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Theorem sbc3or 37759
Description: sbcor 3446 with a 3-disjuncts. This proof is sbc3orgVD 38108 automatically translated and minimized. (Contributed by Alan Sare, 31-Dec-2011.) (Revised by NM, 24-Aug-2018.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
sbc3or ([𝐴 / 𝑥](𝜑𝜓𝜒) ↔ ([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓[𝐴 / 𝑥]𝜒))

Proof of Theorem sbc3or
StepHypRef Expression
1 sbcor 3446 . . 3 ([𝐴 / 𝑥]((𝜑𝜓) ∨ 𝜒) ↔ ([𝐴 / 𝑥](𝜑𝜓) ∨ [𝐴 / 𝑥]𝜒))
2 df-3or 1032 . . . . 5 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∨ 𝜒))
32bicomi 213 . . . 4 (((𝜑𝜓) ∨ 𝜒) ↔ (𝜑𝜓𝜒))
43sbcbii 3458 . . 3 ([𝐴 / 𝑥]((𝜑𝜓) ∨ 𝜒) ↔ [𝐴 / 𝑥](𝜑𝜓𝜒))
5 sbcor 3446 . . . 4 ([𝐴 / 𝑥](𝜑𝜓) ↔ ([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓))
65orbi1i 541 . . 3 (([𝐴 / 𝑥](𝜑𝜓) ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
71, 4, 63bitr3i 289 . 2 ([𝐴 / 𝑥](𝜑𝜓𝜒) ↔ (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
8 df-3or 1032 . 2 (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓[𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
97, 8bitr4i 266 1 ([𝐴 / 𝑥](𝜑𝜓𝜒) ↔ ([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓[𝐴 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 195  wo 382  w3o 1030  [wsbc 3402
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590
This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3or 1032  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-v 3175  df-sbc 3403
This theorem is referenced by:  sbcoreleleq  37766
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