Theorem ifan 4358

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 4313	. . 3 ⊢ (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵))
2		ibar 526	. . . 4 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓)))
3	2	ifbid 4329	. . 3 ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∧ 𝜓), 𝐴, 𝐵))
4	1, 3	eqtr2d 2863	. 2 ⊢ (𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
5		simpl 476	. . . . 5 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
6	5	con3i 152	. . . 4 ⊢ (¬ 𝜑 → ¬ (𝜑 ∧ 𝜓))
7	6	iffalsed 4318	. . 3 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = 𝐵)
8		iffalse 4316	. . 3 ⊢ (¬ 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵)
9	7, 8	eqtr4d 2865	. 2 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵))
10	4, 9	pm2.61i 177	1 ⊢ if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)

Syntax hints:  ¬ wn 3   ∧ wa 386   = wceq 1658  ifcif 4307

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1896  ax-4 1910  ax-5 2011  ax-6 2077  ax-7 2114  ax-9 2175  ax-10 2194  ax-11 2209  ax-12 2222  ax-ext 2804

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 881  df-tru 1662  df-ex 1881  df-nf 1885  df-sb 2070  df-clab 2813  df-cleq 2819  df-clel 2822  df-if 4308

This theorem is referenced by:  itg0  23946  iblre  23960  itgreval  23963  iblss  23971  iblss2  23972  itgle  23976  itgss  23978  itgeqa  23980  iblconst  23984  itgconst  23985  ibladdlem  23986  iblabslem  23994  iblabsr  23996  iblmulc2  23997  itgsplit  24002  itgcn  24009  mrsubrn  31957  itg2gt0cn  34009  ibladdnclem  34010  iblabsnclem  34017  iblmulc2nc  34019  bddiblnc  34024  iblsplit  40977