Theorem rabun2

Description: Abstraction restricted to a union. (Contributed by Stefan O'Rear, 5-Feb-2015)

	Ref	Expression
Assertion	rabun2	⊢ { 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ∣ 𝜑 } = ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∪ { 𝑥 ∈ 𝐵 ∣ 𝜑 } )

Proof

Step	Hyp	Ref	Expression
1		df-rab	⊢ { 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ∧ 𝜑 ) }
2		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
3		df-rab	⊢ { 𝑥 ∈ 𝐵 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) }
4	2 3	uneq12i	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∪ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ) = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∪ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } )
5		elun	⊢ ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) )
6	5	anbi1i	⊢ ( ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ∧ 𝜑 ) ↔ ( ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) ∧ 𝜑 ) )
7		andir	⊢ ( ( ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) ∧ 𝜑 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∨ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
8	6 7	bitri	⊢ ( ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ∧ 𝜑 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∨ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) )
9	8	abbii	⊢ { 𝑥 ∣ ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ∧ 𝜑 ) } = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∨ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) }
10		unab	⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∪ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } ) = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∨ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) ) }
11	9 10	eqtr4i	⊢ { 𝑥 ∣ ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ∧ 𝜑 ) } = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∪ { 𝑥 ∣ ( 𝑥 ∈ 𝐵 ∧ 𝜑 ) } )
12	4 11	eqtr4i	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∪ { 𝑥 ∈ 𝐵 ∣ 𝜑 } ) = { 𝑥 ∣ ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ∧ 𝜑 ) }
13	1 12	eqtr4i	⊢ { 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ∣ 𝜑 } = ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∪ { 𝑥 ∈ 𝐵 ∣ 𝜑 } )