For any three sets A, B, C, prove that :

(i) AX(B intersection C) = (AXB) intersection (AXC)

Then p = (x,y) such that x ∈ A and y∈(B∩C).

The logical breakdown is as follows:

x∈A and (y∈(B∩C))

x∈A and (y∈B and y∈C)

(x∈A and y∈B) and (x∈A and y∈C)

(x,y) ∈ (A x B) and (x,y) ∈(A x C) [ This step is the definition of the Cartesian Product. ]

(x,y) ∈ (AxB) ∩ (AxC)

Since p was an arbitrary element of Ax(B∩C), it follows that Ax(B∩C) ⊆ (AxB)∩(AxC)..................(1)

Now Let p be an arbitrary element of (AxB)∩(AxC)

Then p = (x,y) such that (x,y) ∈ (AxB) and (x,y) ∈ (AxC)

The logical breakdown is as follows:

(x,y) ∈ (AxB) and (x,y) ∈ (AxC)

x∈A and y∈B and x∈A and y∈C

x∈A and (y∈B and y∈C)

(x,y) ∈A X (B∩C)

Since p was an arbitrary element of (AxB)∩(AxC), it follows that (AxB)∩(AxC) ⊆A X (B∩C).................(2)

So from (1) and (2) we get,

Ax(B∩C) = (AxB)∩(AxC)

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