Theorem ifor 4532

Description: Rewrite a disjunction in a conditional 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 4484	. . . 4 ⊢ ((𝜑 ∨ 𝜓) → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴)
2	1	orcs 873	. . 3 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴)
3		iftrue 4484	. . 3 ⊢ (𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = 𝐴)
4	2, 3	eqtr4d 2780	. 2 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
5		iffalse 4487	. . 3 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = if(𝜓, 𝐴, 𝐵))
6		biorf 935	. . . 4 ⊢ (¬ 𝜑 → (𝜓 ↔ (𝜑 ∨ 𝜓)))
7	6	ifbid 4501	. . 3 ⊢ (¬ 𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∨ 𝜓), 𝐴, 𝐵))
8	5, 7	eqtr2d 2778	. 2 ⊢ (¬ 𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
9	4, 8	pm2.61i 182	1 ⊢ if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))

Syntax hints:  ¬ wn 3   ∨ wo 845   = wceq 1541  ifcif 4478

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2708

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 846  df-ex 1782  df-sb 2068  df-clab 2715  df-cleq 2729  df-clel 2815  df-if 4479

This theorem is referenced by:  cantnflem1d  9550  cantnflem1  9551