Theorem 2if2 4537

Description: Resolve two nested conditionals. (Contributed by Alexander van der Vekens, 27-Mar-2018.)

Hypotheses

Ref	Expression
2if2.1	⊢ ((𝜑 ∧ 𝜓) → 𝐷 = 𝐴)
2if2.2	⊢ ((𝜑 ∧ ¬ 𝜓 ∧ 𝜃) → 𝐷 = 𝐵)
2if2.3	⊢ ((𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜃) → 𝐷 = 𝐶)

Assertion

Ref	Expression
2if2	⊢ (𝜑 → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))

Proof of Theorem 2if2

Step	Hyp	Ref	Expression
1		2if2.1	. . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝐷 = 𝐴)
2		iftrue 4487	. . . 4 ⊢ (𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
3	2	adantl 481	. . 3 ⊢ ((𝜑 ∧ 𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
4	1, 3	eqtr4d 2775	. 2 ⊢ ((𝜑 ∧ 𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
5		2if2.2	. . . . . 6 ⊢ ((𝜑 ∧ ¬ 𝜓 ∧ 𝜃) → 𝐷 = 𝐵)
6	5	3expa 1119	. . . . 5 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = 𝐵)
7		iftrue 4487	. . . . . 6 ⊢ (𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐵)
8	7	adantl 481	. . . . 5 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → if(𝜃, 𝐵, 𝐶) = 𝐵)
9	6, 8	eqtr4d 2775	. . . 4 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
10		2if2.3	. . . . . 6 ⊢ ((𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜃) → 𝐷 = 𝐶)
11	10	3expa 1119	. . . . 5 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = 𝐶)
12		iffalse 4490	. . . . . . 7 ⊢ (¬ 𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐶)
13	12	eqcomd 2743	. . . . . 6 ⊢ (¬ 𝜃 → 𝐶 = if(𝜃, 𝐵, 𝐶))
14	13	adantl 481	. . . . 5 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐶 = if(𝜃, 𝐵, 𝐶))
15	11, 14	eqtrd 2772	. . . 4 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
16	9, 15	pm2.61dan 813	. . 3 ⊢ ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜃, 𝐵, 𝐶))
17		iffalse 4490	. . . 4 ⊢ (¬ 𝜓 → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
18	17	adantl 481	. . 3 ⊢ ((𝜑 ∧ ¬ 𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
19	16, 18	eqtr4d 2775	. 2 ⊢ ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
20	4, 19	pm2.61dan 813	1 ⊢ (𝜑 → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   ∧ w3a 1087   = wceq 1542  ifcif 4481

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 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-if 4482

This theorem is referenced by:  pfxccat3  14669  swrdccat  14670  swrdccat3b  14675