Theorem ssequn1 3185

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 139	. . . 4 ⊢ ((𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ 𝐵))
2		pm4.72 775	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)))
3		elun 3156	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
4	3	bibi1i 227	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ 𝐵))
5	1, 2, 4	3bitr4i 211	. . 3 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
6	5	albii 1411	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
7		dfss2 3028	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
8		dfcleq 2089	. 2 ⊢ ((𝐴 ∪ 𝐵) = 𝐵 ↔ ∀𝑥(𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
9	6, 7, 8	3bitr4i 211	1 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∪ 𝐵) = 𝐵)

Syntax hints:   → wi 4   ↔ wb 104   ∨ wo 667  ∀wal 1294   = wceq 1296   ∈ wcel 1445   ∪ cun 3011   ⊆ wss 3013

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077

This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-v 2635  df-un 3017  df-in 3019  df-ss 3026

This theorem is referenced by:  ssequn2  3188  uniop  4106  pwssunim  4135  unisuc  4264  unisucg  4265  rdgisucinc  6188  oasuc  6265  omsuc  6273  undifdc  6714