Theorem isufl 23861

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 6896	. . 3 ⊢ (𝑥 = 𝑋 → (Fil‘𝑥) = (Fil‘𝑋))
2		fveq2 6896	. . . 4 ⊢ (𝑥 = 𝑋 → (UFil‘𝑥) = (UFil‘𝑋))
3	2	rexeqdv 3315	. . 3 ⊢ (𝑥 = 𝑋 → (∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔 ↔ ∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))
4	1, 3	raleqbidv 3329	. 2 ⊢ (𝑥 = 𝑋 → (∀𝑓 ∈ (Fil‘𝑥)∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔 ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))
5		df-ufl 23850	. 2 ⊢ UFL = {𝑥 ∣ ∀𝑓 ∈ (Fil‘𝑥)∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔}
6	4, 5	elab2g 3666	1 ⊢ (𝑋 ∈ 𝑉 → (𝑋 ∈ UFL ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))

Syntax hints:   → wi 4   ↔ wb 205   = wceq 1533   ∈ wcel 2098  ∀wral 3050  ∃wrex 3059   ⊆ wss 3944  ‘cfv 6549  Filcfil 23793  UFilcufil 23847  UFLcufl 23848

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2696

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-sb 2060  df-clab 2703  df-cleq 2717  df-clel 2802  df-ral 3051  df-rex 3060  df-rab 3419  df-v 3463  df-dif 3947  df-un 3949  df-ss 3961  df-nul 4323  df-if 4531  df-sn 4631  df-pr 4633  df-op 4637  df-uni 4910  df-br 5150  df-iota 6501  df-fv 6557  df-ufl 23850

This theorem is referenced by:  ufli  23862  numufl  23863  ssufl  23866  ufldom  23910