3
$\begingroup$

Derivation of the answer provided to my previous stack-question was elegant and answer thus obtained was correct according to Kenneth Rosan.

Proposition in question:

“[You cannot ride the roller coaster] if [you are under 4 feet tall] unless [you are older than 16 years old.]”

Which boils down to formula:

$\neg q$ if r unless s

Correct answer arrived was to formulate the compound proposition by grouping it as follows:

$((\neg q) \mathbf{\text{ if }} r) \mathbf{\text{ unless }} s$ (leads to right answer)

Whereas, I (wrongly) solved this question by grouping it as follows:

($\neg q$) if ($r$ unless $s$) (leads to wrong answer)

Here, both keywords if and unless generates implication ($\to$).

i.e. (Q if P) $\leftrightarrow$ (P $\to$Q) and (P unless Q) $\leftrightarrow$ ($\lnot Q \to P$)

Now, according Stanford reference

When an operand is surrounded by operators of equal precedence, the operand associates to the right.

Given example in reference:

$P \to Q \to R$ should be grouped as ($P \to (Q \to R))$

So, here lies my question:

How to approach such word problem in which operands have equal precedence?

If I follow Stanford reference (or generally excepted precedence) then, the answer thus results is wrong. If we go with other way around then, answer obtained is correct.

$\endgroup$
0
$\begingroup$

English words are not symbolic operators. Rules of the English language, which are both quite complicated and sometimes ambiguous, determine the groupings of words into phrases and/or catenas, the relationships between them, and the meanings of those relationships. These rules can be somewhat modeled in similar ways to a strictly-defined formal grammar, but never perfectly.

The reasons the example sentence groups phrases the way it does would be more on topic on english.stackexchange.com.

$\endgroup$
  • 1
    $\begingroup$ They are not mere words they are keywords. The keywords if and unless are very much defined in Kenneth Rosan and in other standard logic books. Ref. page 3 Also, by no means it is English.SE question. As, I am not concern with dictionary meaning of these keywords. $\endgroup$ – Ubi hatt Apr 19 '20 at 23:39
  • $\begingroup$ All examples in that document are ones where the normal English sense coincides with the truth table of a logical operator. I don't see anything there suggesting defining strings of English words in a way other than their normal sense. Even in mathematical papers, anything explained in words is subject to the rules of English (or the paper's natural language); anything explained in symbols is subject to the operator precedence rules for those symbols. $\endgroup$ – aschepler Apr 20 '20 at 0:03
  • $\begingroup$ Well by that logic everything falls in the domain of English.SE. $\endgroup$ – Ubi hatt Apr 20 '20 at 0:28
  • $\begingroup$ Also, I logically figured out the answer that why in this case we have to follows ((P $\to$ Q) $\to$ R) and not other way around. $\endgroup$ – Ubi hatt Apr 20 '20 at 0:30
  • $\begingroup$ So I can delete this answer if you think it's appropriate and you can edit your question to clarify that you are asking about interpretation of the proposition according to some specific set of rules, instead of according to normal English usage. (By the way, the associations of "if" and "and" in the previous sentence would change with the addition of a comma. But that's about English again.) $\endgroup$ – aschepler Apr 20 '20 at 1:10
0
$\begingroup$

The sentence

You cannot ride the roller coaster if you are under 4 feet tall unless you are older than 16 years old.

should tell us mainly one thing that young kids (younger than 16) shorter than 4 ft. are prohibited from riding. We can encode this idea as follows: $(r\land \neg s)\rightarrow \neg q$. Easily we can infer the sentence $(r\rightarrow \neg q)\lor s$ which is "the right" grouping.

We can consider the other way of grouping sentences, which is $(r\lor s)\rightarrow\neg q$. But by grouping atomic sentences this way we arrive at the situation where if you are older than 16 y.o. you cannot ride the roller coaster, which obviously not the idea we wanted to express with the original sentence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.