Theorem ifcomnan 4137

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		pm3.13 522	. 2 ⊢ (¬ (𝜑 ∧ 𝜓) → (¬ 𝜑 ∨ ¬ 𝜓))
2		iffalse 4095	. . . 4 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, 𝐶))
3		iffalse 4095	. . . . 5 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, 𝐶) = 𝐶)
4	3	ifeq2d 4105	. . . 4 ⊢ (¬ 𝜑 → if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)) = if(𝜓, 𝐵, 𝐶))
5	2, 4	eqtr4d 2659	. . 3 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))
6		iffalse 4095	. . . . 5 ⊢ (¬ 𝜓 → if(𝜓, 𝐵, 𝐶) = 𝐶)
7	6	ifeq2d 4105	. . . 4 ⊢ (¬ 𝜓 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜑, 𝐴, 𝐶))
8		iffalse 4095	. . . 4 ⊢ (¬ 𝜓 → if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)) = if(𝜑, 𝐴, 𝐶))
9	7, 8	eqtr4d 2659	. . 3 ⊢ (¬ 𝜓 → if(𝜑, 𝐴, if(𝜓, 𝐵, 𝐶)) = if(𝜓, 𝐵, if(𝜑, 𝐴, 𝐶)))
10	5, 9	jaoi 394	. 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   ∨ wo 383   ∧ wa 384   = wceq 1483  ifcif 4086

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-rab 2921  df-v 3202  df-un 3579  df-if 4087

This theorem is referenced by:  mdetunilem6  20423