Talk:Scott–Potter set theory

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Scott began with an axiom he declined to name: the atomic formula x∈y implies that y is a set. In symbols:

∀x,y∃a[x∈y→y=a].

Wouldn’t “x∈y implies that y is a set” be more clearly expressed as

∀x,y [x∈y]→∃a y=a?

(more clearly)

∀x,y x∈y .→ ∃a y=a

In particular, if x∈y is always false and the domain of quantification for a (but not x and y) is empty, the informal version and mine are both true, but the formal version in the article is not.

I’m reluctant to correct this, since I’m not familiar with Potter nor the formalism (in particular the domain(s) of quantification), and not sure which is wrong, the explanation or the formalization.

—FlashSheridan 21:12, 27 June 2007 (UTC)