Re: Quantified Statements

Quote:

Originally Posted by

**Bashyboy** d) Some student has not asked any faculty member a question.

Why isn't the answer to this question, "$\displaystyle \exists x \forall y ((S(x) \wedge F(x)) \implies \neg A(x,y))$?

The English claim asserts, in particular, that there is a student. The formula, on the other hand, asserts that there is a non-student (or something else is true). Indeed, recall that $\displaystyle P\Rightarrow Q$ is equivalent to $\displaystyle \neg P\lor Q$. Since S(x) in the premise of the implication, it is, so to speak, under negation.

More formally, the formula is true in an interpretation where the domain contains at least one non-student. If x is instantiated with this person, then the premise of the implication is false regardless of y, so the implication is true.

Correct formulas are $\displaystyle \exists x \forall y\,(S(x) \wedge (F(y))\Rightarrow \neg A(x,y)))$ and $\displaystyle \exists x\,(S(x) \wedge \forall y\,(F(y)\Rightarrow \neg A(x,y)))$.