Theorem unrab

Description: Union of two restricted class abstractions. (Contributed by NM, 25-Mar-2004)

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

Proof

Step	Hyp	Ref	Expression
1		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
2		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) }
3	1 2	uneq12i	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∪ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ) = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∪ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } )
4		df-rab	⊢ { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∨ 𝜓 ) } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∨ 𝜓 ) ) }
5		unab	⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∪ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } ) = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∨ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) }
6		andi	⊢ ( ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∨ 𝜓 ) ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∨ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) )
7	6	abbii	⊢ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∨ 𝜓 ) ) } = { 𝑥 ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ∨ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ) }
8	5 7	eqtr4i	⊢ ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∪ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } ) = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ ( 𝜑 ∨ 𝜓 ) ) }
9	4 8	eqtr4i	⊢ { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∨ 𝜓 ) } = ( { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } ∪ { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) } )
10	3 9	eqtr4i	⊢ ( { 𝑥 ∈ 𝐴 ∣ 𝜑 } ∪ { 𝑥 ∈ 𝐴 ∣ 𝜓 } ) = { 𝑥 ∈ 𝐴 ∣ ( 𝜑 ∨ 𝜓 ) }