Theorem uneqri 1602

Description: Inference from membership to union.

Hypothesis

Ref	Expression
uneqri.1	⊢ ((x ∈ A ∨ x ∈ B) ↔ x ∈ C)

Assertion

Ref	Expression
uneqri	⊢ (A ∪ B) = C

Distinct variable group(s):   x,A   x,B   x,C

Proof of Theorem uneqri

Step	Hyp	Ref	Expression
1		elun 1601	. . 3 ⊢ (x ∈ (A ∪ B) ↔ (x ∈ A ∨ x ∈ B))
2		uneqri.1	. . 3 ⊢ ((x ∈ A ∨ x ∈ B) ↔ x ∈ C)
3	1, 2	bitr 151	. 2 ⊢ (x ∈ (A ∪ B) ↔ x ∈ C)
4	3	cleqri 1101	1 ⊢ (A ∪ B) = C

Syntax hints:   ↔ wb 127   ∨ wo 195   = wceq 1091   ∈ wcel 1092   ∪ cun 1485

This theorem is referenced by:  unidm 1603  unass 1615  un0 1721

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

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