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I came across this question on an exam and am not sure why I am wrong. The question:

Using the following predicates over the domain of people:

$$A(x) = \text{x is an American.}$$

$$B(x) = \text{x likes Burgers.}$$

$$H(x) = \text{x likes Hot Dogs.}$$

Correctly represent the statement, "All Americans like Burgers except those who like Hot Dogs."

So my answer was:

$$\forall x,[A(x)\longrightarrow (B(x)\iff \neg H(x))]$$

The correct answer was:

$$\forall x,[A(x) \land B(x) \longrightarrow \neg H(x)]$$

My logic for my answer is that the statement essentially says all Americans either like burgers or hot dogs. My professor justifies his answer by saying that the statement is still true even if the left hand side is false. Now, where I'm confused is I think both statements are true, and mine gives more information. I think that the 'correct' answer misrepresents the statement in the case where $A(x)$ is true, $B(x)$ is false, and $H(x)$ is false. This results in F -> T, which makes the conditional true. However, by the provided statement, this should never happen. Because, assume $H(x)$ is empty. Then we have all Americans like burgers. So $B(x)$ must be true. Then, if $H(x)$ gains people, $B(x)$ is only false when $H(x)$ is true, so this is a contradiction for the 'correct' statement. However, my answer correctly accounts for this.

If someone could clarify, that would be much appreciated, thank you.

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  • $\begingroup$ If from your propsed solution you remove the bi-conditional and replace it with $\to$ the two are equivalent. See Exportation $\endgroup$ Feb 12 at 16:35
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    $\begingroup$ For some basic information about writing mathematics at this site see, e.g., here, here, here and here. $\endgroup$ Feb 12 at 16:37
  • $\begingroup$ I'm aware how to LaTeX math, but this was like two lines of math, I didn't think it was necessary. $\endgroup$
    – Jackson
    Feb 12 at 16:39
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    $\begingroup$ To the top comment, @MauroALLEGRANZA, yes I have no doubt the two are equivalent if I replace the bi-conditional with an implies. That's not my question. My question is why my logic is incorrect. $\endgroup$
    – Jackson
    Feb 12 at 16:39
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    $\begingroup$ @Jackson It's necessary, or at least an important courtesy to readers. $\endgroup$ Feb 12 at 16:51

2 Answers 2

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The English statement is ambiguous. One construction is, as you say, that all Americans like either burgers or hot dogs but not both. Another is that all Americans like burgers except for those who like hot dogs, who may or may not also like burgers.

If the first meaning is intended, I agree that your answer is better than the professor's for the reasons you stated. If the second meaning is intended, the correct answer is:

$$\forall x ~((A(x) \land \lnot H(x)) \Rightarrow B(x)).$$

This sentence says that any American who doesn't like hot dogs does like burgers. That is not equivalent to either of the solutions proposed in the question.

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  • $\begingroup$ Thanks. This is my thinking as well. I think the statement is ambiguous, and even if it explicitly said what he intended, I still don't believe that the 'correct' answer is right. The second meaning is what is in intended, but since it is ambiguous/poor use of English language, I will bring it up to him. $\endgroup$
    – Jackson
    Feb 12 at 16:43
  • $\begingroup$ @Jackson If the second meaning was intended, then the professor's answer is clearly wrong because it says (among other things) that no American likes both burgers and hot dogs. $\endgroup$ Feb 12 at 16:54
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I am totally with you! While English is notoriously ambiguous, I think the most reasonable interpretation of a statement like: "All P's, with the exception of Q's, are R's" is to say that the P's that are not Q's are R's, but the P's that are Q's are not R's. When we say that the Q's are the exception, I think it is pretty clear that it is meant that being R does not hold for them.

And even if your professor wants to interpret the statement as saying that those P that are not Q's are R's, while for those P's that are Q's we don't know whether they are R's, you would get:

$\forall x (A(x) \to (\neg H(x) \to B(x))$

or, equivalently:

$\forall x (A(x) \to (\neg B(x) \to H(x))$

or:

$\forall x ((A(x) \land \neg H(x)) \to B(x))$

or:

$\forall x ((A(x) \land \neg B(x)) \to H(x))$

but certainly not your professor's:

$\forall x ((A(x) \land B(x)) \to \neg H(x))$

because that would be true if no Americans like either hot dogs or burgers ... meaning that even those who don;t like hot dogs still don;t burgers either. No reasonable interpretation would say that!

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    $\begingroup$ Thank you. I completely agree. Logic makes sense to me again! Haha $\endgroup$
    – Jackson
    Feb 12 at 16:59
  • $\begingroup$ Might be worth establishing that the underlying meaning of the first interpretation is actually "All Americans like Burgers except all those who like Hot Dogs." Otherwise it is "All Americans like Burgers except some of those who like Hot Dogs" which could also be interpreted as "All Americans like Burgers except perhaps some of those who like Hot Dogs", allowing for the possibility of "all Americans like Burgers", which is an interesting conclusion. $\endgroup$ Feb 12 at 17:14
  • $\begingroup$ ... and careful about committing the Existential Fallacy. $\endgroup$ Feb 12 at 17:15

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