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

Theorem ifan 4579
Description: Rewrite a conjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ifan if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)

Proof of Theorem ifan
StepHypRef Expression
1 iftrue 4531 . . 3 (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
2 ibar 528 . . . 4 (𝜑 → (𝜓 ↔ (𝜑𝜓)))
32ifbid 4549 . . 3 (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑𝜓), 𝐴, 𝐵))
41, 3eqtr2d 2778 . 2 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
5 simpl 482 . . . . 5 ((𝜑𝜓) → 𝜑)
65con3i 154 . . . 4 𝜑 → ¬ (𝜑𝜓))
76iffalsed 4536 . . 3 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = 𝐵)
8 iffalse 4534 . . 3 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵)
97, 8eqtr4d 2780 . 2 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
104, 9pm2.61i 182 1 if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wa 395   = wceq 1540  ifcif 4525
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-if 4526
This theorem is referenced by:  psdmvr  22173  itg0  25815  iblre  25829  itgreval  25832  iblss  25840  iblss2  25841  itgle  25845  itgss  25847  itgeqa  25849  iblconst  25853  itgconst  25854  ibladdlem  25855  iblabslem  25863  iblabsr  25865  iblmulc2  25866  itgsplit  25871  bddiblnc  25877  itgcn  25880  mrsubrn  35518  itg2gt0cn  37682  ibladdnclem  37683  iblabsnclem  37690  iblmulc2nc  37692  selvvvval  42595  iblsplit  45981
  Copyright terms: Public domain W3C validator