Theorem unrab 1694

Description: Union of two restricted class abstractions.

Assertion

Ref	Expression
unrab	⊢ ({x ∈ A∣φ} ∪ {x ∈ A∣ψ}) = {x ∈ A∣(φ ∨ ψ)}

Proof of Theorem unrab

Step	Hyp	Ref	Expression
1		unab 1691	. . 3 ⊢ ({x∣(x ∈ A ∧ φ)} ∪ {x∣(x ∈ A ∧ ψ)}) = {x∣((x ∈ A ∧ φ) ∨ (x ∈ A ∧ ψ))}
2		andi 456	. . . . 5 ⊢ ((x ∈ A ∧ (φ ∨ ψ)) ↔ ((x ∈ A ∧ φ) ∨ (x ∈ A ∧ ψ)))
3	2	biabi 1181	. . . 4 ⊢ {x∣(x ∈ A ∧ (φ ∨ ψ))} = {x∣((x ∈ A ∧ φ) ∨ (x ∈ A ∧ ψ))}
4	3	cleqcomi 1105	. . 3 ⊢ {x∣((x ∈ A ∧ φ) ∨ (x ∈ A ∧ ψ))} = {x∣(x ∈ A ∧ (φ ∨ ψ))}
5	1, 4	eqtr 1119	. 2 ⊢ ({x∣(x ∈ A ∧ φ)} ∪ {x∣(x ∈ A ∧ ψ)}) = {x∣(x ∈ A ∧ (φ ∨ ψ))}
6		df-rab 1208	. . 3 ⊢ {x ∈ A∣φ} = {x∣(x ∈ A ∧ φ)}
7		df-rab 1208	. . 3 ⊢ {x ∈ A∣ψ} = {x∣(x ∈ A ∧ ψ)}
8	6, 7	uneq12i 1609	. 2 ⊢ ({x ∈ A∣φ} ∪ {x ∈ A∣ψ}) = ({x∣(x ∈ A ∧ φ)} ∪ {x∣(x ∈ A ∧ ψ)})
9		df-rab 1208	. 2 ⊢ {x ∈ A∣(φ ∨ ψ)} = {x∣(x ∈ A ∧ (φ ∨ ψ))}
10	5, 8, 9	3eqtr4 1126	1 ⊢ ({x ∈ A∣φ} ∪ {x ∈ A∣ψ}) = {x ∈ A∣(φ ∨ ψ)}

Syntax hints:   ∨ wo 195   ∧ wa 196  {cab 1090   = wceq 1091   ∈ wcel 1092  {crab 1204   ∪ cun 1485

This theorem is referenced by:  kmlem3 3582

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-rab 1208  df-v 1349  df-un 1490

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