# Why can't we use implication for the existential quantifier?

I'm not quite sure that I really understand WHY I need to use implication for universal quantification, and conjunction for existential quantification.

Let $F$ be the domain of fruits and

$$A(x) : \text{is an apple}$$

$$D(x) : \text{is delicious}$$

Let's say: $$\forall{x} \in F, A(x) \implies D(x)$$ Is correct and means all apples are delicous.

Whereas, $$\forall{x} \in F, A(x) \land D(x)$$ is incorrect because this would be saying that all fruits are apples and delicious which is wrong.

But when it comes to the existential quantifier: $$\exists{x} \in F, A(x) \land D(x)$$ Is correct and means there is some apple that is delicious.

Also, $$\exists{x} \in F, A(x) \implies D(x)$$ Is incorrect, but I cannot tell why. To me it says there is some fruit that if it is an apple, it is delicious.

I cannot tell the difference in this case, and why the second case is incorrect?

To me it says there is some fruit that if it is an apple, it is delicious.

This is absolutely correct. There exists a fruit such that if it is an apple, then it is delicious. Let $x$ be such a fruit. We have two cases for what $x$ may be here:

• $x$ is an apple. Then $x$ is delicious. This is the $x$ you are searching for.
• $x$ is not an apple. Now the statement "if $x$ is an apple, then $x$ is delicious" automatically holds true. Since $x$ is not an apple, the conclusion doesn't matter. The statement is vacuously true.

So the statement $\exists{x} \in F, A(x) \implies D(x)$ fails to capture precisely your desired values of $x$, i.e., apples which are delicious, because it also includes other fruits.

• Just to add a specific and rather extreme example: Consider a universe where nothing is delicious but there is at least one fruit that isn't an apple. Then that one fruit is an example showing that $(\exists x\in F)\,(A(x)\implies D(x))$ is true. (Note that this works whether or not this universe contains any apples.) – Andreas Blass Oct 19 '15 at 19:12
• Thank you! I have a follow-up question though. For ∀x in F, A(x) -> D(x) wouldn't then it be vacuously true for every fruit? So wouldn't I not be capturing the desired values for that one either? – Adam Thompson Oct 19 '15 at 19:12
• Andreas's way of thinking about it is better; think about which universes you want your statement to hold true in, rather than which values of $x$ are "captured" by a quantifier. In the existentially quantified statement, surely you don't want it to hold true in a universe that doesn't contain any apples, or one in which no apple is delicious. Similarly, in the universally quantified statement, surely you don't want it to be false if all apples are delicious, but there are other fruits in the universe. That's why it's a good idea for the implication to be vacuously true if the premise is true. – Dustan Levenstein Oct 19 '15 at 19:21
• Would it be correct to basically rationalize it by saying that you can use the implication with the universal quantifier because we don't care that it becomes vacuously true for non-Apple fruits? We just care that we don't have a universe where there are any apples that are not delicious? – Adam Thompson Oct 19 '15 at 19:27
• Exactly. ${}{}{}$ – Dustan Levenstein Oct 19 '15 at 19:28

I'm not quite sure that I really understand WHY I need to use implication for universal quantification, and conjunction for existential quantification.

Modifying your analysis a bit, let $A$ be the set of apples, and $D$ the set of delicious things.

$\forall x: [x\in A \implies x\in D]$ means all apples are delicious. Often written $\forall x\in A :x\in D$

$\forall x: [x\in A \land x\in D]$ means everything is a delicious apple.

$\exists x:[x\in A \land x\in D]$ means there exists at least one delicious apple. Often written $\exists x\in A:x\in D$, or equivalently $\exists x\in D: x\in A$

What does $\exists x:[x\in A \implies x\in D]$ mean? It is equivalent to $\exists x:[x\notin A \lor x\in D]$.

For a given $x$ then, either of the following possibilities that will satisfy this condition:

1. $x\in A \land x\in D$, i.e. there exists at least one delicious apple (as above)

2. $x\notin A\land x\in D$, i.e. there exists at least one non-apple that is delicious

3. $x\notin A\land x\notin D$, i.e. there exists at least one non-apple that is not delicious

So, the implication allows for more possibilities than the conjunction. In particular, the implication allows for the possibility that there are no apples. The conjunction does not.

Furthermore, $\exists x: [x\in A \implies x\in D]$ is a set theoretic variation of the so-called Drinker's Paradox. Here's where it gets crazy! For any set $A$ and any proposition $P$, we can prove using ordinary set theory that $$\exists x: [x\in A \implies P]$$

You could even prove, for example, that $$\exists x: [x\in A \implies x\notin A]$$ So, to avoid confusion, you would probably want to avoid such constructs in mathematics. For a formal development, see The Drinker's Paradox: A Tale of Three Paradoxes at my blog.

$\forall x\in F (A(x)\to D(x))$ versus $\exists x\in F(A(x)\wedge D(x))$

$\forall x\in F (A(x)\to D(x))$ says "any fruit, if it is an apple, then it is delicious," or simply, "apples are delicious fruit".

$\forall x\in F (A(x)\wedge D(x))$ says "any fruit, is an apple and is delicious", or simply "all fruit are delicious apples."

Which says what you mean?

$\exists x\in F(A(x)\wedge D(x))$ says "some fruits, is an apple and is delicious," or simply "there is a delicious apple".

$\exists x\in F (A(x)\to D(x))$ says "there is some fruit, that if it were an apple then it would be delicious". Witness this mouldy orange.

Which says what you mean?

It's not that $\exists x\, (F(x) \wedge (A(x) \to D(x)))$ is incorrect -- it isn't, for your interpretations of $F$, $A$ and $D$. It's just not usually what you mean to say: it's weaker (more broad, true of more things $x$) than $\exists x (F(x) \wedge (A(x) \wedge D(x)))$ – it can also be true of some pear $x$, for example.

The general rule (or, rule of thumb) that "$\forall$ goes with $\to$, and $\exists$ goes with $\wedge$" arises from how these two statements are rendered in first order logic:

1. All $A$s are $B$s
2. Some $A$s are $B$s

The analysis of these statements and how to reason with groups of them goes all the way back to (most famously) Aristotle.

1. is rendered as $\forall x \,(A(x) \to B(x))$,
2. is rendered as $\exists x\,(A(x) \wedge B(x))$.