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MIT Opencourseware Notes 6.042J says that $$\exists x \forall y \:P(x, y) \implies \forall y \exists x \: P(x, y)$$ is a valid assertion. I am confused because of a counter model that I thought of: if the domain is the integers and $P(x,y)$ means $x > y,$ then isn't the hypothesis wrong? Because in all $y$, which is the set of the integers, how can there exist an $x$ such that $x > y$?

Then, in the bit after that, it says that $$∀y∃x\:P(x,y)⟹∃x∀y\:P(x,y)$$ is not valid, because under the above interpretation (the counter model), "the conclusion asserts that there is an integer that is bigger than all integers, which is certainly false." I am confused on the difference between "does for any integer $y$ exist a bigger integer $x$" and "does there exist an integer $x$ that is bigger than any integer $y$".

ryang
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photon
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    An assertion of the form "$A\implies B$" does not tell you that $A$ is true. Indeed, if $A$ is false then the implication is vacuously true. – lulu Aug 17 '22 at 13:54
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    The left side of the implication being wrong does not negate the implication itself. The implication is broken only when the left side is true but the right side is false. In that sense, your counterexample is not the counterexample for the entire implication. –  Aug 17 '22 at 13:54
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    And indeed the implication is true. If $P$ is such proposition that there exists $x$ such that all $P(x,y)$ are true, then for every $y$ you can find at least one $x$ (namely the same $x$ that exists in the previous part) and $P(x,y)$ will be true. –  Aug 17 '22 at 13:56
  • Thanks, I get it now. – photon Aug 17 '22 at 13:57
  • @StinkingBishop In the bit after that, it says that ∀y∃x:P(x,y)⟹∃x∀y:P(x,y) is not valid, because under this interpretation (the counter model that I had in the Question) "the conclusion asserts that there is an integer that is bigger than all integers, which is certainly false." – photon Aug 17 '22 at 14:07
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    First check if the left side is true. Does for every integer $y$ exist a bigger integer $x$? Yes. Now, does there exist an integer $x$ that is bigger than any integer $y$? No. Thus, you have a failed implication, where the premise is true but the conclusion is false. That implication does not hold. –  Aug 17 '22 at 14:50
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    So yes, the example you had is a counterexample for this new implication, but not for the original implication. "True implies false" is false but "false implies true" is true. –  Aug 17 '22 at 14:52
  • "Does for every integer y exist a bigger integer x? Yes". I am unsure as to why this is true. Doesn't this mean that there is an integer that is bigger than all integers? In the notes it said that for y = n, x can just be n+1. But it is for ALL integers that there exists a bigger integer X. I am confused on this. – photon Aug 17 '22 at 15:28
  • Basically I am confused on the difference between "does for any integer y exist a bigger integer x" and "does there exist an integer x that is bigger than any integer y". Because to me it seems like both are saying the same thing... For the first statement here, if for any integer a bigger integer x exists is true then I don't see why the second one wouldn't be true. – photon Aug 17 '22 at 15:32
  • (nevermind I got it) – photon Aug 17 '22 at 17:08
  • Closely related: Interpretation of quantifiers – MJD Aug 17 '22 at 18:08

2 Answers2

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As you point out, if the domain is integers and $P(x,y)$ is defined as $x > y,$ then the statement $\exists x \forall y \,P(x, y)$ is false. It asserts that there is a single, specific integer -- whose value we could describe exactly right now if we had enough time and materials -- that is greater than any other integer. But of course if we could describe such an integer, there would be (by definition) the sum of $1$ plus that integer, which would be greater than the integer we just described. So the greatest integer just doesn't exist.

But you must always remember that in mathematical logic, implication $(\implies)$ is treated as follows:

$$ \mathbf{true} \implies \mathbf{false} \quad\mathrm{is}\quad \mathbf{false}. $$ $$ \mathbf{true} \implies \mathbf{true} \quad\mathrm{is}\quad \mathbf{true}. $$ $$ \mathbf{false} \implies \mathbf{false} \quad\mathrm{is}\quad \mathbf{true}. $$ $$ \mathbf{false} \implies \mathbf{true} \quad\mathrm{is}\quad \mathbf{true}. $$

There is only one way for this kind of implication to be false: in the case where we assert that a true statement implies a false one. Every other case is a true implication.

In particular, once you have determined that $\exists x \forall y \,P(x, y)$ is false, you know that $$ \exists x \forall y \,P(x, y) \implies \forall y \exists x \, P(x, y) $$ You don't even need to ask (at this point) whether $\forall y \exists x \, P(x, y)$ is true.

But in your particular example, it turns out that $\forall y \exists x \, P(x, y)$ is true.

The difference is that for the first statement, we had to name one integer once and for all time that would be greater than every other. But in the second example, because the $\forall y$ comes first, the $\exists x$ gets to choose a (possibly) different choice of $x$ for each value of $y.$

Therefore we know that $$ \forall y \exists x \, P(x, y) \implies \exists x \forall y \,P(x, y) $$ is false in this example, because you have already determined that $\exists x \forall y \,P(x, y)$ is false and we know that $\mathbf{true} \implies \mathbf{false}$ is always a false statement.

So your example, far from contradicting the statements in the course notes, is actually a good example of how those statements are true. Of course to actually prove that the statements are true you need more than a single example, but the first step toward understanding a proof is to understand what it is that you are proving. So it is important that you can see how this example would work.

Another way to think of $\forall y \exists x$ versus $\exists x \forall y$ is in terms of game strategy. We have two players, $x$ and $y.$ Whoever appears first in the sequence of $\forall$ and $\exists$ has the first move.

The sequence $\forall y \exists x \, P(x,y)$ means "no matter what player $y$ does as the first move, player $x$ can make a move afterward such that $P(x,y)$ will be true."

The sequence $\exists x \forall y \, P(x,y)$ means "there is a move that player $x$ can make first such that, no matter what player $y$ does afterward, $P(x,y) will be true."

If we're thinking about our own strategy for playing the game, we think of ourselves as $x$ (because we just need to come up with one "best" move) and our opponent as $y$ (because we need to be prepared for whatever our opponent might do).

In the "choose the greatest number game" we clearly want to move last, corresponding to $\forall y \exists x$. Moving first ($\exists x \forall y$) is a guaranteed loss.

David K
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For the former, we may pick a new value $x$ for each $y$ that satisfies $P(x,y)$. But, in the latter, we have to choose a single $x$, such that $P(x,y)$ for each $y$. So, if the latter is true, the former is as well, since we can choose the same value of $x$ on each instance of $y$ occurred.

It should be clear the implication only goes one way.

Will Sherwood
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