Theorem ifan

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

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

Proof

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