Slate - Intersection is associative

Proposition 1.1.9. Let U be a set, R, S, T ⊆ U. Then:

(R ∩ S) ∩ T = R ∩ (S ∩ T)

Proof.

Let x ∈ (R ∩ S) ∩ T. Then x ∈ R ∩ (S ∩ T):
We show that x ∈ R and x ∈ S ∩ T:
x ∈ (R ∩ S) ∩ T ⇒^def x ∈ R ∩ S and x ∈ T
x ∈ R ∩ S ⇒^def x ∈ R and x ∈ S
We have x ∈ S ∩ T by definition.
Let x ∈ R ∩ (S ∩ T). Then x ∈ (R ∩ S) ∩ T:
We show that x ∈ R ∩ S and x ∈ T:
x ∈ R ∩ (S ∩ T) ⇒^def x ∈ R and x ∈ S ∩ T
x ∈ S ∩ T ⇒^def x ∈ S and x ∈ T
We have x ∈ R ∩ S by definition.

References.