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.
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.