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Theorem ifor 4585
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 4537 . . . 4 ((𝜑𝜓) → if((𝜑𝜓), 𝐴, 𝐵) = 𝐴)
21orcs 875 . . 3 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = 𝐴)
3 iftrue 4537 . . 3 (𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = 𝐴)
42, 3eqtr4d 2778 . 2 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
5 iffalse 4540 . . 3 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = if(𝜓, 𝐴, 𝐵))
6 biorf 936 . . . 4 𝜑 → (𝜓 ↔ (𝜑𝜓)))
76ifbid 4554 . . 3 𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑𝜓), 𝐴, 𝐵))
85, 7eqtr2d 2776 . 2 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
94, 8pm2.61i 182 1 if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wo 847   = wceq 1537  ifcif 4531
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-if 4532
This theorem is referenced by:  cantnflem1d  9726  cantnflem1  9727
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