Metamath Proof Explorer


Theorem ineq1d

Description: Equality deduction for intersection of two classes. (Contributed by NM, 10-Apr-1994)

Ref Expression
Hypothesis ineq1d.1 ( 𝜑𝐴 = 𝐵 )
Assertion ineq1d ( 𝜑 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )

Proof

Step Hyp Ref Expression
1 ineq1d.1 ( 𝜑𝐴 = 𝐵 )
2 ineq1 ( 𝐴 = 𝐵 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )
3 1 2 syl ( 𝜑 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )