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Theorem ifordc 3572
Description: Rewrite a disjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ifordc (DECID 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))

Proof of Theorem ifordc
StepHypRef Expression
1 exmiddc 836 . 2 (DECID 𝜑 → (𝜑 ∨ ¬ 𝜑))
2 iftrue 3539 . . . . 5 ((𝜑𝜓) → if((𝜑𝜓), 𝐴, 𝐵) = 𝐴)
32orcs 735 . . . 4 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = 𝐴)
4 iftrue 3539 . . . 4 (𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = 𝐴)
53, 4eqtr4d 2213 . . 3 (𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
6 iffalse 3542 . . . 4 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = if(𝜓, 𝐴, 𝐵))
7 biorf 744 . . . . 5 𝜑 → (𝜓 ↔ (𝜑𝜓)))
87ifbid 3555 . . . 4 𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑𝜓), 𝐴, 𝐵))
96, 8eqtr2d 2211 . . 3 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
105, 9jaoi 716 . 2 ((𝜑 ∨ ¬ 𝜑) → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
111, 10syl 14 1 (DECID 𝜑 → if((𝜑𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wo 708  DECID wdc 834   = wceq 1353  ifcif 3534
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-11 1506  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-dc 835  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-if 3535
This theorem is referenced by:  nninfwlpoimlemg  7166
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