Comments on How to find dualized quantifier pairs in substitutionally quantified analysis?
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How to find dualized quantifier pairs in substitutionally quantified analysis?
Classically, $\exists$ and $\forall$ are dual. These don't seem to be the quantifiers Newton had quite in mind when differentiating. Suppose quantifiers attached separately to the x- and h-terms, "at least zero-many/much x or h." In this context, the h-terms are of a different sortal or typical aspect than the x-terms, and it's not that they themselves become 0 in the limit, but we consider the existential case where we toggle between "at least zero" and... what?
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Are there many and varying such dualities? Like (at least zero, at most zero), (at least zero, exactly zero) (roughly equivalent to the preceding, at least in outcome), (almost zero-much, almost all) (where -much is stronger than -many, passing from discrete cardinal magnitude to continuous material), etc.
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Is there a limit $(f: r \rightarrow 0) \circ (rh) = (0h)$? Is that one way to vanish the h-terms at the end of the derivative, by evaluating the model of the formula in which the h-subdomain is empty?
- (For Newton, then: "kinematically," the h-variable could be any physical thing that admits of continuous determination in this sense. The h-terms, or Newton's counterparts, were not automatically overloaded with specific physical sense, like "h-much metal" or "h-many units of energy/force," but are simply from any sort "not merely numerical variables" in an at least two-sorted logic. So considering their empty subdomain is considering the case of the formula where its merely numerical factors "exist.")
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| User | Comment | Date |
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| Ripheus24 | (no comment) | May 23, 2026 at 19:03 |
I confess I don't fully understand this question (I await further clarification), but in general dualization is not that complicated. If $P$ is the dual of $Q$ it just means that it acts on $\bot$ the same way that $Q$ acts on $\top$, and vice versa. So whenever you have $P$ (say, a quantifier), it's easy to find its dual: just define $Q(x_1,x_2,…) := ¬P(¬x_1,¬x_2,…)$.
E.g. let's say we want to find the dual of "there exist countably many $x$ such that…" ($\exists^{\aleph_0} x…$). The dual would be $¬\exists^{\aleph_0}x ¬…$ "there do not exist countably many $x$ such that not…". This expression expands easily to "for all $x$ (except perhaps either a finite or uncountable number of different $x$), …". Thus we have found the dual of the quantifier $\exists^{\aleph_0} x$!
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