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I can't figure out how to interpret this. Is my understanding of the statement correct?

"There exists an x such that for all y, if p(x) is true then x = y"

i. P(0) = true, if y = 1 then x != y, formula is not true. p(1) = true, if y = 0 then, x != y, this formula is not true.

ii. P(0) = true, same as above. P(1) = false, since the premise is false the statement should be true. But of course x != y if y = 0

Same for iii.

Am I sort of making sense?

Shiny_and_Chrome
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  • You are sort of making sense. A little clearer writing and you can completely make sense. – Sarvesh Ravichandran Iyer Jan 11 '17 at 09:36
  • Your reading of the formula is correct. – Mauro ALLEGRANZA Jan 11 '17 at 09:41
  • You have to consider in turn each sub-question... (i) both $P(0)$ and $P(1)$ are true. Thus, what happens to the formula "There exists an $x$ such that for all $y$, if $P(x)$ is true then $x = y$" ? – Mauro ALLEGRANZA Jan 11 '17 at 09:42
  • There are only two possbile value for the "initial" $x$ : $0$ and $1$. Consider then in order: for $x=0$ is it true that "for all $y$, if $P(0)$ is true then $0=y$" ? – Mauro ALLEGRANZA Jan 11 '17 at 09:44
  • Again, two possible choices for $y$ : $0$ and $1$. With $y=1$ we have $P(0) \to 0=1$; antecedent T and consequen F: thus the conditional is F. So it is not true for all $y$... – Mauro ALLEGRANZA Jan 11 '17 at 09:46
  • Now repeat with $x=1$... With $y=0$ we have $P(1) \to 1=0$ and the conclusion is the same. Final conclusion : it is not true that "there is an $x$ ..." – Mauro ALLEGRANZA Jan 11 '17 at 09:48
  • @MauroALLEGRANZA Ok, so to your first comment, since $P(0)$ and $P(1)$ are true, it implies x = y for all values of y, but of course this is not true, so the statement is false

    So, in all cases (i, ii, iii) the formula is false. The only way it can be true is if $P(0) = P(1) = 0$, which makes the formula a tautology.

     – Shiny_and_Chrome Jan 11 '17 at 09:49
  • The statement is false but the "summary" reason you are stating is not so clear.. It is false because there is no value of $x$ such that $P(x) \to (x=y)$ holds for every value of $y$. – Mauro ALLEGRANZA Jan 11 '17 at 09:53
  • OK, there are x such that P(x)→(x=y)P(x)→(x=y) holds for every value of y.

    But, we're talking about interpretations, so, for example, in (iii) it is given that $P(0) = false, and P(1) = true$ which makes the formula false under that interpretation.

    But you can have the case where the premise of the predicate statement always be false, so that the conclusion is false, but that still makes the statement vacuously true.

     – Shiny_and_Chrome Jan 11 '17 at 10:01
  • from Wikipedia: $\forall x:P(x)\Rightarrow Q(x)$, where it is the case that ${\displaystyle \forall x:\neg P(x)}$

    https://en.wikipedia.org/wiki/Vacuous_truth

     – Shiny_and_Chrome Jan 11 '17 at 10:02
  • Not correct; for (ii) $P(1)$ is false; thus $P(1) \to (1=y)$ is true for every $y$, because a conditional with antecedent F is T. And so, we have (at least) one value for $x$ such that $\forall y (P(x) \to (x=y))$ holds. – Mauro ALLEGRANZA Jan 11 '17 at 10:04
  • OK, I think I got it now. thank you. – Shiny_and_Chrome Jan 11 '17 at 10:06

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