Theorem ifor 4135

Description: Rewrite a disjunction in an if statement as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)

Assertion

Ref	Expression
ifor	⊢ if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))

Proof of Theorem ifor

Step	Hyp	Ref	Expression
1		iftrue 4092	. . . 4 ⊢ ((𝜑 ∨ 𝜓) → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴)
2	1	orcs 409	. . 3 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴)
3		iftrue 4092	. . 3 ⊢ (𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = 𝐴)
4	2, 3	eqtr4d 2659	. 2 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
5		iffalse 4095	. . 3 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = if(𝜓, 𝐴, 𝐵))
6		biorf 420	. . . 4 ⊢ (¬ 𝜑 → (𝜓 ↔ (𝜑 ∨ 𝜓)))
7	6	ifbid 4108	. . 3 ⊢ (¬ 𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∨ 𝜓), 𝐴, 𝐵))
8	5, 7	eqtr2d 2657	. 2 ⊢ (¬ 𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
9	4, 8	pm2.61i 176	1 ⊢ if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))

Syntax hints:  ¬ wn 3   ∨ wo 383   = 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-if 4087

This theorem is referenced by:  cantnflem1d  8585  cantnflem1  8586