Fill in the blanks to complete the proof:Recall that permutations are defined as bijections on a set X. In order to show that the two permutations στ and τσ are equal, it’s enough to show that they are the same function. In other words, we just need to show that στ (x) = < 1 > for all x ∈ X. We’ll define A = {a1, a2, . . . , aj} and B = {b1, b2, . . . , bk}. By hypothesis A and B are disjoint, so A < 2 > B = < 3 > . Given an arbitrary x ∈ X, there are three possibilities: (i) x ∈ A and x /∈ B; (ii) x ∈ < 4 > and x /∈ < 6 > ; (iii) x /∈ < 7 > and x /∈ < 8 > . (i) In this case, since x /∈ B it follows that τ (x) = x. We then have στ (x) = σ(τ (x)) = σ(x). Furthermore, since x ∈ A it follows that σ(x) ∈ A, so σ(x) ∈/ B. We then have τσ(x) = τ (σ(x)) = σ(x). It follows that στ (x) = τσ(x). (ii) In this case, since x /∈ < 9 > it follows that < 10 > (x) = x. We then have τσ(x) = < 11 > = < 12 > (x). Furthermore, since x ∈ < 13 > it follows that < 14 > (x) ∈ < 15 > , so < 16 > (x) ∈/ < 17 > . We then have στ (x) = < 18 > = < 19 > (x). It follows that στ (x) = τσ(x). (iii) In this case, since x /∈ A it follows that < 20 > (x) = x. Similarly since x /∈ < 21 > it follows that < 22 > (x) = x.We then have τσ(x) = < 23 > and στ (x) = < 24 > . It follows that στ (x) = τσ(x). In all three cases we have στ (x) = < 25 > , so therefore στ = τσ   The blanks are the spots with numbers surrounded by these symbols < >