Theorem ssequn1 3142

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		pm4.72 769	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)))
3		elun 3113	. . . . 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 1399	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ (𝐴 ∪ 𝐵) ↔ 𝑥 ∈ 𝐵))
7		dfss2 2988	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
8		dfcleq 2075	. 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 661  ∀wal 1282   = wceq 1284   ∈ wcel 1433   ∪ cun 2971   ⊆ wss 2973

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 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-un 2977  df-in 2979  df-ss 2986

This theorem is referenced by:  ssequn2  3145  uniop  4010  pwssunim  4039  unisuc  4168  unisucg  4169  rdgisucinc  5995  oasuc  6067  omsuc  6074