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Metamath Proof Explorer

Theorem ineq2d

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

Ref Expression
Hypothesis ineq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion ineq2d ( 𝜑 → ( 𝐶 ∩ 𝐴 ) = ( 𝐶 ∩ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 ineq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 ineq2 ⊢ ( 𝐴 = 𝐵 → ( 𝐶 ∩ 𝐴 ) = ( 𝐶 ∩ 𝐵 ) )
3 1 2 syl ⊢ ( 𝜑 → ( 𝐶 ∩ 𝐴 ) = ( 𝐶 ∩ 𝐵 ) )