Theorem ifordc 3610

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

Assertion

Ref	Expression
ifordc	⊢ (DECID 𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))

Proof of Theorem ifordc

Step	Hyp	Ref	Expression
1		exmiddc 837	. 2 ⊢ (DECID 𝜑 → (𝜑 ∨ ¬ 𝜑))
2		iftrue 3575	. . . . 5 ⊢ ((𝜑 ∨ 𝜓) → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴)
3	2	orcs 736	. . . 4 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴)
4		iftrue 3575	. . . 4 ⊢ (𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = 𝐴)
5	3, 4	eqtr4d 2240	. . 3 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
6		iffalse 3578	. . . 4 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = if(𝜓, 𝐴, 𝐵))
7		biorf 745	. . . . 5 ⊢ (¬ 𝜑 → (𝜓 ↔ (𝜑 ∨ 𝜓)))
8	7	ifbid 3591	. . . 4 ⊢ (¬ 𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∨ 𝜓), 𝐴, 𝐵))
9	6, 8	eqtr2d 2238	. . 3 ⊢ (¬ 𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
10	5, 9	jaoi 717	. 2 ⊢ ((𝜑 ∨ ¬ 𝜑) → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
11	1, 10	syl 14	1 ⊢ (DECID 𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 709  DECID wdc 835   = wceq 1372  ifcif 3570

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-11 1528  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-dc 836  df-tru 1375  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-if 3571

This theorem is referenced by:  nninfwlpoimlemg  7259