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Theorem ifan 4330
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 4285 . . 3 (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
2 ibar 520 . . . 4 (𝜑 → (𝜓 ↔ (𝜑𝜓)))
32ifbid 4301 . . 3 (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑𝜓), 𝐴, 𝐵))
41, 3eqtr2d 2841 . 2 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
5 simpl 470 . . . . 5 ((𝜑𝜓) → 𝜑)
65con3i 151 . . . 4 𝜑 → ¬ (𝜑𝜓))
76iffalsed 4290 . . 3 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = 𝐵)
8 iffalse 4288 . . 3 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵)
97, 8eqtr4d 2843 . 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 1637  ifcif 4279
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2068  ax-7 2104  ax-9 2165  ax-10 2185  ax-11 2201  ax-12 2214  ax-ext 2784
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2061  df-clab 2793  df-cleq 2799  df-clel 2802  df-if 4280
This theorem is referenced by:  itg0  23760  iblre  23774  itgreval  23777  iblss  23785  iblss2  23786  itgle  23790  itgss  23792  itgeqa  23794  iblconst  23798  itgconst  23799  ibladdlem  23800  iblabslem  23808  iblabsr  23810  iblmulc2  23811  itgsplit  23816  itgcn  23823  mrsubrn  31733  itg2gt0cn  33777  ibladdnclem  33778  iblabsnclem  33785  iblmulc2nc  33787  bddiblnc  33792  iblsplit  40661
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