Meaning of the quantified expression in case of different truth values for its variables.

Question

I get confused when using conditionals. This question is from the Rosen book :

Q. Use quantifiers and predicates with more than one variable to express the statements:

There is a student in this class who has been in every room of at least one building on campus

The solution given is $$∃x∃y∀z(P (z, y) → Q(x, z))$$ where P (z, y) is “z is in y” and Q(x, z) is “x has been in z”; the domain for x con- sists of all students in the class, the domain for y consists of all buildings on campus, and the domain of z consists of all rooms.

Please help me understand what exactly P (z, y) → Q(x, z) means. I understand it means 'If a room z belongs to a building y, then x has been to room z'. I understand the quantification part. But I dont understand the implification part.

For example above statement is false when P is true and Q is false, but what does that mean? Does it mean that in that case x may have been in room z but z does not belong to building y? I am really confused.

Answer 1

Let's break down/rephrase the initial statement a bit:

There is a student in this class who has been in every room of at least one building on campus

This means that:

• There is a student in this class we can say something about

• What we can say is that they've been in certain rooms

• "of at least one building" means that there exists a building on campus that the rest of the thing-we-can-say-about-the-student applies to

• "every room" means that we're talking about all the rooms that meet some qualification.

So to focus on the bit that's giving you trouble, let's imagine pulling the things that become the $x$ and $y$ out of the sentence:

You know that student $x$, right? You've seen how big building $y$ over there is? Student $x$ has been in every room in $y$.

We're now only going to translate that last sentence. Still, with this shorter sentence, we can't say

$$ \forall z(Q(x,z)) $$

Because that isn't what we know; we know nothing about the rooms not in building $y$. We can only assert $Q(x,z)$ for those rooms $z$ which are in building $y$. So we need a bit that says "If $P(z, y)$ then $Q(x, z)$".

And that's exactly what asserting an implication gives us: it gives us a way to limit our statement so that it's no longer saying something about all rooms everywhere, but only about the rooms in that building. (That is, only about those rooms for which $P(z,y)$ is true) Therefore, the shorter sentence is translated as:

$$ \forall z(P(z,y) \to Q(x,z)) $$

And you can see that the statement is only false when $P(z,y)$ is true but $Q(x,z)$ is false; that is, the statement is false only when there is a room that is in building $y$ but that our student hasn't been in. This is exactly where we would want our translation to be false.

Answer 2

The solution given is

$$∃x∃y∀z(P(z,y)→Q(x,z))$$

where P (z, y) is “z is in y” and Q(x, z) is “x has been in z”

For example above statement is false when P is true and Q is false, but what does that mean? Does it mean that in that case x may have been in room z but z does not belong to building y? I am really confused.

You got that just the other way around: If $P (z, y)$ is true, and $Q(x, z)$ is false, then $z$ belongs to building $y$, but $x$ has not been in room z