Theorem isufl 22518

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.)

Assertion

Ref	Expression
isufl	⊢ (𝑋 ∈ 𝑉 → (𝑋 ∈ UFL ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))

Distinct variable group:   𝑓,𝑔,𝑋

Allowed substitution hints:   𝑉(𝑓,𝑔)

Proof of Theorem isufl

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		fveq2 6645	. . 3 ⊢ (𝑥 = 𝑋 → (Fil‘𝑥) = (Fil‘𝑋))
2		fveq2 6645	. . . 4 ⊢ (𝑥 = 𝑋 → (UFil‘𝑥) = (UFil‘𝑋))
3	2	rexeqdv 3365	. . 3 ⊢ (𝑥 = 𝑋 → (∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔 ↔ ∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))
4	1, 3	raleqbidv 3354	. 2 ⊢ (𝑥 = 𝑋 → (∀𝑓 ∈ (Fil‘𝑥)∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔 ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))
5		df-ufl 22507	. 2 ⊢ UFL = {𝑥 ∣ ∀𝑓 ∈ (Fil‘𝑥)∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔}
6	4, 5	elab2g 3616	1 ⊢ (𝑋 ∈ 𝑉 → (𝑋 ∈ UFL ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))

Syntax hints:   → wi 4   ↔ wb 209   = wceq 1538   ∈ wcel 2111  ∀wral 3106  ∃wrex 3107   ⊆ wss 3881  ‘cfv 6324  Filcfil 22450  UFilcufil 22504  UFLcufl 22505

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3an 1086  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-nfc 2938  df-ral 3111  df-rex 3112  df-v 3443  df-un 3886  df-in 3888  df-ss 3898  df-sn 4526  df-pr 4528  df-op 4532  df-uni 4801  df-br 5031  df-iota 6283  df-fv 6332  df-ufl 22507

This theorem is referenced by:  ufli  22519  numufl  22520  ssufl  22523  ufldom  22567