Theorem ifcomnan 4515

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		pm3.13 992	. 2 ⊢ (¬ (𝜑 ∧ 𝜓) → (¬ 𝜑 ∨ ¬ 𝜓))
2		iffalse 4468	. . . 4 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, 𝐶))
3		iffalse 4468	. . . . 5 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, 𝐶) = 𝐶)
4	3	ifeq2d 4479	. . . 4 ⊢ (¬ 𝜑 → if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)) = if(𝜓, 𝐵, 𝐶))
5	2, 4	eqtr4d 2781	. . 3 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))
6		iffalse 4468	. . . . 5 ⊢ (¬ 𝜓 → if(𝜓, 𝐵, 𝐶) = 𝐶)
7	6	ifeq2d 4479	. . . 4 ⊢ (¬ 𝜓 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜑, 𝐴, 𝐶))
8		iffalse 4468	. . . 4 ⊢ (¬ 𝜓 → if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)) = if(𝜑, 𝐴, 𝐶))
9	7, 8	eqtr4d 2781	. . 3 ⊢ (¬ 𝜓 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))
10	5, 9	jaoi 854	. 2 ⊢ ((¬ 𝜑 ∨ ¬ 𝜓) → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))
11	1, 10	syl 17	1 ⊢ (¬ (𝜑 ∧ 𝜓) → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 396   ∨ wo 844   = wceq 1539  ifcif 4459

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-rab 3073  df-v 3434  df-un 3892  df-if 4460

This theorem is referenced by:  mdetunilem6  21766