Theorem ifor

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

	Ref	Expression
Assertion	ifor	⊢ if ( ( 𝜑 ∨ 𝜓 ) , 𝐴 , 𝐵 ) = if ( 𝜑 , 𝐴 , if ( 𝜓 , 𝐴 , 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		iftrue	⊢ ( ( 𝜑 ∨ 𝜓 ) → if ( ( 𝜑 ∨ 𝜓 ) , 𝐴 , 𝐵 ) = 𝐴 )
2	1	orcs	⊢ ( 𝜑 → if ( ( 𝜑 ∨ 𝜓 ) , 𝐴 , 𝐵 ) = 𝐴 )
3		iftrue	⊢ ( 𝜑 → if ( 𝜑 , 𝐴 , if ( 𝜓 , 𝐴 , 𝐵 ) ) = 𝐴 )
4	2 3	eqtr4d	⊢ ( 𝜑 → if ( ( 𝜑 ∨ 𝜓 ) , 𝐴 , 𝐵 ) = if ( 𝜑 , 𝐴 , if ( 𝜓 , 𝐴 , 𝐵 ) ) )
5		iffalse	⊢ ( ¬ 𝜑 → if ( 𝜑 , 𝐴 , if ( 𝜓 , 𝐴 , 𝐵 ) ) = if ( 𝜓 , 𝐴 , 𝐵 ) )
6		biorf	⊢ ( ¬ 𝜑 → ( 𝜓 ↔ ( 𝜑 ∨ 𝜓 ) ) )
7	6	ifbid	⊢ ( ¬ 𝜑 → if ( 𝜓 , 𝐴 , 𝐵 ) = if ( ( 𝜑 ∨ 𝜓 ) , 𝐴 , 𝐵 ) )
8	5 7	eqtr2d	⊢ ( ¬ 𝜑 → if ( ( 𝜑 ∨ 𝜓 ) , 𝐴 , 𝐵 ) = if ( 𝜑 , 𝐴 , if ( 𝜓 , 𝐴 , 𝐵 ) ) )
9	4 8	pm2.61i	⊢ if ( ( 𝜑 ∨ 𝜓 ) , 𝐴 , 𝐵 ) = if ( 𝜑 , 𝐴 , if ( 𝜓 , 𝐴 , 𝐵 ) )