# Predicate Logic Problem

1. Nov 10, 2006

### robert

I have the sentence: No American who hasn't met any Canadian's knows Canada. The teacher gave the correct answer as being:

Vx-Ex((Ax ^ Cy ^ Mxy) -> -Kxc)

Would this version also work?:

-ExEy(Ax ^ Kxc ^ Cy ^ Mxy)

or is it supposed to be:

-ExEy(Ax ^ Kxc ^ Cy ^ -Mxy)

After thinking about it, I think it may be the second. Which means I got it wrong on the assignment.

2. Nov 10, 2006

### AKG

A correct answer must be logically equivalent to:

$$\neg (\exists x)(Ax\ \wedge \ (\forall y)(Cy\ \rightarrow \ \neg Mxy)\ \wedge \ Kxc)$$

$$(\forall x)\neg (\exists y)[(Ax \wedge Cy \wedge Mxy) \rightarrow \neg Kxc]$$

which is logically equivalent to:

$$(\forall x)(\forall y)(Ax \wedge Cy \wedge Mxy \wedge Kxc)$$

which says: "Everybody is American, everybody is Canadian, everybody has met everyone, and everyone knows Canada" which is obviously not what the original English sentence says. Anyways, neither of your answers are logically equivalent to the answer your teacher supposedly gave, nor the answer I gave at the start of this post.

3. Nov 13, 2006

### fopc

Let U = the set of all Americans (presumably U.S. citizens).
Let M(x) denote: x has never met a Canadian.
Let K(x) denote: x does not know Canada.

For any x [C(x)) -> K(x)] <=>
There does not exit an x [C(x) and ~K(x)]

4. Nov 13, 2006