Theorem isbasisg 23141

Description: Express the predicate "the set 𝐵 is a basis for a topology". (Contributed by NM, 17-Jul-2006.)

Assertion

Ref	Expression
isbasisg	⊢ (𝐵 ∈ 𝐶 → (𝐵 ∈ TopBases ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥 ∩ 𝑦) ⊆ ∪ (𝐵 ∩ 𝒫 (𝑥 ∩ 𝑦))))

Distinct variable group:   𝑥,𝑦,𝐵

Allowed substitution hints:   𝐶(𝑥, 𝑦)

Proof of Theorem isbasisg

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ineq1 4169	. . . . . 6 ⊢ (𝑧 = 𝐵 → (𝑧 ∩ 𝒫 (𝑥 ∩ 𝑦)) = (𝐵 ∩ 𝒫 (𝑥 ∩ 𝑦)))
2	1	unieqd 4890	. . . . 5 ⊢ (𝑧 = 𝐵 → ∪ (𝑧 ∩ 𝒫 (𝑥 ∩ 𝑦)) = ∪ (𝐵 ∩ 𝒫 (𝑥 ∩ 𝑦)))
3	2	sseq2d 3972	. . . 4 ⊢ (𝑧 = 𝐵 → ((𝑥 ∩ 𝑦) ⊆ ∪ (𝑧 ∩ 𝒫 (𝑥 ∩ 𝑦)) ↔ (𝑥 ∩ 𝑦) ⊆ ∪ (𝐵 ∩ 𝒫 (𝑥 ∩ 𝑦))))
4	3	raleqbi1dv 3336	. . 3 ⊢ (𝑧 = 𝐵 → (∀𝑦 ∈ 𝑧 (𝑥 ∩ 𝑦) ⊆ ∪ (𝑧 ∩ 𝒫 (𝑥 ∩ 𝑦)) ↔ ∀𝑦 ∈ 𝐵 (𝑥 ∩ 𝑦) ⊆ ∪ (𝐵 ∩ 𝒫 (𝑥 ∩ 𝑦))))
5	4	raleqbi1dv 3336	. 2 ⊢ (𝑧 = 𝐵 → (∀𝑥 ∈ 𝑧 ∀𝑦 ∈ 𝑧 (𝑥 ∩ 𝑦) ⊆ ∪ (𝑧 ∩ 𝒫 (𝑥 ∩ 𝑦)) ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥 ∩ 𝑦) ⊆ ∪ (𝐵 ∩ 𝒫 (𝑥 ∩ 𝑦))))
6		df-bases 23140	. 2 ⊢ TopBases = {𝑧 ∣ ∀𝑥 ∈ 𝑧 ∀𝑦 ∈ 𝑧 (𝑥 ∩ 𝑦) ⊆ ∪ (𝑧 ∩ 𝒫 (𝑥 ∩ 𝑦))}
7	5, 6	elab2g 3642	1 ⊢ (𝐵 ∈ 𝐶 → (𝐵 ∈ TopBases ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥 ∩ 𝑦) ⊆ ∪ (𝐵 ∩ 𝒫 (𝑥 ∩ 𝑦))))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570   ∈ wcel 2146  ∀wral 3082   ∩ cin 3907   ⊆ wss 3908  𝒫 cpw 4567  ∪ cuni 4877  TopBasesctb 23139

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-in 3915  df-ss 3925  df-uni 4878  df-bases 23140

This theorem is used by:  isbasis2g  23142  basis1  23144  basdif0  23147  baspartn  23148  basqtop  23905