Metamath Proof Explorer


Theorem inindi

Description: Intersection distributes over itself. (Contributed by NM, 6-May-1994)

Ref Expression
Assertion inindi ( 𝐴 ∩ ( 𝐵𝐶 ) ) = ( ( 𝐴𝐵 ) ∩ ( 𝐴𝐶 ) )

Proof

Step Hyp Ref Expression
1 inidm ( 𝐴𝐴 ) = 𝐴
2 1 ineq1i ( ( 𝐴𝐴 ) ∩ ( 𝐵𝐶 ) ) = ( 𝐴 ∩ ( 𝐵𝐶 ) )
3 in4 ( ( 𝐴𝐴 ) ∩ ( 𝐵𝐶 ) ) = ( ( 𝐴𝐵 ) ∩ ( 𝐴𝐶 ) )
4 2 3 eqtr3i ( 𝐴 ∩ ( 𝐵𝐶 ) ) = ( ( 𝐴𝐵 ) ∩ ( 𝐴𝐶 ) )