Metamath Proof Explorer


Theorem eliun

Description: Membership in indexed union. (Contributed by NM, 3-Sep-2003)

Ref Expression
Assertion eliun ( 𝐴 𝑥𝐵 𝐶 ↔ ∃ 𝑥𝐵 𝐴𝐶 )

Proof

Step Hyp Ref Expression
1 elex ( 𝐴 𝑥𝐵 𝐶𝐴 ∈ V )
2 elex ( 𝐴𝐶𝐴 ∈ V )
3 2 rexlimivw ( ∃ 𝑥𝐵 𝐴𝐶𝐴 ∈ V )
4 eleq1 ( 𝑦 = 𝐴 → ( 𝑦𝐶𝐴𝐶 ) )
5 4 rexbidv ( 𝑦 = 𝐴 → ( ∃ 𝑥𝐵 𝑦𝐶 ↔ ∃ 𝑥𝐵 𝐴𝐶 ) )
6 df-iun 𝑥𝐵 𝐶 = { 𝑦 ∣ ∃ 𝑥𝐵 𝑦𝐶 }
7 5 6 elab2g ( 𝐴 ∈ V → ( 𝐴 𝑥𝐵 𝐶 ↔ ∃ 𝑥𝐵 𝐴𝐶 ) )
8 1 3 7 pm5.21nii ( 𝐴 𝑥𝐵 𝐶 ↔ ∃ 𝑥𝐵 𝐴𝐶 )