Theorem ssequn1 3374

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		pm4.72 832	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)))
3		elun 3345	. . . . 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 1516	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
7		ssalel 3212	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
8		dfcleq 2223	. 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 713  ∀wal 1393   = wceq 1395   ∈ wcel 2200   ∪ cun 3195   ⊆ wss 3197

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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-un 3201  df-in 3203  df-ss 3210

This theorem is referenced by:  ssequn2  3377  uniop  4341  pwssunim  4372  unisuc  4501  unisucg  4502  rdgisucinc  6521  oasuc  6600  omsuc  6608  undifdc  7074