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Theorem ifan 3701
 Description: Rewrite a conjunction in an if statement as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ifan if((φ ψ), A, B) = if(φ, if(ψ, A, B), B)

Proof of Theorem ifan
StepHypRef Expression
1 iftrue 3668 . . 3 (φ → if(φ, if(ψ, A, B), B) = if(ψ, A, B))
2 ibar 490 . . . 4 (φ → (ψ ↔ (φ ψ)))
32ifbid 3680 . . 3 (φ → if(ψ, A, B) = if((φ ψ), A, B))
41, 3eqtr2d 2386 . 2 (φ → if((φ ψ), A, B) = if(φ, if(ψ, A, B), B))
5 simpl 443 . . . . 5 ((φ ψ) → φ)
65con3i 127 . . . 4 φ → ¬ (φ ψ))
7 iffalse 3669 . . . 4 (¬ (φ ψ) → if((φ ψ), A, B) = B)
86, 7syl 15 . . 3 φ → if((φ ψ), A, B) = B)
9 iffalse 3669 . . 3 φ → if(φ, if(ψ, A, B), B) = B)
108, 9eqtr4d 2388 . 2 φ → if((φ ψ), A, B) = if(φ, if(ψ, A, B), B))
114, 10pm2.61i 156 1 if((φ ψ), A, B) = if(φ, if(ψ, A, B), B)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ∧ wa 358   = 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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