Theorem isufl 22523

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 6672	. . 3 ⊢ (𝑥 = 𝑋 → (Fil‘𝑥) = (Fil‘𝑋))
2		fveq2 6672	. . . 4 ⊢ (𝑥 = 𝑋 → (UFil‘𝑥) = (UFil‘𝑋))
3	2	rexeqdv 3418	. . 3 ⊢ (𝑥 = 𝑋 → (∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔 ↔ ∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))
4	1, 3	raleqbidv 3403	. 2 ⊢ (𝑥 = 𝑋 → (∀𝑓 ∈ (Fil‘𝑥)∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔 ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))
5		df-ufl 22512	. 2 ⊢ UFL = {𝑥 ∣ ∀𝑓 ∈ (Fil‘𝑥)∃𝑔 ∈ (UFil‘𝑥)𝑓 ⊆ 𝑔}
6	4, 5	elab2g 3670	1 ⊢ (𝑋 ∈ 𝑉 → (𝑋 ∈ UFL ↔ ∀𝑓 ∈ (Fil‘𝑋)∃𝑔 ∈ (UFil‘𝑋)𝑓 ⊆ 𝑔))

Syntax hints:   → wi 4   ↔ wb 208   = wceq 1537   ∈ wcel 2114  ∀wral 3140  ∃wrex 3141   ⊆ wss 3938  ‘cfv 6357  Filcfil 22455  UFilcufil 22509  UFLcufl 22510

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2795

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-ral 3145  df-rex 3146  df-rab 3149  df-v 3498  df-dif 3941  df-un 3943  df-in 3945  df-ss 3954  df-nul 4294  df-if 4470  df-sn 4570  df-pr 4572  df-op 4576  df-uni 4841  df-br 5069  df-iota 6316  df-fv 6365  df-ufl 22512

This theorem is referenced by:  ufli  22524  numufl  22525  ssufl  22528  ufldom  22572