Metamath Proof Explorer


Theorem inindir

Description: Intersection distributes over itself. (Contributed by NM, 17-Aug-2004)

Ref Expression
Assertion inindir ( ( 𝐴𝐵 ) ∩ 𝐶 ) = ( ( 𝐴𝐶 ) ∩ ( 𝐵𝐶 ) )

Proof

Step Hyp Ref Expression
1 inidm ( 𝐶𝐶 ) = 𝐶
2 1 ineq2i ( ( 𝐴𝐵 ) ∩ ( 𝐶𝐶 ) ) = ( ( 𝐴𝐵 ) ∩ 𝐶 )
3 in4 ( ( 𝐴𝐵 ) ∩ ( 𝐶𝐶 ) ) = ( ( 𝐴𝐶 ) ∩ ( 𝐵𝐶 ) )
4 2 3 eqtr3i ( ( 𝐴𝐵 ) ∩ 𝐶 ) = ( ( 𝐴𝐶 ) ∩ ( 𝐵𝐶 ) )