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Theorem ifan 4517
 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 4472 . . 3 (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
2 ibar 531 . . . 4 (𝜑 → (𝜓 ↔ (𝜑𝜓)))
32ifbid 4488 . . 3 (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑𝜓), 𝐴, 𝐵))
41, 3eqtr2d 2857 . 2 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
5 simpl 485 . . . . 5 ((𝜑𝜓) → 𝜑)
65con3i 157 . . . 4 𝜑 → ¬ (𝜑𝜓))
76iffalsed 4477 . . 3 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = 𝐵)
8 iffalse 4475 . . 3 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵)
97, 8eqtr4d 2859 . 2 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
104, 9pm2.61i 184 1 if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ∧ wa 398   = wceq 1533  ifcif 4466 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-ext 2793 This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1777  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-if 4467 This theorem is referenced by:  itg0  24379  iblre  24393  itgreval  24396  iblss  24404  iblss2  24405  itgle  24409  itgss  24411  itgeqa  24413  iblconst  24417  itgconst  24418  ibladdlem  24419  iblabslem  24427  iblabsr  24429  iblmulc2  24430  itgsplit  24435  itgcn  24442  mrsubrn  32760  itg2gt0cn  34946  ibladdnclem  34947  iblabsnclem  34954  iblmulc2nc  34956  bddiblnc  34961  iblsplit  42249
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