Theorem unrab 2273

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 2270	. . 3 ⊢ ({x∣(x ∈ A ⋀ φ)} ∪ {x∣(x ∈ A ⋀ ψ)}) = {x∣((x ∈ A ⋀ φ) ⋁ (x ∈ A ⋀ ψ))}
2		andi 606	. . . 4 ⊢ ((x ∈ A ⋀ (φ ⋁ ψ)) ↔ ((x ∈ A ⋀ φ) ⋁ (x ∈ A ⋀ ψ)))
3	2	abbii 1578	. . 3 ⊢ {x∣(x ∈ A ⋀ (φ ⋁ ψ))} = {x∣((x ∈ A ⋀ φ) ⋁ (x ∈ A ⋀ ψ))}
4	1, 3	eqtr4 1501	. 2 ⊢ ({x∣(x ∈ A ⋀ φ)} ∪ {x∣(x ∈ A ⋀ ψ)}) = {x∣(x ∈ A ⋀ (φ ⋁ ψ))}
5		df-rab 1655	. . 3 ⊢ {x ∈ A∣φ} = {x∣(x ∈ A ⋀ φ)}
6		df-rab 1655	. . 3 ⊢ {x ∈ A∣ψ} = {x∣(x ∈ A ⋀ ψ)}
7	5, 6	uneq12i 2185	. 2 ⊢ ({x ∈ A∣φ} ∪ {x ∈ A∣ψ}) = ({x∣(x ∈ A ⋀ φ)} ∪ {x∣(x ∈ A ⋀ ψ)})
8		df-rab 1655	. 2 ⊢ {x ∈ A∣(φ ⋁ ψ)} = {x∣(x ∈ A ⋀ (φ ⋁ ψ))}
9	4, 7, 8	3eqtr4 1508	1 ⊢ ({x ∈ A∣φ} ∪ {x ∈ A∣ψ}) = {x ∈ A∣(φ ⋁ ψ)}

Syntax hints:   ⋁ wo 222   ⋀ wa 223   = wceq 958   ∈ wcel 960  {cab 1466  {crab 1651   ∪ cun 2048

This theorem is referenced by:  kmlem3 4777

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-rab 1655  df-v 1815  df-un 2053

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