Theorem ifan 4544

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

Assertion

Ref	Expression
ifan	⊢ if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)

Proof of Theorem ifan

Step	Hyp	Ref	Expression
1		iftrue 4496	. . 3 ⊢ (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
2		ibar 528	. . . 4 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓)))
3	2	ifbid 4514	. . 3 ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∧ 𝜓), 𝐴, 𝐵))
4	1, 3	eqtr2d 2766	. 2 ⊢ (𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
5		simpl 482	. . . . 5 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
6	5	con3i 154	. . . 4 ⊢ (¬ 𝜑 → ¬ (𝜑 ∧ 𝜓))
7	6	iffalsed 4501	. . 3 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = 𝐵)
8		iffalse 4499	. . 3 ⊢ (¬ 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵)
9	7, 8	eqtr4d 2768	. 2 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
10	4, 9	pm2.61i 182	1 ⊢ if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)

Syntax hints:  ¬ wn 3   ∧ wa 395   = wceq 1540  ifcif 4490

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-if 4491

This theorem is referenced by:  psdmvr  22062  itg0  25687  iblre  25701  itgreval  25704  iblss  25712  iblss2  25713  itgle  25717  itgss  25719  itgeqa  25721  iblconst  25725  itgconst  25726  ibladdlem  25727  iblabslem  25735  iblabsr  25737  iblmulc2  25738  itgsplit  25743  bddiblnc  25749  itgcn  25752  mrsubrn  35500  itg2gt0cn  37664  ibladdnclem  37665  iblabsnclem  37672  iblmulc2nc  37674  selvvvval  42566  iblsplit  45957