Theorem ssequn1 1628

Description: A relationship between subclass and union. Theorem 26 of [Suppes] p. 27.

Assertion

Ref	Expression
ssequn1	⊢ (A ⊆ B ↔ (A ∪ B) = B)

Proof of Theorem ssequn1

Step	Hyp	Ref	Expression
1		df-un 1490	. . 3 ⊢ (A ∪ B) = {x∣(x ∈ A ∨ x ∈ B)}
2	1	cleq2i 1111	. 2 ⊢ (B = (A ∪ B) ↔ B = {x∣(x ∈ A ∨ x ∈ B)})
3		cleqcom 1103	. 2 ⊢ ((A ∪ B) = B ↔ B = (A ∪ B))
4		pm4.72 485	. . . 4 ⊢ ((x ∈ A → x ∈ B) ↔ (x ∈ B ↔ (x ∈ A ∨ x ∈ B)))
5	4	bial 695	. . 3 ⊢ (∀x(x ∈ A → x ∈ B) ↔ ∀x(x ∈ B ↔ (x ∈ A ∨ x ∈ B)))
6		dfss2 1497	. . 3 ⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))
7		cleqab 1174	. . 3 ⊢ (B = {x∣(x ∈ A ∨ x ∈ B)} ↔ ∀x(x ∈ B ↔ (x ∈ A ∨ x ∈ B)))
8	5, 6, 7	3bitr4 158	. 2 ⊢ (A ⊆ B ↔ B = {x∣(x ∈ A ∨ x ∈ B)})
9	2, 3, 8	3bitr4r 159	1 ⊢ (A ⊆ B ↔ (A ∪ B) = B)

Syntax hints:   → wi 2   ↔ wb 127   ∨ wo 195  ∀wal 672  {cab 1090   = wceq 1091   ∈ wcel 1092   ∪ cun 1485   ⊆ wss 1487

This theorem is referenced by:  ssequn2 1631  ssundif 1764  pwssun 1917  unop 1931  unisuc 2299  ordssun 2330  ordequn 2331  onuninsuc 2356  onun 2358  kmlem10 3589

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-un 1490  df-in 1491  df-ss 1492

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