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Theorem 2if2 4397
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 4350 . . . 4 (𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
32adantl 474 . . 3 ((𝜑𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
41, 3eqtr4d 2811 . 2 ((𝜑𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
5 2if2.2 . . . . . 6 ((𝜑 ∧ ¬ 𝜓𝜃) → 𝐷 = 𝐵)
653expa 1098 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = 𝐵)
7 iftrue 4350 . . . . . 6 (𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐵)
87adantl 474 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → if(𝜃, 𝐵, 𝐶) = 𝐵)
96, 8eqtr4d 2811 . . . 4 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
10 2if2.3 . . . . . 6 ((𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜃) → 𝐷 = 𝐶)
11103expa 1098 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = 𝐶)
12 iffalse 4353 . . . . . . 7 𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐶)
1312eqcomd 2778 . . . . . 6 𝜃𝐶 = if(𝜃, 𝐵, 𝐶))
1413adantl 474 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐶 = if(𝜃, 𝐵, 𝐶))
1511, 14eqtrd 2808 . . . 4 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
169, 15pm2.61dan 800 . . 3 ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜃, 𝐵, 𝐶))
17 iffalse 4353 . . . 4 𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
1817adantl 474 . . 3 ((𝜑 ∧ ¬ 𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
1916, 18eqtr4d 2811 . 2 ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
204, 19pm2.61dan 800 1 (𝜑𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 387  w3a 1068   = wceq 1507  ifcif 4344
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-8 2052  ax-9 2059  ax-ext 2744
This theorem depends on definitions:  df-bi 199  df-an 388  df-or 834  df-3an 1070  df-ex 1743  df-sb 2016  df-clab 2753  df-cleq 2765  df-clel 2840  df-if 4345
This theorem is referenced by:  pfxccat3  13930  swrdccat3OLD  13931  swrdccat  13932  swrdccatOLD  13933  swrdccat3aOLD  13937  swrdccat3b  13939  swrdccat3bOLD  13940
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