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Theorem 2if2 4478
 Description: Resolve two nested conditionals. (Contributed by Alexander van der Vekens, 27-Mar-2018.)
Hypotheses
Ref Expression
2if2.1 ((𝜑𝜓) → 𝐷 = 𝐴)
2if2.2 ((𝜑 ∧ ¬ 𝜓𝜃) → 𝐷 = 𝐵)
2if2.3 ((𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜃) → 𝐷 = 𝐶)
Assertion
Ref Expression
2if2 (𝜑𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))

Proof of Theorem 2if2
StepHypRef Expression
1 2if2.1 . . 3 ((𝜑𝜓) → 𝐷 = 𝐴)
2 iftrue 4431 . . . 4 (𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
32adantl 485 . . 3 ((𝜑𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
41, 3eqtr4d 2836 . 2 ((𝜑𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
5 2if2.2 . . . . . 6 ((𝜑 ∧ ¬ 𝜓𝜃) → 𝐷 = 𝐵)
653expa 1115 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = 𝐵)
7 iftrue 4431 . . . . . 6 (𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐵)
87adantl 485 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → if(𝜃, 𝐵, 𝐶) = 𝐵)
96, 8eqtr4d 2836 . . . 4 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
10 2if2.3 . . . . . 6 ((𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜃) → 𝐷 = 𝐶)
11103expa 1115 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = 𝐶)
12 iffalse 4434 . . . . . . 7 𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐶)
1312eqcomd 2804 . . . . . 6 𝜃𝐶 = if(𝜃, 𝐵, 𝐶))
1413adantl 485 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐶 = if(𝜃, 𝐵, 𝐶))
1511, 14eqtrd 2833 . . . 4 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
169, 15pm2.61dan 812 . . 3 ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜃, 𝐵, 𝐶))
17 iffalse 4434 . . . 4 𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
1817adantl 485 . . 3 ((𝜑 ∧ ¬ 𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
1916, 18eqtr4d 2836 . 2 ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
204, 19pm2.61dan 812 1 (𝜑𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 399   ∧ w3a 1084   = wceq 1538  ifcif 4425 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3an 1086  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-if 4426 This theorem is referenced by:  pfxccat3  14090  swrdccat  14091  swrdccat3b  14096
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