 18 '13 at 15:49