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Theorem sbc3or 39055
 Description: sbcor 3512 with a 3-disjuncts. This proof is sbc3orgVD 39400 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 3512 . . 3 ([𝐴 / 𝑥]((𝜑𝜓) ∨ 𝜒) ↔ ([𝐴 / 𝑥](𝜑𝜓) ∨ [𝐴 / 𝑥]𝜒))
2 df-3or 1055 . . . . 5 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∨ 𝜒))
32bicomi 214 . . . 4 (((𝜑𝜓) ∨ 𝜒) ↔ (𝜑𝜓𝜒))
43sbcbii 3524 . . 3 ([𝐴 / 𝑥]((𝜑𝜓) ∨ 𝜒) ↔ [𝐴 / 𝑥](𝜑𝜓𝜒))
5 sbcor 3512 . . . 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 290 . 2 ([𝐴 / 𝑥](𝜑𝜓𝜒) ↔ (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
8 df-3or 1055 . 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 1053  [wsbc 3468 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631 This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3or 1055  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-v 3233  df-sbc 3469 This theorem is referenced by:  sbcoreleleq  39062
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