Theorem basgen 14994

Description: Given a topology 𝐽, show that a subset 𝐵 satisfying the third antecedent is a basis for it. Lemma 2.3 of [Munkres] p. 81 using abbreviations. (Contributed by NM, 22-Jul-2006.) (Revised by Mario Carneiro, 2-Sep-2015.)

Assertion

Ref	Expression
basgen	⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)

Proof of Theorem basgen

Step	Hyp	Ref	Expression
1		tgss 14977	. . . 4 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
2	1	3adant3 1044	. . 3 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
3		tgtop 14982	. . . 4 ⊢ (𝐽 ∈ Top → (topGen‘𝐽) = 𝐽)
4	3	3ad2ant1 1045	. . 3 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐽) = 𝐽)
5	2, 4	sseqtrd 3278	. 2 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ 𝐽)
6		simp3 1026	. 2 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → 𝐽 ⊆ (topGen‘𝐵))
7	5, 6	eqssd 3257	1 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)

Syntax hints:   → wi 4   ∧ w3a 1005   = wceq 1398   ∈ wcel 2205   ⊆ wss 3213  ‘cfv 5354  topGenctg 13488  Topctop 14911

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-13 2207  ax-14 2208  ax-ext 2216  ax-sep 4230  ax-pow 4289  ax-pr 4324  ax-un 4556

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1812  df-eu 2085  df-mo 2086  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-ral 2527  df-rex 2528  df-v 2817  df-sbc 3045  df-un 3217  df-in 3219  df-ss 3226  df-pw 3673  df-sn 3697  df-pr 3698  df-op 3700  df-uni 3917  df-br 4112  df-opab 4174  df-mpt 4175  df-id 4416  df-xp 4757  df-rel 4758  df-cnv 4759  df-co 4760  df-dm 4761  df-iota 5314  df-fun 5356  df-fv 5362  df-topgen 13494  df-top 14912

This theorem is referenced by:  basgen2  14995