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Metamath Proof Explorer

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 ) )