Theorem ssequn2 3116

Description: A relationship between subclass and union. (Contributed by NM, 13-Jun-1994.)

Assertion

Ref	Expression
ssequn2	⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)

Proof of Theorem ssequn2

Step	Hyp	Ref	Expression
1		ssequn1 3113	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∪ 𝐵) = 𝐵)
2		uncom 3087	. . 3 ⊢ (𝐴 ∪ 𝐵) = (𝐵 ∪ 𝐴)
3	2	eqeq1i 2047	. 2 ⊢ ((𝐴 ∪ 𝐵) = 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)
4	1, 3	bitri 173	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐴) = 𝐵)

Syntax hints:   ↔ wb 98   = wceq 1243   ∪ cun 2915   ⊆ wss 2917

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-un 2922  df-in 2924  df-ss 2931

This theorem is referenced by:  unabs  3167  pwssunim  4021  pwundifss  4022  oneluni  4168  relresfld  4847  relcoi1  4849  fsnunf  5362