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According to Stanford:

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

If I follow the above rule,

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

means

  • $ Q \mathbf{\text{ if }} (R \mathbf{\text{ unless }} S),$

but the correct answer is

  • $(Q \mathbf{\text{ if }} R) \mathbf{\text{ unless }} S.$

How to approach word problems whose operands have equal precedence?

Both the keywords if and unless generate implication:
(Q if P) $\leftrightarrow$ (P $\to$Q)
(P unless Q) $\leftrightarrow$ ($\lnot Q \to P$).

ryang
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Ubi.B
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3 Answers3

3

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

Operator precedence is totally irrelevant to (reading or writing) verbal sentences, as they apply only to symbolic logic!

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

The quoted sentence is sloppily written, but can mean only either of the following, and common sense disambiguates it as Option 1 rather than Option 2:

  1. Unless you are older than 16 years old, if you are under 4 feet tall, then you cannot ride the roller coaster.

  2. If you are under 4 feet tall unless you are older than 16 years old, then you cannot ride the roller coaster.

ryang
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1

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.

aschepler
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    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. – Ubi.B Apr 19 '20 at 23:39
  • 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. – aschepler Apr 20 '20 at 00:03
  • Well by that logic everything falls in the domain of English.SE. – Ubi.B Apr 20 '20 at 00:28
  • 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. – Ubi.B Apr 20 '20 at 00:30
  • 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.) – aschepler Apr 20 '20 at 01:10
  • I know that English.SE doesn't entertain domain related terminology. I have already requested Mod to delete or close my question. Also, at the least you would have routed me to Philosophy.SE. Anyways, here is the logical explanation- If you pay attention to the proposition then, it boils down to (IF r THEN q) UNLESS s. Then you can expand further and just step before final result you have to use associative property. By this approach you don't break standard precedence order and you get desired result. Try it. – Ubi.B Apr 20 '20 at 01:34
1

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.

anfauglit
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