Theorem isufl

Description: Define the (strong) ultrafilter lemma, parameterized over base sets. A set X satisfies the ultrafilter lemma if every filter on X is a subset of some ultrafilter. (Contributed by Mario Carneiro, 26-Aug-2015)

	Ref	Expression
Assertion	isufl	$$⊢X\in V\to \left(X\in \mathrm{UFL}\leftrightarrow \forall f\in \mathrm{Fil}\left(X\right)\exists g\in \mathrm{UFil}\left(X\right)f\subseteq g\right)$$

Proof

Step	Hyp	Ref	Expression
1		fveq2	$$⊢x=X\to \mathrm{Fil}\left(x\right)=\mathrm{Fil}\left(X\right)$$
2		fveq2	$$⊢x=X\to \mathrm{UFil}\left(x\right)=\mathrm{UFil}\left(X\right)$$
3	2	rexeqdv	$$⊢x=X\to \left(\exists g\in \mathrm{UFil}\left(x\right)f\subseteq g\leftrightarrow \exists g\in \mathrm{UFil}\left(X\right)f\subseteq g\right)$$
4	1 3	raleqbidv	$$⊢x=X\to \left(\forall f\in \mathrm{Fil}\left(x\right)\exists g\in \mathrm{UFil}\left(x\right)f\subseteq g\leftrightarrow \forall f\in \mathrm{Fil}\left(X\right)\exists g\in \mathrm{UFil}\left(X\right)f\subseteq g\right)$$
5		df-ufl	$$⊢\mathrm{UFL}=\left\{x|\forall f\in \mathrm{Fil}\left(x\right)\exists g\in \mathrm{UFil}\left(x\right)f\subseteq g\right\}$$
6	4 5	elab2g	$$⊢X\in V\to \left(X\in \mathrm{UFL}\leftrightarrow \forall f\in \mathrm{Fil}\left(X\right)\exists g\in \mathrm{UFil}\left(X\right)f\subseteq g\right)$$