How to translate a first-order formula to natural language

Question

I have to do a simple translation of a sentence in first-order logic to natural language. The sentence is $$\exists x \forall y (\text{soldier}(x) \land \text{general}(x,y)),$$

where $\text{soldier}(x) $ means "$x$ is a soldier" and $\text{general}(x,y)$ means "$x$ is the general of $y$".

Now this formula, as with a couple of minutes thinking, means in natural language $$\text{there is someone who is everyone's general.}$$

My question is: To what does that formula translate, if we replace $\land$ with $\rightarrow $ ?

Somehow it seems to me that in natural language it would mean the same, but that can't be, since $\exists x \forall y (\text{soldier}(x) \rightarrow \text{general}(x,y))$ is a different formula than the one above.

Answer 1

The sentence $$\exists x \forall y \; (\text{soldier}(x) \land \text{general}(x,y))$$ means there is someone who is everyone's general and also a soldier, while $$\exists x \forall y \; (\text{soldier}(x) \rightarrow \text{general}(x,y))$$ means if everyone is a soldier, one of them is everyone's general.

In the second case, if there is any $x$ who isn't a soldier, for every $y$ and this $x$ the formula under quantificators is true (since implication is true whenever the premise is false). So, your sentence is true whenever there is someone who is not a soldier; on the other hand, if everyone is a soldier, then sentence reduces to $$\exists x \forall y \; (\text{general}(x,y)).$$

Answer 2

(NB: Assuming the domain is a subset of "people". )

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The following statement reads "There exists someone who is a soldier, and they are everyone's general." $$\exists x~\forall y~(\operatorname{soldier}(x)\wedge\operatorname{general}(x,y))$$ I assert that this is true if and only if, indeed, there is a soldier who is general of everyone in the domain.

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The following statement reads "There exists someone who, if they are a soldier, then they are everyone's general." $$\exists x~\forall y~(\operatorname{soldier}(x)\to\operatorname{general}(x,y))$$ I assert that this is true if and only if either (a) everyone(in the domain) is a soldier and one of them is general of them all, or (b) there is someone who is not a soldier.

Answer 3

$$\exists x \forall y \; (\text{soldier}(x) \rightarrow \text{general}(x,y))$$ means that there is someone ($x$) such that if if he is a soldier, then he is general of everyone.

This may be true either because there is a soldier who is the general of everyone, or because there is at least one non-soldier.

It may be easier to unravel the meaning if you remember that $p\to q$ is the same as $\neg p\lor q$, so what you have is equivalent to

$$\exists x \forall y \; (\neg\text{soldier}(x) \lor \text{general}(x,y))$$

and because $\forall y(p\lor q(y))$ is the same as $p\lor\forall y\,q(y)$ when $p$ does not depend on $y$, this is the same as

$$\exists x \; (\neg\text{soldier}(x) \lor \forall y\, \text{general}(x,y))$$

and $\exists$ distributes over $\lor$ so this is again the same as

$$(\exists x \; \neg\text{soldier}(x)) \lor \exists x \forall y\, \text{general}(x,y)$$

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While this is all impeccable from a strictly technical point of view, it can feel a bit counterintuitive. That basic source of this strangeness is that you have an existentially quantified variable to the left of a $\to$ (where the quantifier ranges over the entire implication), which is not really a situation that $\to$ is designed to be useful for.

One famous example of this is the so-called drinker's paradox.