Theorem ifcomnan 4542

Description: Commute the conditions in two nested conditionals if both conditions are not simultaneously true. (Contributed by SO, 15-Jul-2018.)

Assertion

Ref	Expression
ifcomnan	⊢ (¬ (𝜑 ∧ 𝜓) → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))

Proof of Theorem ifcomnan

Step	Hyp	Ref	Expression
1		pm3.13 1010	. 2 ⊢ (¬ (𝜑 ∧ 𝜓) → (¬ 𝜑 ∨ ¬ 𝜓))
2		iffalse 4494	. . . 4 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, 𝐶))
3		iffalse 4494	. . . . 5 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, 𝐶) = 𝐶)
4	3	ifeq2d 4506	. . . 4 ⊢ (¬ 𝜑 → if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)) = if(𝜓, 𝐵, 𝐶))
5	2, 4	eqtr4d 2801	. . 3 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))
6		iffalse 4494	. . . . 5 ⊢ (¬ 𝜓 → if(𝜓, 𝐵, 𝐶) = 𝐶)
7	6	ifeq2d 4506	. . . 4 ⊢ (¬ 𝜓 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜑, 𝐴, 𝐶))
8		iffalse 4494	. . . 4 ⊢ (¬ 𝜓 → if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)) = if(𝜑, 𝐴, 𝐶))
9	7, 8	eqtr4d 2801	. . 3 ⊢ (¬ 𝜓 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))
10	5, 9	jaoi 871	. 2 ⊢ ((¬ 𝜑 ∨ ¬ 𝜓) → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))
11	1, 10	syl 18	1 ⊢ (¬ (𝜑 ∧ 𝜓) → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∨ wo 861   = wceq 1569  ifcif 4485

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 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-un 4154  df-if 4486

This theorem is used by:  mdetunilem6  22947