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Theorem ifan 4112
 Description: Rewrite a conjunction in an if statement 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 4070 . . 3 (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
2 ibar 525 . . . 4 (𝜑 → (𝜓 ↔ (𝜑𝜓)))
32ifbid 4086 . . 3 (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑𝜓), 𝐴, 𝐵))
41, 3eqtr2d 2656 . 2 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
5 simpl 473 . . . . 5 ((𝜑𝜓) → 𝜑)
65con3i 150 . . . 4 𝜑 → ¬ (𝜑𝜓))
76iffalsed 4075 . . 3 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = 𝐵)
8 iffalse 4073 . . 3 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵)
97, 8eqtr4d 2658 . 2 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
104, 9pm2.61i 176 1 if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ∧ wa 384   = wceq 1480  ifcif 4064 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-if 4065 This theorem is referenced by:  itg0  23486  iblre  23500  itgreval  23503  iblss  23511  iblss2  23512  itgle  23516  itgss  23518  itgeqa  23520  iblconst  23524  itgconst  23525  ibladdlem  23526  iblabslem  23534  iblabsr  23536  iblmulc2  23537  itgsplit  23542  itgcn  23549  mrsubrn  31171  itg2gt0cn  33136  ibladdnclem  33137  iblabsnclem  33144  iblmulc2nc  33146  bddiblnc  33151  iblsplit  39519
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