Theorem ssequn1 3434

Description: A relationship between subclass and union. Theorem 26 of [Suppes] p. 27. (Contributed by NM, 30-Aug-1993.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

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

Proof of Theorem ssequn1

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		bicom 191	. . . 4 ⊢ ((x ∈ B ↔ (x ∈ A ∨ x ∈ B)) ↔ ((x ∈ A ∨ x ∈ B) ↔ x ∈ B))
2		pm4.72 846	. . . 4 ⊢ ((x ∈ A → x ∈ B) ↔ (x ∈ B ↔ (x ∈ A ∨ x ∈ B)))
3		elun 3221	. . . . 5 ⊢ (x ∈ (A ∪ B) ↔ (x ∈ A ∨ x ∈ B))
4	3	bibi1i 305	. . . 4 ⊢ ((x ∈ (A ∪ B) ↔ x ∈ B) ↔ ((x ∈ A ∨ x ∈ B) ↔ x ∈ B))
5	1, 2, 4	3bitr4i 268	. . 3 ⊢ ((x ∈ A → x ∈ B) ↔ (x ∈ (A ∪ B) ↔ x ∈ B))
6	5	albii 1566	. 2 ⊢ (∀x(x ∈ A → x ∈ B) ↔ ∀x(x ∈ (A ∪ B) ↔ x ∈ B))
7		dfss2 3263	. 2 ⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))
8		dfcleq 2347	. 2 ⊢ ((A ∪ B) = B ↔ ∀x(x ∈ (A ∪ B) ↔ x ∈ B))
9	6, 7, 8	3bitr4i 268	1 ⊢ (A ⊆ B ↔ (A ∪ B) = B)

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∨ wo 357  ∀wal 1540   = wceq 1642   ∈ wcel 1710   ∪ cun 3208   ⊆ wss 3258

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-ss 3260

This theorem is used by:  ssequn2  3437  undif  3631  unsneqsn  3888  dflec2  6211