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 487	. . 3 ⊢ ((𝜑 ∧ 𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = 𝐴)
4	1, 3	eqtr4d 2798	. 2 ⊢ ((𝜑 ∧ 𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
5		2if2.2	. . . . . 6 ⊢ ((𝜑 ∧ ¬ 𝜓 ∧ 𝜃) → 𝐷 = 𝐵)
6	5	3expa 1136	. . . . 5 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = 𝐵)
7		iftrue 4487	. . . . . 6 ⊢ (𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐵)
8	7	adantl 487	. . . . 5 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → if(𝜃, 𝐵, 𝐶) = 𝐵)
9	6, 8	eqtr4d 2798	. . . 4 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
10		2if2.3	. . . . . 6 ⊢ ((𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜃) → 𝐷 = 𝐶)
11	10	3expa 1136	. . . . 5 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = 𝐶)
12		iffalse 4490	. . . . . . 7 ⊢ (¬ 𝜃 → if(𝜃, 𝐵, 𝐶) = 𝐶)
13	12	eqcomd 2766	. . . . . 6 ⊢ (¬ 𝜃 → 𝐶 = if(𝜃, 𝐵, 𝐶))
14	13	adantl 487	. . . . 5 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐶 = if(𝜃, 𝐵, 𝐶))
15	11, 14	eqtrd 2795	. . . 4 ⊢ (((𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜃) → 𝐷 = if(𝜃, 𝐵, 𝐶))
16	9, 15	pm2.61dan 825	. . 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 487	. . 3 ⊢ ((𝜑 ∧ ¬ 𝜓) → if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)) = if(𝜃, 𝐵, 𝐶))
19	16, 18	eqtr4d 2798	. 2 ⊢ ((𝜑 ∧ ¬ 𝜓) → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))
20	4, 19	pm2.61dan 825	1 ⊢ (𝜑 → 𝐷 = if(𝜓, 𝐴, if(𝜃, 𝐵, 𝐶)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∧ w3a 1103   = wceq 1569  ifcif 4481

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-if 4482

This theorem is used by:  pfxccat3  14883  swrdccat  14884  swrdccat3b  14889