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Theorem 2if2 4169
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 4125 . . . 4 (𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
32adantl 481 . . 3 ((𝜑𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
41, 3eqtr4d 2688 . 2 ((𝜑𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
5 2if2.2 . . . . . 6 ((𝜑 ∧ ¬ 𝜓𝜃) → 𝐷 = 𝐵)
653expa 1284 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = 𝐵)
7 iftrue 4125 . . . . . 6 (𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐵)
87adantl 481 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → if(𝜃, 𝐵, 𝐶) = 𝐵)
96, 8eqtr4d 2688 . . . 4 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
10 2if2.3 . . . . . 6 ((𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜃) → 𝐷 = 𝐶)
11103expa 1284 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = 𝐶)
12 iffalse 4128 . . . . . . 7 𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐶)
1312eqcomd 2657 . . . . . 6 𝜃𝐶 = if(𝜃, 𝐵, 𝐶))
1413adantl 481 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐶 = if(𝜃, 𝐵, 𝐶))
1511, 14eqtrd 2685 . . . 4 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
169, 15pm2.61dan 849 . . 3 ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜃, 𝐵, 𝐶))
17 iffalse 4128 . . . 4 𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
1817adantl 481 . . 3 ((𝜑 ∧ ¬ 𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
1916, 18eqtr4d 2688 . 2 ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
204, 19pm2.61dan 849 1 (𝜑𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 383  w3a 1054   = wceq 1523  ifcif 4119
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1056  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-if 4120
This theorem is referenced by:  swrdccat3  13538  swrdccat  13539  swrdccat3a  13540  swrdccat3b  13542  pfxccat3  41751
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