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Theorem elif 3649
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 2833 . 2 (𝐴 ∈ if(𝜑, 𝐵, 𝐶) → 𝐴 ∈ V)
2 elex 2833 . . . 4 (𝐴𝐵𝐴 ∈ V)
32adantl 277 . . 3 ((𝜑𝐴𝐵) → 𝐴 ∈ V)
4 elex 2833 . . . 4 (𝐴𝐶𝐴 ∈ V)
54adantl 277 . . 3 ((¬ 𝜑𝐴𝐶) → 𝐴 ∈ V)
63, 5jaoi 728 . 2 (((𝜑𝐴𝐵) ∨ (¬ 𝜑𝐴𝐶)) → 𝐴 ∈ V)
7 eleq1 2301 . . . . . 6 (𝑥 = 𝐴 → (𝑥𝐵𝐴𝐵))
87anbi1d 469 . . . . 5 (𝑥 = 𝐴 → ((𝑥𝐵𝜑) ↔ (𝐴𝐵𝜑)))
9 eleq1 2301 . . . . . 6 (𝑥 = 𝐴 → (𝑥𝐶𝐴𝐶))
109anbi1d 469 . . . . 5 (𝑥 = 𝐴 → ((𝑥𝐶 ∧ ¬ 𝜑) ↔ (𝐴𝐶 ∧ ¬ 𝜑)))
118, 10orbi12d 805 . . . 4 (𝑥 = 𝐴 → (((𝑥𝐵𝜑) ∨ (𝑥𝐶 ∧ ¬ 𝜑)) ↔ ((𝐴𝐵𝜑) ∨ (𝐴𝐶 ∧ ¬ 𝜑))))
12 df-if 3636 . . . 4 if(𝜑, 𝐵, 𝐶) = {𝑥 ∣ ((𝑥𝐵𝜑) ∨ (𝑥𝐶 ∧ ¬ 𝜑))}
1311, 12elab2g 2973 . . 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 776 . . 3 (((𝐴𝐵𝜑) ∨ (𝐴𝐶 ∧ ¬ 𝜑)) ↔ ((𝜑𝐴𝐵) ∨ (¬ 𝜑𝐴𝐶)))
1713, 16bitrdi 196 . 2 (𝐴 ∈ V → (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ ((𝜑𝐴𝐵) ∨ (¬ 𝜑𝐴𝐶))))
181, 6, 17pm5.21nii 716 1 (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ ((𝜑𝐴𝐵) ∨ (¬ 𝜑𝐴𝐶)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wa 104  wb 105  wo 720   = wceq 1402  wcel 2209  Vcvv 2821  ifcif 3635
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 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220
This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-if 3636
This theorem is referenced by:  rabsnif  3774  iftrueb01  7572  pw1if  7574  eupth2lem1  16613
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