Theorem ssequn1 3154

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 138	. . . 4 ⊢ ((𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ 𝐵))
2		pm4.72 770	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)))
3		elun 3125	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
4	3	bibi1i 226	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ 𝐵))
5	1, 2, 4	3bitr4i 210	. . 3 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
6	5	albii 1400	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
7		dfss2 2999	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
8		dfcleq 2077	. 2 ⊢ ((𝐴 ∪ 𝐵) = 𝐵 ↔ ∀𝑥(𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
9	6, 7, 8	3bitr4i 210	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∪ 𝐵) = 𝐵)

Syntax hints:   → wi 4   ↔ wb 103   ∨ wo 662  ∀wal 1283   = wceq 1285   ∈ wcel 1434   ∪ cun 2982   ⊆ wss 2984

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065

This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-v 2614  df-un 2988  df-in 2990  df-ss 2997

This theorem is referenced by:  ssequn2  3157  uniop  4046  pwssunim  4075  unisuc  4204  unisucg  4205  rdgisucinc  6082  oasuc  6157  omsuc  6165  undifdc  6561