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Theorem 2if2 4280
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 4236 . . . 4 (𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
32adantl 473 . . 3 ((𝜑𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
41, 3eqtr4d 2797 . 2 ((𝜑𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
5 2if2.2 . . . . . 6 ((𝜑 ∧ ¬ 𝜓𝜃) → 𝐷 = 𝐵)
653expa 1112 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = 𝐵)
7 iftrue 4236 . . . . . 6 (𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐵)
87adantl 473 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → if(𝜃, 𝐵, 𝐶) = 𝐵)
96, 8eqtr4d 2797 . . . 4 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
10 2if2.3 . . . . . 6 ((𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜃) → 𝐷 = 𝐶)
11103expa 1112 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = 𝐶)
12 iffalse 4239 . . . . . . 7 𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐶)
1312eqcomd 2766 . . . . . 6 𝜃𝐶 = if(𝜃, 𝐵, 𝐶))
1413adantl 473 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐶 = if(𝜃, 𝐵, 𝐶))
1511, 14eqtrd 2794 . . . 4 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
169, 15pm2.61dan 867 . . 3 ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜃, 𝐵, 𝐶))
17 iffalse 4239 . . . 4 𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
1817adantl 473 . . 3 ((𝜑 ∧ ¬ 𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
1916, 18eqtr4d 2797 . 2 ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
204, 19pm2.61dan 867 1 (𝜑𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 383  w3a 1072   = wceq 1632  ifcif 4230
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1074  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-if 4231
This theorem is referenced by:  swrdccat3  13692  swrdccat  13693  swrdccat3a  13694  swrdccat3b  13696  pfxccat3  41936
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