Theorem ifordc 3651

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 844	. 2 ⊢ (DECID 𝜑 → (𝜑 ∨ ¬ 𝜑))
2		iftrue 3614	. . . . 5 ⊢ ((𝜑 ∨ 𝜓) → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴)
3	2	orcs 743	. . . 4 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴)
4		iftrue 3614	. . . 4 ⊢ (𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = 𝐴)
5	3, 4	eqtr4d 2267	. . 3 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
6		iffalse 3617	. . . 4 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = if(𝜓, 𝐴, 𝐵))
7		biorf 752	. . . . 5 ⊢ (¬ 𝜑 → (𝜓 ↔ (𝜑 ∨ 𝜓)))
8	7	ifbid 3631	. . . 4 ⊢ (¬ 𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∨ 𝜓), 𝐴, 𝐵))
9	6, 8	eqtr2d 2265	. . 3 ⊢ (¬ 𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)))
10	5, 9	jaoi 724	. 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 716  DECID wdc 842   = wceq 1398  ifcif 3607

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 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-11 1555  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-dc 843  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-if 3608

This theorem is referenced by:  nninfwlpoimlemg  7417