Theorem ifandc 3646

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 842	. 2 ⊢ (DECID 𝜑 ↔ (𝜑 ∨ ¬ 𝜑))
2		iftrue 3610	. . . 4 ⊢ (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
3		ibar 301	. . . . 5 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓)))
4	3	ifbid 3627	. . . 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 3615	. . . 4 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = 𝐵)
9		iffalse 3613	. . . 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 723	. 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 715  DECID wdc 841   = wceq 1397  ifcif 3605

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-11 1554  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-dc 842  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-if 3606

This theorem is referenced by:  isumss  11957