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Theorem elif 3614
Description: Membership in a conditional operator. (Contributed by NM, 14-Feb-2005.)
Assertion
Ref Expression
elif (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ ((𝜑𝐴𝐵) ∨ (¬ 𝜑𝐴𝐶)))

Proof of Theorem elif
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 elex 2811 . 2 (𝐴 ∈ if(𝜑, 𝐵, 𝐶) → 𝐴 ∈ V)
2 elex 2811 . . . 4 (𝐴𝐵𝐴 ∈ V)
32adantl 277 . . 3 ((𝜑𝐴𝐵) → 𝐴 ∈ V)
4 elex 2811 . . . 4 (𝐴𝐶𝐴 ∈ V)
54adantl 277 . . 3 ((¬ 𝜑𝐴𝐶) → 𝐴 ∈ V)
63, 5jaoi 721 . 2 (((𝜑𝐴𝐵) ∨ (¬ 𝜑𝐴𝐶)) → 𝐴 ∈ V)
7 eleq1 2292 . . . . . 6 (𝑥 = 𝐴 → (𝑥𝐵𝐴𝐵))
87anbi1d 465 . . . . 5 (𝑥 = 𝐴 → ((𝑥𝐵𝜑) ↔ (𝐴𝐵𝜑)))
9 eleq1 2292 . . . . . 6 (𝑥 = 𝐴 → (𝑥𝐶𝐴𝐶))
109anbi1d 465 . . . . 5 (𝑥 = 𝐴 → ((𝑥𝐶 ∧ ¬ 𝜑) ↔ (𝐴𝐶 ∧ ¬ 𝜑)))
118, 10orbi12d 798 . . . 4 (𝑥 = 𝐴 → (((𝑥𝐵𝜑) ∨ (𝑥𝐶 ∧ ¬ 𝜑)) ↔ ((𝐴𝐵𝜑) ∨ (𝐴𝐶 ∧ ¬ 𝜑))))
12 df-if 3603 . . . 4 if(𝜑, 𝐵, 𝐶) = {𝑥 ∣ ((𝑥𝐵𝜑) ∨ (𝑥𝐶 ∧ ¬ 𝜑))}
1311, 12elab2g 2950 . . 3 (𝐴 ∈ V → (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ ((𝐴𝐵𝜑) ∨ (𝐴𝐶 ∧ ¬ 𝜑))))
14 ancom 266 . . . 4 ((𝐴𝐵𝜑) ↔ (𝜑𝐴𝐵))
15 ancom 266 . . . 4 ((𝐴𝐶 ∧ ¬ 𝜑) ↔ (¬ 𝜑𝐴𝐶))
1614, 15orbi12i 769 . . 3 (((𝐴𝐵𝜑) ∨ (𝐴𝐶 ∧ ¬ 𝜑)) ↔ ((𝜑𝐴𝐵) ∨ (¬ 𝜑𝐴𝐶)))
1713, 16bitrdi 196 . 2 (𝐴 ∈ V → (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ ((𝜑𝐴𝐵) ∨ (¬ 𝜑𝐴𝐶))))
181, 6, 17pm5.21nii 709 1 (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ ((𝜑𝐴𝐵) ∨ (¬ 𝜑𝐴𝐶)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wa 104  wb 105  wo 713   = wceq 1395  wcel 2200  Vcvv 2799  ifcif 3602
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211
This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-if 3603
This theorem is referenced by:  iftrueb01  7404  pw1if  7406
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