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My attempt for translating

$a^m \mid b^n$ implies $a^{m'}\mid b^{n'}$ only if $m'n \le mn'.$

to logical symbols is

$$a^m \mid b^n \implies a^{m'}\mid b^{n'} \ \ (m'n \le mn').$$

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    $\begingroup$ "A only if B" is $A \Rightarrow B$. Thus, the statement can be better symbolized with : $m′n ≤ mn′ \Rightarrow (a^m |b^n \Rightarrow \ldots)$ $\endgroup$
    – Mauro ALLEGRANZA
    Oct 3 at 9:05
  • $\begingroup$ @MauroALLEGRANZA No it's the other way round. It should be: $ \left( a^m \mid b^n \implies a^{m'} \mid b^{n'}\right) \implies (m'n \leq mn').$ $\endgroup$
    – Adam Rubinson
    Oct 3 at 9:19
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    $\begingroup$ Ok, there is some ambiguity... Maybe we have "$a^m∣b^n$ implies $[a^{m'}∣b^{n′} $ only if $(m′n≤mn′)]$", in which case is: "$a^m∣b^n$ implies $[(m′n≤mn′) \Rightarrow a^{m'}∣b^{n′}]$. $\endgroup$
    – Mauro ALLEGRANZA
    Oct 3 at 9:23

1 Answer 1

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$$a^m \mid b^n \implies a^{m'}\mid b^{n'} \ \ (m'n \le mn').\tag1$$

This is ambiguous; which of these is $$P(a,b,m,n)\implies Q(a,b,m,n) \;\;\Big(R(m,n)\Big)$$ intended to mean?

  • $R\;\text{ and }\;(P\implies Q)$
  • $R\implies(P\implies Q)$
  • $P\implies(R\implies Q)$
  • $P\implies(R\;\text{ and }\;Q)$

In $(1),$ do I read the parenthetical portion as “where $m'n \le mn'$ holds” or “if $m'n \le mn'$ holds”?

$a^m \mid b^n$ implies $a^{m'}\mid b^{n'}$ only if $m'n \le mn'.$

This is a little ambiguous; $$(a^m \mid b^n \implies a^{m'}\mid b^{n'})\;\text{ only if }\;m'n \le mn'$$ certainly means $$(a^m \mid b^n \implies a^{m'}\mid b^{n'})\implies m'n \le mn'.$$

Possibly, the word “only” was mistakenly inserted into the quoted line; $$(a^m \mid b^n \implies a^{m'}\mid b^{n'})\;\text{ if }\;m'n \le mn'$$ means $$m'n \le mn'\implies (a^m \mid b^n \implies a^{m'}\mid b^{n'})$$ or, equivalently, $$(m'n \le mn'\;\text{ and }\;a^m \mid b^n) \implies a^{m'}\mid b^{n'}$$ or, more explicitly, $$\forall m,n,a,b\;\Big((m'n \le mn'\;\text{ and }\;a^m \mid b^n) \implies a^{m'}\mid b^{n'}\Big).$$

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  • $\begingroup$ @MauroALLEGRANZA 1. Your first comment's "No" doesn't make sense, since it is not disagreeing with what I wrote. $\quad$ 2. My four bullets do not specifically pertain to the OP's example, and the point is that that general form is ambiguous and might—depending on its particular context—mean any of the four bullets. $\endgroup$
    – ryang
    Oct 3 at 16:51

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