Verified answer

Let

(x,y) ∈ (AxB)∩(BxA)

=> {(x,y) ∈ (AxB)} and {(x,y) ∈ (BxA)}

=> {x ∈ A and y ∈ B} and {x ∈ B and y ∈ A}

=> {x ∈ A and x ∈ B} and {y ∈ A and y ∈ B}

=> {x ∈ (A∩B)} and {y ∈ (A∩B)}

=> (x,y) ∈ (A∩B)x(A∩B)

=> (AxB)∩(BxA) ⊆ (A∩B)x(A∩B) ... (1)

Now let,

(x,y) ∈ (A∩B)x(A∩B)

=> x ∈ (A∩B) and y ∈ (A∩B)

=> (x ∈ A and x ∈ B) and (y ∈ A and y ∈ B)

=> (x ∈ A and y ∈ B) and (x ∈ B and y ∈ A)

=> (x,y) ∈ (AxB) and (x,y) ∈ (BxA)

=> (x,y) ∈ (AxB)∩(BxA)

=> (A∩B)x(A∩B) ⊆ (AxB)∩(BxA)...(2)

(1) and (2) implies,

(A∩B)x(A∩B) = (AxB)∩(BxA)