Why is this true? $(\exists x, Px \to r) \iff (\forall x, Px) \to r$

Question

I can totally understand the forward direction: if there is an $x$ such that $Px$ implies $r$, then clearly having $Px$ true for all $x$ will imply $r$. But the other direction doesn't make any sense to me.

If I give a more natural example, assume that "If it rains every day this week, then the pot will be full of water". Here, $r$ is "the pot is full of water", $x$ ranges over the days of this week, and $Px$ denotes that it rains on a particular day. The statement in the title would suggest that there is a single day of the week such that if it rains on that day, then the pot will be full, and that's just not true.

So what am I missing? Because apparently the statement in the title is true, it can be proven, so where does my understanding break down?

Answer 1

$$(\forall x, Px) \to r\implies(\exists x, Px \to r) $$

Clearer: $$(\forall x\, Px) \to R\implies\exists x\, (Px \to R) $$

assume that

• if it rains every day this week, then the pot will be full

then

• there is a single day this week such that if it rains on that day, then the pot will be full

Since $R$ is not a function of $x$ (this point is crucial for the title's equivalence to hold), this is less ambiguous:

assumption (A)

• if it rains every day this week, then the pot will be full next week

consequence (C)

• there is some day this week such that if it rains on that day, then the pot will be full next week

(A) says: either the pot will be full next week or some day this week isn't rainy. In particular: on some day this week, (either the pot will be full next week or some day this week isn't rainy). Thus: on some day this week, either the pot will be full next week or that day isn't rainy. This is exactly what (C) says.

Therefore, (C) is a consequence of (A).

Answer 2

Thanks to Daniel Schepler for the push in the right direction. This is indeed the Drinker Paradox, which comes from looking at implication as a causal relationship, instead of as just a way of categorising patterns in pairs of boolean values.

Answer 3

One way of seeing that $(\exists x, Px \to r)$ and $(\forall x, Px) \to r$ are equivalent is to show that their negations are equivalent. The negation of $(\exists x, Px \to r)$ is equivalent to $\forall x, \lnot (Px \to r)$ which is itself equivalent to $\forall x, (Px \land \lnot r)$; while the negation of $(\forall x, Px) \to r$ is equivalent to $(\forall x, Px) \land \lnot r$. These two statements are more easily seen to be equivalent.

Answer 4

Suppose $(\forall x, Px)\to r$

Assume $\neg(\exists x, Px\to r)$, that is: no $a$ will satisfy $Pa\to r$.

Now if every $a$ satisfied $Pa$ , then $r$ holds, so there are $a$ which satisfy $Pa\to r$ (a contradiction). And if some $a$ does not satisfy $Pa$, then that satisfies $Pa\to r$, (also a contradiction).

Either way, the assumption is contradicted.