Theorem isufl 23937

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 6907	. . 3 ⊢ (𝑥 = 𝑋 → (Fil‘𝑥) = (Fil‘𝑋))
2		fveq2 6907	. . . 4 ⊢ (𝑥 = 𝑋 → (UFil‘𝑥) = (UFil‘𝑋))
3	2	rexeqdv 3325	. . 3 ⊢ (𝑥 = 𝑋 → (∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔 ↔ ∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))
4	1, 3	raleqbidv 3344	. 2 ⊢ (𝑥 = 𝑋 → (∀𝑓 ∈ (Fil‘𝑥)∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔 ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))
5		df-ufl 23926	. 2 ⊢ UFL = {𝑥 ∣ ∀𝑓 ∈ (Fil‘𝑥)∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔}
6	4, 5	elab2g 3683	1 ⊢ (𝑋 ∈ 𝑉 → (𝑋 ∈ UFL ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))

Syntax hints:   → wi 4   ↔ wb 206   = wceq 1537   ∈ wcel 2106  ∀wral 3059  ∃wrex 3068   ⊆ wss 3963  ‘cfv 6563  Filcfil 23869  UFilcufil 23923  UFLcufl 23924

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-uni 4913  df-br 5149  df-iota 6516  df-fv 6571  df-ufl 23926

This theorem is referenced by:  ufli  23938  numufl  23939  ssufl  23942  ufldom  23986