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 Emeritus Sci Advisor PF Gold P: 9,212 OK, I think I see what I did wrong. $(P\land Q)\Rightarrow R$ is equivalent to $\lnot R\Rightarrow (\lnot P\land Q)\lor(P\land\lnot Q)$, and you can't just drop one of the terms on the right in the last expression, which is essentially what I did. So let's go back to the implication that I want to rewrite: $$(I\subset\tau)\ \land\ \bigg(\bigcup_{i\in I}i\supset K\bigg)\Rightarrow\exists I_0(I_0\subset I,\ |I_0|<\infty,\ \bigcup_{i\in I_0}i\supset K)$$ It implies several different things, one of which is $$\bigg(\forall I_0(I_0\subset I,\ |I_0|<\infty) \bigcap_{i\in I_0}F_i\cup K\neq\emptyset\bigg)$$ $$\Rightarrow \bigg((I\not\subset\tau)\land\bigg(\bigcup_{i\in I}i\supset K\bigg)\bigg)\lor\bigg((I\subset\tau)\land\bigg(\bigcap_{i\in I}F_i\cup K\neq\emptyset\bigg)\bigg)$$ And this implies $$(I\subset\tau)\land\bigg(\forall I_0(I_0\subset I,\ |I_0|<\infty) \bigcap_{i\in I_0}F_i\cup K\neq\emptyset\bigg) \Rightarrow \bigcap_{i\in I}F_i\cup K\neq\emptyset$$ Alternatively: $$\forall I\bigg((I\subset\tau)\land\bigg(\forall I_0(I_0\subset I,\ |I_0|<\infty) \bigcap_{i\in I_0}F_i\cup K\neq\emptyset\bigg)\bigg)\quad \bigcap_{i\in I}F_i\cup K\neq\emptyset$$ This is the expression I want. I thought it would be possible to obtain it in a way that corresponds to $A\Rightarrow B$ if and only if $\lnot B\Rightarrow\lnot A$, but it was more difficult than that.