Metamath Proof Explorer


Theorem ifcl

Description: Membership (closure) of a conditional operator. (Contributed by NM, 4-Apr-2005)

Ref Expression
Assertion ifcl ( ( 𝐴𝐶𝐵𝐶 ) → if ( 𝜑 , 𝐴 , 𝐵 ) ∈ 𝐶 )

Proof

Step Hyp Ref Expression
1 eleq1 ( 𝐴 = if ( 𝜑 , 𝐴 , 𝐵 ) → ( 𝐴𝐶 ↔ if ( 𝜑 , 𝐴 , 𝐵 ) ∈ 𝐶 ) )
2 eleq1 ( 𝐵 = if ( 𝜑 , 𝐴 , 𝐵 ) → ( 𝐵𝐶 ↔ if ( 𝜑 , 𝐴 , 𝐵 ) ∈ 𝐶 ) )
3 1 2 ifboth ( ( 𝐴𝐶𝐵𝐶 ) → if ( 𝜑 , 𝐴 , 𝐵 ) ∈ 𝐶 )