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Theorem 2if2 4136
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 4092 . . . 4 (𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
32adantl 482 . . 3 ((𝜑𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
41, 3eqtr4d 2659 . 2 ((𝜑𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
5 2if2.2 . . . . . 6 ((𝜑 ∧ ¬ 𝜓𝜃) → 𝐷 = 𝐵)
653expa 1265 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = 𝐵)
7 iftrue 4092 . . . . . 6 (𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐵)
87adantl 482 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → if(𝜃, 𝐵, 𝐶) = 𝐵)
96, 8eqtr4d 2659 . . . 4 (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
10 2if2.3 . . . . . 6 ((𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜃) → 𝐷 = 𝐶)
11103expa 1265 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = 𝐶)
12 iffalse 4095 . . . . . . 7 𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐶)
1312eqcomd 2628 . . . . . 6 𝜃𝐶 = if(𝜃, 𝐵, 𝐶))
1413adantl 482 . . . . 5 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐶 = if(𝜃, 𝐵, 𝐶))
1511, 14eqtrd 2656 . . . 4 (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
169, 15pm2.61dan 832 . . 3 ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜃, 𝐵, 𝐶))
17 iffalse 4095 . . . 4 𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
1817adantl 482 . . 3 ((𝜑 ∧ ¬ 𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
1916, 18eqtr4d 2659 . 2 ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
204, 19pm2.61dan 832 1 (𝜑𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 384  w3a 1037   = wceq 1483  ifcif 4086
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-if 4087
This theorem is referenced by:  swrdccat3  13492  swrdccat  13493  swrdccat3a  13494  swrdccat3b  13496  pfxccat3  41426
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