Theorem ifcomnan 4546

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 4498	. . . 4 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, 𝐶))
3		iffalse 4498	. . . . 5 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, 𝐶) = 𝐶)
4	3	ifeq2d 4510	. . . 4 ⊢ (¬ 𝜑 → if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)) = if(𝜓, 𝐵, 𝐶))
5	2, 4	eqtr4d 2803	. . 3 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))
6		iffalse 4498	. . . . 5 ⊢ (¬ 𝜓 → if(𝜓, 𝐵, 𝐶) = 𝐶)
7	6	ifeq2d 4510	. . . 4 ⊢ (¬ 𝜓 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜑, 𝐴, 𝐶))
8		iffalse 4498	. . . 4 ⊢ (¬ 𝜓 → if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)) = if(𝜑, 𝐴, 𝐶))
9	7, 8	eqtr4d 2803	. . 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 1570  ifcif 4489

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-un 3911  df-if 4490

This theorem is used by:  mdetunilem6  22824