Theorem unab

Description: Union of two class abstractions. (Contributed by NM, 29-Sep-2002) (Proof shortened by Andrew Salmon, 26-Jun-2011)

	Ref	Expression
Assertion	unab	⊢ ( { 𝑥 ∣ 𝜑 } ∪ { 𝑥 ∣ 𝜓 } ) = { 𝑥 ∣ ( 𝜑 ∨ 𝜓 ) }

Proof

Step	Hyp	Ref	Expression
1		sbor	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∨ [ 𝑦 / 𝑥 ] 𝜓 ) )
2		df-clab	⊢ ( 𝑦 ∈ { 𝑥 ∣ ( 𝜑 ∨ 𝜓 ) } ↔ [ 𝑦 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) )
3		df-clab	⊢ ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ↔ [ 𝑦 / 𝑥 ] 𝜑 )
4		df-clab	⊢ ( 𝑦 ∈ { 𝑥 ∣ 𝜓 } ↔ [ 𝑦 / 𝑥 ] 𝜓 )
5	3 4	orbi12i	⊢ ( ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ∨ 𝑦 ∈ { 𝑥 ∣ 𝜓 } ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∨ [ 𝑦 / 𝑥 ] 𝜓 ) )
6	1 2 5	3bitr4ri	⊢ ( ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ∨ 𝑦 ∈ { 𝑥 ∣ 𝜓 } ) ↔ 𝑦 ∈ { 𝑥 ∣ ( 𝜑 ∨ 𝜓 ) } )
7	6	uneqri	⊢ ( { 𝑥 ∣ 𝜑 } ∪ { 𝑥 ∣ 𝜓 } ) = { 𝑥 ∣ ( 𝜑 ∨ 𝜓 ) }