Theorem ifan 3702

Description: Rewrite a conjunction in an if statement 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 3669	. . 3 ⊢ (φ → if(φ, if(ψ, A, B), B) = if(ψ, A, B))
2		ibar 490	. . . 4 ⊢ (φ → (ψ ↔ (φ ∧ ψ)))
3	2	ifbid 3681	. . 3 ⊢ (φ → if(ψ, A, B) = if((φ ∧ ψ), A, B))
4	1, 3	eqtr2d 2386	. 2 ⊢ (φ → if((φ ∧ ψ), A, B) = if(φ, if(ψ, A, B), B))
5		simpl 443	. . . . 5 ⊢ ((φ ∧ ψ) → φ)
6	5	con3i 127	. . . 4 ⊢ (¬ φ → ¬ (φ ∧ ψ))
7		iffalse 3670	. . . 4 ⊢ (¬ (φ ∧ ψ) → if((φ ∧ ψ), A, B) = B)
8	6, 7	syl 15	. . 3 ⊢ (¬ φ → if((φ ∧ ψ), A, B) = B)
9		iffalse 3670	. . 3 ⊢ (¬ φ → if(φ, if(ψ, A, B), B) = B)
10	8, 9	eqtr4d 2388	. 2 ⊢ (¬ φ → if((φ ∧ ψ), A, B) = if(φ, if(ψ, A, B), B))
11	4, 10	pm2.61i 156	1 ⊢ if((φ ∧ ψ), A, B) = if(φ, if(ψ, A, B), B)

Syntax hints:  ¬ wn 3   ∧ wa 358   = wceq 1642   ifcif 3663

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-if 3664

This theorem is referenced by: (None)