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In my question Proof Verification: Properties of Divisors, I had originally said something along the lines of the following:

Since $b \nmid (a+c)$ we have, by definition, $a+c \neq bq_2 \text{ for } q_2 \in \mathbb{Z}$.

But was told that I might have meant the following:

Since $b \nmid (a+c)$ we have, by definition, $ \nexists q_2\in \mathbb{Z} \text{ such that } a+c = bq_2$.

Is one of the previous more valid than the other and if so, why is it more appropriate to use? (i.e. Is there a difference between these two statements?)

Mark S
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    Logically, $\forall q_2 \not P(q_2)$ is equivalent to $\not\exists q_2 P(q_2)$; the negation of a propositional function's existential quantification is a universal quantification of that propositional function's negation – J. W. Tanner Feb 23 '20 at 18:29
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    Actually, your sentence 'for $q_2\in\Bbb Z$' hides an implicit quantifier but it's not clear which one you mean. You want 'for all $q_2\in\Bbb Z$'. – Berci Feb 23 '20 at 18:41
  • As per previous comments, there is an implicit quantifier, but which must be obvious. Start from definition: $b | (a+c) \text { iff there is some } q_2 \in \mathbb Z \text { such that } a+c=bq_2$. Now, we have only to negate the definition.. – Mauro ALLEGRANZA Feb 24 '20 at 15:33
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    We have : $\lnot (b | (a+c)) \text { iff } \lnot \exists q_2 \in \mathbb Z ((a+c)=bq_2)$ i.e. $\forall q_2 \in \mathbb Z ((a+c) \ne bq_2)$. – Mauro ALLEGRANZA Feb 24 '20 at 15:36

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Your first claim was not correct.

To correct it, you should indicate that the statement is universally true; like so:

Since $b \nmid (a+c)$ we have, by definition, $a+c \neq bq_2 \text{ for }\textbf{any } q_2 \in \mathbb{Z}$.

$$b\nmid(a+c)\iff \forall q_2\in\Bbb Z~.(a+c)\neq bq_2$$

Which is now equivalent to the second claim.

Since $b \nmid (a+c)$ we have, by definition, $ \nexists q_2\in \mathbb{Z} \text{ such that } a+c = bq_2$.

$$b\nmid(a+c)\iff \lnot\exists q_2\in\Bbb Z~.(a+c) = bq_2$$

Graham Kemp
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