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Theorem ifor 4502
 Description: Rewrite a disjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ifor if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))

Proof of Theorem ifor
StepHypRef Expression
1 iftrue 4456 . . . 4 ((𝜑𝜓) → if((𝜑𝜓), 𝐴, 𝐵) = 𝐴)
21orcs 872 . . 3 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = 𝐴)
3 iftrue 4456 . . 3 (𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = 𝐴)
42, 3eqtr4d 2862 . 2 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
5 iffalse 4459 . . 3 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = if(𝜓, 𝐴, 𝐵))
6 biorf 934 . . . 4 𝜑 → (𝜓 ↔ (𝜑𝜓)))
76ifbid 4472 . . 3 𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑𝜓), 𝐴, 𝐵))
85, 7eqtr2d 2860 . 2 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
94, 8pm2.61i 185 1 if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ∨ wo 844   = wceq 1538  ifcif 4450 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-ext 2796 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-if 4451 This theorem is referenced by:  cantnflem1d  9148  cantnflem1  9149
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