Theorem isufl 21937

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 6332	. . 3 ⊢ (𝑥 = 𝑋 → (Fil‘𝑥) = (Fil‘𝑋))
2		fveq2 6332	. . . 4 ⊢ (𝑥 = 𝑋 → (UFil‘𝑥) = (UFil‘𝑋))
3	2	rexeqdv 3294	. . 3 ⊢ (𝑥 = 𝑋 → (∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔 ↔ ∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))
4	1, 3	raleqbidv 3301	. 2 ⊢ (𝑥 = 𝑋 → (∀𝑓 ∈ (Fil‘𝑥)∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔 ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))
5		df-ufl 21926	. 2 ⊢ UFL = {𝑥 ∣ ∀𝑓 ∈ (Fil‘𝑥)∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔}
6	4, 5	elab2g 3504	1 ⊢ (𝑋 ∈ 𝑉 → (𝑋 ∈ UFL ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))

Syntax hints:   → wi 4   ↔ wb 196   = wceq 1631   ∈ wcel 2145  ∀wral 3061  ∃wrex 3062   ⊆ wss 3723  ‘cfv 6031  Filcfil 21869  UFilcufil 21923  UFLcufl 21924

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-9 2154  ax-10 2174  ax-11 2190  ax-12 2203  ax-13 2408  ax-ext 2751

This theorem depends on definitions:  df-bi 197  df-an 383  df-or 835  df-3an 1073  df-tru 1634  df-ex 1853  df-nf 1858  df-sb 2050  df-clab 2758  df-cleq 2764  df-clel 2767  df-nfc 2902  df-ral 3066  df-rex 3067  df-rab 3070  df-v 3353  df-dif 3726  df-un 3728  df-in 3730  df-ss 3737  df-nul 4064  df-if 4226  df-sn 4317  df-pr 4319  df-op 4323  df-uni 4575  df-br 4787  df-iota 5994  df-fv 6039  df-ufl 21926

This theorem is referenced by:  ufli  21938  numufl  21939  ssufl  21942  ufldom  21986