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The following two sentences in the language of $\mathbb{N}$ are logically equivalent, in the sense that first-order logic alone is enough to get from one to the other.

  1. For all $a,b;$ if there exists $k$ such that $ak=b$, then $a\mid b$.
  2. For all $a,b$ and all $k$, if $ak=b$, then $a\mid b$.

Many similar examples abound in mathematics. Is there any reason to prefer one phrasing over the other?

goblin GONE
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1 Answers1

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The first variant seems to occur more naturally and match the ways we think as it compares two predicates $\Phi(a,b) \equiv \exists k(ak=b)$ and $\Psi(a,b)\equiv a|b$. It also has the advantage that we can in fact formulate the equivalence $$ \forall a\forall b(\exists k(ak=b)\leftrightarrow a|b).$$ On the other hand, the second variant has the advantage that all quantors have been "pulled out". Moreover, there is no existential quantifier so that we may infer $$ ak=b\to a|b$$ for arbitrary values of $k,a,b$, which can be advantageous for formal deductions.

Hagen von Eitzen
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