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Theorem ifor 3702
 Description: Rewrite a disjunction in an if statement as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ifor if((φ ψ), A, B) = if(φ, A, if(ψ, A, B))

Proof of Theorem ifor
StepHypRef Expression
1 iftrue 3668 . . . 4 ((φ ψ) → if((φ ψ), A, B) = A)
21orcs 383 . . 3 (φ → if((φ ψ), A, B) = A)
3 iftrue 3668 . . 3 (φ → if(φ, A, if(ψ, A, B)) = A)
42, 3eqtr4d 2388 . 2 (φ → if((φ ψ), A, B) = if(φ, A, if(ψ, A, B)))
5 iffalse 3669 . . 3 φ → if(φ, A, if(ψ, A, B)) = if(ψ, A, B))
6 biorf 394 . . . 4 φ → (ψ ↔ (φ ψ)))
76ifbid 3680 . . 3 φ → if(ψ, A, B) = if((φ ψ), A, B))
85, 7eqtr2d 2386 . 2 φ → if((φ ψ), A, B) = if(φ, A, if(ψ, A, B)))
94, 8pm2.61i 156 1 if((φ ψ), A, B) = if(φ, A, if(ψ, A, B))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ∨ wo 357   = wceq 1642   ifcif 3662 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-if 3663 This theorem is referenced by: (None)
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