Theorem ssequn1 3107

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 128	. . . 4 ⊢ ((x ∈ B ↔ (x ∈ A ∨ x ∈ B)) ↔ ((x ∈ A ∨ x ∈ B) ↔ x ∈ B))
2		pm4.72 735	. . . 4 ⊢ ((x ∈ A → x ∈ B) ↔ (x ∈ B ↔ (x ∈ A ∨ x ∈ B)))
3		elun 3078	. . . . 5 ⊢ (x ∈ (A ∪ B) ↔ (x ∈ A ∨ x ∈ B))
4	3	bibi1i 217	. . . 4 ⊢ ((x ∈ (A ∪ B) ↔ x ∈ B) ↔ ((x ∈ A ∨ x ∈ B) ↔ x ∈ B))
5	1, 2, 4	3bitr4i 201	. . 3 ⊢ ((x ∈ A → x ∈ B) ↔ (x ∈ (A ∪ B) ↔ x ∈ B))
6	5	albii 1356	. 2 ⊢ (∀x(x ∈ A → x ∈ B) ↔ ∀x(x ∈ (A ∪ B) ↔ x ∈ B))
7		dfss2 2928	. 2 ⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))
8		dfcleq 2031	. 2 ⊢ ((A ∪ B) = B ↔ ∀x(x ∈ (A ∪ B) ↔ x ∈ B))
9	6, 7, 8	3bitr4i 201	1 ⊢ (A ⊆ B ↔ (A ∪ B) = B)

Syntax hints:   → wi 4   ↔ wb 98   ∨ wo 628  ∀wal 1240   = wceq 1242   ∈ wcel 1390   ∪ cun 2909   ⊆ wss 2911

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-un 2916  df-in 2918  df-ss 2925

This theorem is referenced by:  ssequn2  3110  uniop  3983  pwssunim  4012  unisuc  4116  unisucg  4117  rdgisucinc  5912  oasuc  5983  omsuc  5990