Theorem ifandc 3599

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 836	. 2 ⊢ (DECID 𝜑 ↔ (𝜑 ∨ ¬ 𝜑))
2		iftrue 3566	. . . 4 ⊢ (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
3		ibar 301	. . . . 5 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓)))
4	3	ifbid 3582	. . . 4 ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∧ 𝜓), 𝐴, 𝐵))
5	2, 4	eqtr2d 2230	. . 3 ⊢ (𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
6		simpl 109	. . . . . 6 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
7	6	con3i 633	. . . . 5 ⊢ (¬ 𝜑 → ¬ (𝜑 ∧ 𝜓))
8	7	iffalsed 3571	. . . 4 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = 𝐵)
9		iffalse 3569	. . . 4 ⊢ (¬ 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵)
10	8, 9	eqtr4d 2232	. . 3 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
11	5, 10	jaoi 717	. 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 709  DECID wdc 835   = wceq 1364  ifcif 3561

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 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-11 1520  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-dc 836  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-if 3562

This theorem is referenced by:  isumss  11556