Theorem ifandc 3650

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

Assertion

Ref	Expression
ifandc	⊢ (DECID 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))

Proof of Theorem ifandc

Step	Hyp	Ref	Expression
1		df-dc 843	. 2 ⊢ (DECID 𝜑 ↔ (𝜑 ∨ ¬ 𝜑))
2		iftrue 3614	. . . 4 ⊢ (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
3		ibar 301	. . . . 5 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓)))
4	3	ifbid 3631	. . . 4 ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∧ 𝜓), 𝐴, 𝐵))
5	2, 4	eqtr2d 2265	. . 3 ⊢ (𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
6		simpl 109	. . . . . 6 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
7	6	con3i 637	. . . . 5 ⊢ (¬ 𝜑 → ¬ (𝜑 ∧ 𝜓))
8	7	iffalsed 3619	. . . 4 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = 𝐵)
9		iffalse 3617	. . . 4 ⊢ (¬ 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵)
10	8, 9	eqtr4d 2267	. . 3 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
11	5, 10	jaoi 724	. 2 ⊢ ((𝜑 ∨ ¬ 𝜑) → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
12	1, 11	sylbi 121	1 ⊢ (DECID 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ∨ 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:  isumss  12013