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Metamath Proof Explorer

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 )