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I was reading the book "Advanced Calculus" by Loomis and Sternberg and I was following everything perfectly until I came across a paragraph which I have attached here.

attached paragraph

My first confusion is that I am unable to understand what the authors are trying to say in this paragraph about hanging quantifiers and how does the first ambiguous statement become true and second become false?. My second confusion arises when I came across two statements: $$'[(\forall x)(x<y)](\exists y)'$$ and $$'(\forall x)[(x<y)(\exists y)]'$$ My problem is that a quantifier has been used after the predicate $(x<y)$ in both the statements. How is this possible? And how does first is false and second is true? I have never seen such a way of writing logical statement.

My current understanding:

I understand that the order of quantifiers are important.

"Every $x$ is less than some $y$" is False.

"Some $y$ is greater than every $x$" is False.

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    $\begingroup$ It means simply that in natural language we can move around "quntifiers" in a statement, like e.g. "some y is greater than every x" causing troubles in corerctly understand the sentence. This is due to the fact that the order of quantifiers matters: $\forall x \exists y (x < y)$ is true while $\exists y \forall x (x < y)$ is false. $\endgroup$ – Mauro ALLEGRANZA Apr 4 '18 at 12:35
  • $\begingroup$ @MauroALLEGRANZA and why in the last sentence the authors are saying that 'every x is less than some y' is true? I understand 'some y is greater than every x' is false. $\endgroup$ – Vivek Apr 4 '18 at 12:39
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A formula like $∀x[(x < y)(∃y)]$, which is wrongly written according to every mathematical logic textbook, is proposed by the authors as a way to mimick more closely the way we use quantifiers in natural language.

The case of "every $x$ is less than some $y$" is ambiguous beacuse it is not clear which is the "inner" quantifier (the first applied one).

The use of parentheses in the formulas determine the scope of quantifiers.

In the formula $∀x[(x < y)(∃y)]$ the scope of the existential quantifier is the subformula $(x < y)$, and thus we have to read it as $∀x[(∃y)(x < y)]$, which is true in $\mathbb N$.

In the formula $[∀x(x < y)](∃y)$ the scope of the existential quantifier is the subformula $∀x(x < y)$, and thus we have to read it as $∃y[(∀x)(x < y)]$, which is false in $\mathbb N$.

In the case of "every $x$ is less than some $y$", we can read it as follows : for every $x$ we can find some $y$ (one is enough; choose : $x+1$) such that $(x < y)$.

In this reading, the right-hand existential quantifier is read as the "inner" one.

The second example has the same ambiguity. If we use the same "rule" above and read the right-hand universal quantifier as the "inner" one, we have : there is some $y$ such that for every $x$ : $(y > x)$.

The conclusion is :

use "hanging" quantifiers [quantifiers put at the end of the sentence, instead that in front of it] only if [you] are sure the reader will understand their order of application.

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    $\begingroup$ so you mean 'every x is less than some y' is same as $\forall x \exists y (x < y)$ and 'some y is greater than every x' is same as $\exists y \forall x (x < y)$? $\endgroup$ – Vivek Apr 4 '18 at 12:53
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    $\begingroup$ @Vivek - this is the more "natural" way to resolve the ambiguity of natural language corerctly discussed by the authors. $\endgroup$ – Mauro ALLEGRANZA Apr 4 '18 at 12:57
  • $\begingroup$ @MauroALLEGRANZA I understood what you are saying, just clarify one more thing, I am unable to justify to myself that 'every x is less than some y' is true. Don't you think the same? Let some y=7, every x is not less than 7, for example x=8,9,10... $\endgroup$ – user521346 Apr 4 '18 at 13:19
  • $\begingroup$ @MauroALLEGRANZA tell me if my following conclusion is right, "every x is less than some y", 'inner' quantifier is $\exists y$ and in "some y is greater than every x", 'inner' quantifier is $\forall x$. Am I right? $\endgroup$ – user521346 Apr 4 '18 at 14:18
  • $\begingroup$ @Simba - correct. But the ambiguity of natural language is "neutral" between the two: this is the gist of the (little bit convoluted) discussion in the book. $\endgroup$ – Mauro ALLEGRANZA Apr 4 '18 at 14:21

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