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No-data

  • one year ago

Can someone explain this in plain english, or spanish: \[\forall x \in \emptyset : P(x)\] is TRUE regardless of the value of P(x)

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  1. binarymimic
    • one year ago
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    this is vacuously true for all elements, x, in the empty set, P(x) is true there is no x to falsify the claim of P(x) so you can conclude the statement is true

  2. No-data
    • one year ago
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    I've been doing some research and I found that this is a shorthand for \[x\in \emptyset \Rightarrow P(x) \] Which is TRUE, but I don't see the relationship.

  3. No-data
    • one year ago
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    But that is like a convention because there is no x to make TRUE the statement.

  4. binarymimic
    • one year ago
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    doesn't matter the only thing that causes a condition to be false is if "True implies False" in this case, the conditional statement is always false, so it can imply anything, and the entire statement will always evaluate to true

  5. binarymimic
    • one year ago
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    in other words, the only thing that causes a conditional statement to be false is if the conditional is true AND the consequent is false

  6. No-data
    • one year ago
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    I understand what you say about the conditional statement, but what I'm not sure is how you can go from the universal quantifier to a conditional statement.

  7. No-data
    • one year ago
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    \[x\in \emptyset \Rightarrow P(x)\] totally makes sense to me. What I can't see is how is this equivalent to the universal quantification.

  8. ParthKohli
    • one year ago
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    Sorry, can anyone just tell me what \(\rm P(x)\) stands for?

  9. ParthKohli
    • one year ago
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    I do know that conditional is not equivalent to \(\rm if\cdots then\) statements.

  10. ParthKohli
    • one year ago
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    Oh, I get it.

  11. No-data
    • one year ago
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    \[P(x)\] is a predicate with a parameter x. Once you know the value of the parameter x you can say it becomes a proposition and you can say if its TRUE or FALSE. For example \[x > 0\]

  12. No-data
    • one year ago
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    if you give x the value -1 you can say that the proposition is FALSE.

  13. ParthKohli
    • one year ago
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    \[\forall x \in \emptyset :\rm P(x)\]This is true, means that \(\rm P(x)\) is true for all \(x\) in empty set. If a conditional is true, then: * The first value is false. * If not false, then the second value must be true. In other words \(1\implies 0\) is false.

  14. ParthKohli
    • one year ago
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    So \(x \in \emptyset \implies P(x)\) is true. If \(x \not \in \emptyset\), then the first part of the conditional is false. So the statement is true.

  15. ParthKohli
    • one year ago
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    If \(x \in \emptyset\), we know that \(P(x)\) is true so the conditional is true.

  16. ParthKohli
    • one year ago
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    So the statement is always true, and it is given in your original statement that it is true.

  17. ParthKohli
    • one year ago
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    Two statements having the same truth tables are equivalent.

  18. binarymimic
    • one year ago
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    what i was trying to say earlier was that a universal quantifier states a case for any x in the empty set so an equivalent statement is exactly as you have written you can state the case for any x in the empty set as \[(x \in \emptyset) \rightarrow P(x)\]

  19. No-data
    • one year ago
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    Then can I turn every universal quantification into an implication?

  20. binarymimic
    • one year ago
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    the only difference is in the notation. \[\forall x \in \emptyset, P(x)\] just says "for any x in the empty set, P(x) is true" whereas \[(x \in \emptyset) \rightarrow P(x)\] just says "if x is in the empty set, then P(x) is true"

  21. binarymimic
    • one year ago
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    yes you can

  22. binarymimic
    • one year ago
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    generally speaking \[\forall x \in X, P(x)\] is equivalent to \[(x \in X) \rightarrow P(x)\]

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