Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  ifor Structured version   Visualization version   GIF version

Theorem ifor 4085
 Description: Rewrite a disjunction in an if statement 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 4042 . . . 4 ((𝜑𝜓) → if((𝜑𝜓), 𝐴, 𝐵) = 𝐴)
21orcs 408 . . 3 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = 𝐴)
3 iftrue 4042 . . 3 (𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = 𝐴)
42, 3eqtr4d 2647 . 2 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
5 iffalse 4045 . . 3 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = if(𝜓, 𝐴, 𝐵))
6 biorf 419 . . . 4 𝜑 → (𝜓 ↔ (𝜑𝜓)))
76ifbid 4058 . . 3 𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑𝜓), 𝐴, 𝐵))
85, 7eqtr2d 2645 . 2 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
94, 8pm2.61i 175 1 if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ∨ wo 382   = wceq 1475  ifcif 4036 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-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-if 4037 This theorem is referenced by:  cantnflem1d  8468  cantnflem1  8469
 Copyright terms: Public domain W3C validator