Translating statements (discrete math)

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The discussion focuses on translating logical statements related to discrete math, specifically concerning the phrases "Some excuses are unsatisfactory" and "All clear explanations are satisfactory." The correct translation for the first statement is identified as ∃x(R(x) ∨ ~Q(x)), while the initial attempt was ∃x(R(x) → ~Q(x)), which was deemed incorrect due to its logical implications. Participants clarify that the presence of an "if...then" structure is not necessary for the translation, as shown in the second statement, which translates to ∀x(P(x) → Q(x). The conversation highlights the nuances of logical operators and the importance of accurately representing the original statements. Ultimately, the distinctions in translations underscore the complexity of logical expressions in discrete mathematics.
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Homework Statement


Let:
P(x) = "x is a clear explanation"
Q(x) = "x is satisfactory"
R(x) = "x is an excuse"
x be the domain of all English texts

Translate:
1. Some excuses are unsatisfactory
2. All clear explanations are satisfactory

Homework Equations


∃ for "some"

The Attempt at a Solution


(1) ∃x(R(x) → ~Q(x))

I don't understand why this is not the correct translation.
The answer is ∃x(R(x) ∨ ~Q(x)), and I understand that the truth tables for these two are not equivalent, but when I read my answer, it makes sense: "There exists an x such that if x is an excuse, then x is unsatisfactory."I tried reasoning that there is no "If..then" which is why AND was used instead, but for problem (2), the answer is ∀x(P(x)→Q(x)) even when the statement doesn't contain at "If...then".
 
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The book answer you have quoted is definitely wrong. It is true as long as there is one or more excuse, even if all excuses are satisfactory. So it doesn't match the statement.

I would translate the phrase as ##\exists x(R(x)\wedge\neg Q(x))##, which is different from both. However it would be the same as the book answer if they accidentally typeset an OR (##\vee##) instead of an AND (##\wedge##).

Your translation doesn't work because it is true as long as there is some x that is not an excuse, even if all x that are excuses are satisfactory. For instance if a = 'I like chocolate' then ##R(a)\to\neg Q(a)## is vacuously true because the antecedent (the item before the arrow) is false. 'I like chocolate' is not an excuse.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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