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windsylph

  • 2 years ago

Logic/Discrete Math: Let x be a tool, and P(x) be the statement "an x is in excellent condition". How do you express "No tool is in excellent condition" using quantifiers, logical connectives, x, and P(x)? Should the negation sign be before the quantifier or the predicate? *update: new question on the last comment

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  1. anonymoustwo44
    • 2 years ago
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    It will be: |dw:1339138738460:dw| For all x tools, the negotion of P(x) will be true. which will then mean no tool will be in excellent condition since we've stated that for all x, negotion of p(x) is true. hence whatever tool that may be, p(x) is false.

  2. blockcolder
    • 2 years ago
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    Another way is this: \[\neg \exists x (P(x))\] Essentially, our answers are equivalent by De Morgan's Law for Quantifiers.

  3. anonymoustwo44
    • 2 years ago
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    negotion of p(x) is "x is not in excellent condition". now all we have to state is that for any tool x, it will not be in good condition so that no x is in excellent condition which will lead us to the statement "For all x, x is not in excellent condition" which will give us the answer: \[\forall x,negotion of P(x)\]

  4. windsylph
    • 2 years ago
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    Thanks, all. How about this: Let x be a student, y, be a television show, and Q(x,y) be the statement "a student has been in a television show." How do you express "No student has ever been on a television show"?

  5. anonymoustwo44
    • 2 years ago
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    negotion of Q(x,y) will be "a student has not been in a television show". now to show that no student has ever been on a television show, we are to state that for all students and for any tv show, no student has ever been on a television show, which is: |dw:1339141822205:dw|

  6. windsylph
    • 2 years ago
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    Thank you very much!

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