Theorem ssequn2 1631

Description: A relationship between subclass and union.

Assertion

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

Proof of Theorem ssequn2

Step	Hyp	Ref	Expression
1		ssequn1 1628	. 2 ⊢ (A ⊆ B ↔ (A ∪ B) = B)
2		uncom 1604	. . 3 ⊢ (A ∪ B) = (B ∪ A)
3	2	cleq1i 1108	. 2 ⊢ ((A ∪ B) = B ↔ (B ∪ A) = B)
4	1, 3	bitr 151	1 ⊢ (A ⊆ B ↔ (B ∪ A) = B)

Syntax hints:   ↔ wb 127   = wceq 1091   ∪ cun 1485   ⊆ wss 1487

This theorem is referenced by:  pwssun 1917  ordssun 2330  ordequn 2331  onelun 2352  mapdom2 3389  fodomb 3615  chj0 5377

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-un 1490  df-in 1491  df-ss 1492

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