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Theorem ifan 4167
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 4125 . . 3 (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
2 ibar 524 . . . 4 (𝜑 → (𝜓 ↔ (𝜑𝜓)))
32ifbid 4141 . . 3 (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑𝜓), 𝐴, 𝐵))
41, 3eqtr2d 2686 . 2 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
5 simpl 472 . . . . 5 ((𝜑𝜓) → 𝜑)
65con3i 150 . . . 4 𝜑 → ¬ (𝜑𝜓))
76iffalsed 4130 . . 3 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = 𝐵)
8 iffalse 4128 . . 3 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵)
97, 8eqtr4d 2688 . 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 383   = wceq 1523  ifcif 4119
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-if 4120
This theorem is referenced by:  itg0  23591  iblre  23605  itgreval  23608  iblss  23616  iblss2  23617  itgle  23621  itgss  23623  itgeqa  23625  iblconst  23629  itgconst  23630  ibladdlem  23631  iblabslem  23639  iblabsr  23641  iblmulc2  23642  itgsplit  23647  itgcn  23654  mrsubrn  31536  itg2gt0cn  33595  ibladdnclem  33596  iblabsnclem  33603  iblmulc2nc  33605  bddiblnc  33610  iblsplit  40500
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