Translating statements (discrete math)

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SUMMARY

The discussion focuses on translating logical statements from discrete mathematics into formal expressions. The correct translations for the statements are established as ∃x(R(x) ∨ ~Q(x)) for "Some excuses are unsatisfactory" and ∀x(P(x) → Q(x)) for "All clear explanations are satisfactory." The confusion arises from the interpretation of logical implications and the use of conjunction versus disjunction in the translations. The participants clarify that the original attempt at translation does not accurately reflect the logical structure required by the statements.

PREREQUISITES
  • Understanding of predicate logic and quantifiers
  • Familiarity with logical operators: AND (∧), OR (∨), and NOT (¬)
  • Basic knowledge of discrete mathematics concepts
  • Ability to interpret and construct truth tables
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  • Study the differences between logical implications and equivalences in predicate logic
  • Learn about the use of quantifiers in formal logic
  • Explore truth table constructions for various logical expressions
  • Review common pitfalls in translating natural language statements into formal logic
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Students of discrete mathematics, educators teaching logic, and anyone interested in formal logic translations and their applications in mathematical reasoning.

Joseph1739
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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.
 

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