Theorem ssequn1 3307

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 140	. . . 4 ⊢ ((𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ 𝐵))
2		pm4.72 827	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)))
3		elun 3278	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
4	3	bibi1i 228	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ 𝐵))
5	1, 2, 4	3bitr4i 212	. . 3 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
6	5	albii 1470	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
7		dfss2 3146	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
8		dfcleq 2171	. 2 ⊢ ((𝐴 ∪ 𝐵) = 𝐵 ↔ ∀𝑥(𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
9	6, 7, 8	3bitr4i 212	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∪ 𝐵) = 𝐵)

Syntax hints:   → wi 4   ↔ wb 105   ∨ wo 708  ∀wal 1351   = wceq 1353   ∈ wcel 2148   ∪ cun 3129   ⊆ wss 3131

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2741  df-un 3135  df-in 3137  df-ss 3144

This theorem is referenced by:  ssequn2  3310  uniop  4257  pwssunim  4286  unisuc  4415  unisucg  4416  rdgisucinc  6388  oasuc  6467  omsuc  6475  undifdc  6925