Theorem uneq1 1605

Description: Equality theorem for union of two classes.

Assertion

Ref	Expression
uneq1	⊢ (A = B → (A ∪ C) = (B ∪ C))

Proof of Theorem uneq1

Step	Hyp	Ref	Expression
1		eleq2 1150	. . . 4 ⊢ (A = B → (x ∈ A ↔ x ∈ B))
2	1	orbi1d 467	. . 3 ⊢ (A = B → ((x ∈ A ∨ x ∈ C) ↔ (x ∈ B ∨ x ∈ C)))
3		elun 1601	. . 3 ⊢ (x ∈ (A ∪ C) ↔ (x ∈ A ∨ x ∈ C))
4		elun 1601	. . 3 ⊢ (x ∈ (B ∪ C) ↔ (x ∈ B ∨ x ∈ C))
5	2, 3, 4	3bitr4g 428	. 2 ⊢ (A = B → (x ∈ (A ∪ C) ↔ x ∈ (B ∪ C)))
6	5	cleqrd 1100	1 ⊢ (A = B → (A ∪ C) = (B ∪ C))

Syntax hints:   → wi 2   ∨ wo 195   = wceq 1091   ∈ wcel 1092   ∪ cun 1485

This theorem is referenced by:  uneq2 1606  uneq1i 1607  uneq1d 1610  uneq12 1613  unineq 1680  prprc 1839  unexb 1950  suceq 2288  unxpdom 3650  sshjvalt 5321

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-un 1490

metamath.org