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Theorem ifan 4540
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 4493 . . 3 (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
2 ibar 530 . . . 4 (𝜑 → (𝜓 ↔ (𝜑𝜓)))
32ifbid 4510 . . 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 484 . . . . 5 ((𝜑𝜓) → 𝜑)
65con3i 154 . . . 4 𝜑 → ¬ (𝜑𝜓))
76iffalsed 4498 . . 3 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = 𝐵)
8 iffalse 4496 . . 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 397   = wceq 1542  ifcif 4487
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2708
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-ex 1783  df-sb 2069  df-clab 2715  df-cleq 2729  df-clel 2815  df-if 4488
This theorem is referenced by:  itg0  25147  iblre  25161  itgreval  25164  iblss  25172  iblss2  25173  itgle  25177  itgss  25179  itgeqa  25181  iblconst  25185  itgconst  25186  ibladdlem  25187  iblabslem  25195  iblabsr  25197  iblmulc2  25198  itgsplit  25203  bddiblnc  25209  itgcn  25212  mrsubrn  34110  itg2gt0cn  36136  ibladdnclem  36137  iblabsnclem  36144  iblmulc2nc  36146  iblsplit  44214
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