Theorem ifan 4476

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 4431	. . 3 ⊢ (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
2		ibar 532	. . . 4 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓)))
3	2	ifbid 4447	. . 3 ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∧ 𝜓), 𝐴, 𝐵))
4	1, 3	eqtr2d 2834	. 2 ⊢ (𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
5		simpl 486	. . . . 5 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
6	5	con3i 157	. . . 4 ⊢ (¬ 𝜑 → ¬ (𝜑 ∧ 𝜓))
7	6	iffalsed 4436	. . 3 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = 𝐵)
8		iffalse 4434	. . 3 ⊢ (¬ 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵)
9	7, 8	eqtr4d 2836	. 2 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
10	4, 9	pm2.61i 185	1 ⊢ if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)

Syntax hints:  ¬ wn 3   ∧ wa 399   = wceq 1538  ifcif 4425

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-if 4426

This theorem is referenced by:  itg0  24383  iblre  24397  itgreval  24400  iblss  24408  iblss2  24409  itgle  24413  itgss  24415  itgeqa  24417  iblconst  24421  itgconst  24422  ibladdlem  24423  iblabslem  24431  iblabsr  24433  iblmulc2  24434  itgsplit  24439  bddiblnc  24445  itgcn  24448  mrsubrn  32873  itg2gt0cn  35112  ibladdnclem  35113  iblabsnclem  35120  iblmulc2nc  35122  iblsplit  42608