Theorem iunun

Description: Separate a union in an indexed union. (Contributed by NM, 27-Dec-2004) (Proof shortened by Mario Carneiro, 17-Nov-2016)

	Ref	Expression
Assertion	iunun	⊢ ∪ 𝑥 ∈ 𝐴 ( 𝐵 ∪ 𝐶 ) = ( ∪ 𝑥 ∈ 𝐴 𝐵 ∪ ∪ 𝑥 ∈ 𝐴 𝐶 )

Proof

Step	Hyp	Ref	Expression
1		r19.43	⊢ ( ∃ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 ∨ 𝑦 ∈ 𝐶 ) ↔ ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 ∨ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ) )
2		elun	⊢ ( 𝑦 ∈ ( 𝐵 ∪ 𝐶 ) ↔ ( 𝑦 ∈ 𝐵 ∨ 𝑦 ∈ 𝐶 ) )
3	2	rexbii	⊢ ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ ( 𝐵 ∪ 𝐶 ) ↔ ∃ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 ∨ 𝑦 ∈ 𝐶 ) )
4		eliun	⊢ ( 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐵 ↔ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 )
5		eliun	⊢ ( 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐶 ↔ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 )
6	4 5	orbi12i	⊢ ( ( 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐵 ∨ 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐶 ) ↔ ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 ∨ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ) )
7	1 3 6	3bitr4i	⊢ ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ ( 𝐵 ∪ 𝐶 ) ↔ ( 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐵 ∨ 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐶 ) )
8		eliun	⊢ ( 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 ( 𝐵 ∪ 𝐶 ) ↔ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ ( 𝐵 ∪ 𝐶 ) )
9		elun	⊢ ( 𝑦 ∈ ( ∪ 𝑥 ∈ 𝐴 𝐵 ∪ ∪ 𝑥 ∈ 𝐴 𝐶 ) ↔ ( 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐵 ∨ 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐶 ) )
10	7 8 9	3bitr4i	⊢ ( 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 ( 𝐵 ∪ 𝐶 ) ↔ 𝑦 ∈ ( ∪ 𝑥 ∈ 𝐴 𝐵 ∪ ∪ 𝑥 ∈ 𝐴 𝐶 ) )
11	10	eqriv	⊢ ∪ 𝑥 ∈ 𝐴 ( 𝐵 ∪ 𝐶 ) = ( ∪ 𝑥 ∈ 𝐴 𝐵 ∪ ∪ 𝑥 ∈ 𝐴 𝐶 )