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Theorem sbc3or 39209
Description: sbcor 3608 with a 3-disjuncts. This proof is sbc3orgVD 39554 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 3608 . . 3 ([𝐴 / 𝑥]((𝜑𝜓) ∨ 𝜒) ↔ ([𝐴 / 𝑥](𝜑𝜓) ∨ [𝐴 / 𝑥]𝜒))
2 df-3or 1073 . . . . 5 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∨ 𝜒))
32bicomi 214 . . . 4 (((𝜑𝜓) ∨ 𝜒) ↔ (𝜑𝜓𝜒))
43sbcbii 3620 . . 3 ([𝐴 / 𝑥]((𝜑𝜓) ∨ 𝜒) ↔ [𝐴 / 𝑥](𝜑𝜓𝜒))
5 sbcor 3608 . . . 4 ([𝐴 / 𝑥](𝜑𝜓) ↔ ([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓))
65orbi1i 543 . . 3 (([𝐴 / 𝑥](𝜑𝜓) ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
71, 4, 63bitr3i 290 . 2 ([𝐴 / 𝑥](𝜑𝜓𝜒) ↔ (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
8 df-3or 1073 . 2 (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓[𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
97, 8bitr4i 267 1 ([𝐴 / 𝑥](𝜑𝜓𝜒) ↔ ([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓[𝐴 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 196  wo 382  w3o 1071  [wsbc 3564
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1859  ax-4 1874  ax-5 1976  ax-6 2042  ax-7 2078  ax-9 2136  ax-10 2156  ax-11 2171  ax-12 2184  ax-13 2379  ax-ext 2728
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3or 1073  df-tru 1623  df-ex 1842  df-nf 1847  df-sb 2035  df-clab 2735  df-cleq 2741  df-clel 2744  df-v 3330  df-sbc 3565
This theorem is referenced by:  sbcoreleleq  39216
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