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This is a problem from "Mathematical Logic for the Humanities" fro mhttp://jiblm.org/guides/index.php?category=jiblmjournal:

Suppose that a homework problem asks you to write the converse of the conditional “If I cash in my chips, then I got a royal flush.” Suppose that a particular student’s response is “If I get a royal flush, then I will cash in my chips.” This student’s response is not correct. Briefly ex- plain why, by identifying the error and offering a correction. [Hint: the truth values of any of the statements involved aren’t what matters.]

My answer is that the error lies in the student changing the "got" in the original conclusion into a "get" in the hypothesis of the converse, thereby changing the meaning of the statement altogether. The correct converse should say, "If I got a royal flush, then I cashed in my chips."

Is this correct? Is there a simple test I can run the converse against to verify that it's the correct converse of a statement?

matto
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  • Well, in a similar spirit, I'd have said "If I got a royal flush, then I will cash in my chips." – lulu Dec 17 '23 at 17:19
  • I don't really see a difference between "got" and "get" here – IraeVid Dec 18 '23 at 15:00
  • @IraeVid I believe this exemplifies my confusion at what is the correct way to think about negating a statement when time is a factor. – matto Dec 20 '23 at 01:02

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The unmistakably correct answer that is obtained by "simply swapping the hypothesis and conclusion" (see text after Definition 50) is "If I got a royal flush, then I cash in [sic] my chips".

Since the phrase is in a natural language and is open to interpretation, Problem 53 is the kind of hair-splitting that drives students away from the course, regardless of their abilities. Deducting points from the student's answer for it being incorrect would be hypocritical, as when proving theorems in mathematics we rather freely interchange equivalent statements with each other — this helps shorten the texts. The pair of statements in question might or might not be equivalent, depending on context; and in a situation that immediately comes to mind, they are equivalent. However their mathematical content depends on formalization, which is swept under the rug here (it would be even more tiresome for the students).

In a Mathematical Logic course one surely needs to pay attention to the content of the statements, and the student's supposed answer warrants a comment that their $\tilde Q\to \tilde P$ may not be really equivalent to $Q \to P$. Though I don't see a good way for the lecturer to deliver a convincing explanation if they are asked to elaborate on this problem, and in any case it would have to detract from mathematics.

On the same page, instead of adding a single word ("Write a different statement..."), the author decides to hide the crux of the whole Problem 57 in brackets — as if the student should have invented it by themselves via double-think. A statement is always equivalent to itself — this is not a "too simple to be simple" situation. Sorry for the vent...

Amateur_Algebraist
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Actually, the problem deals more with negating a (simple) implication properly than wording it correctly.

Let's rename the two statements "Cashing in the chips" and "Having a royal flush" as $p$ and $q$ respectively, so that the first proposition is written as $p \rightarrow q$. Then, its contraposition is given by $p \rightarrow q \equiv \lnot q \rightarrow \lnot p$, which can be worded as "If I haven't got any royal flush, then (it means that) I didn't cash in my chips".

The statements of this problem have been a bit poorly chosen, in that $p$ and $q$ are not really simultaneous and our brains feel unnatural to conceive this causal relation in a chronologically reversed and negated order. confusion may come from the fact that the You can replace $p$ by "It's rainy" and $q$ by "It's cloudy" in order to convince yourself.

Abezhiko
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  • You have an equality between the negation of a conditional and its contrapositive, which is not correct. Typo? I think you don't want the negation before the brackets, i.e. the conditional is equivalent (use tribar or biconditional) to its contrapositive. – Mariusz Popieluch Dec 18 '23 at 13:50
  • @MariuszPopieluch Thank you for spotting the typos, my bad. It is now corrected. – Abezhiko Dec 18 '23 at 13:56
  • @Abezhiko This is not about contraposition. The manual (p16, Def50) states: "Let $P$ and $Q$ be statements. The conditional $Q \to P$ is called the converse of the conditional $P \to Q$." OP's problem is number 53. – Amateur_Algebraist Dec 19 '23 at 05:14