“…In order to find subfile indices that m th user get from all (b r − 1) z transmissions, we have to vary the value of e s such that e s = a s , ∀s ∈[z].The m th user is able to receive subfile indices given bybr es = 1, es = as ∀s ∈ [z] {C 1,e1 ∩ C 2,e2 ∩ • • • ∩ C z,ez }.In addition to this using Y m , the m th user gets the subfile indices shown in the sequence of expressions numbered (14) to (20), as shown at the top of the page 13. Notice that the last expression in(17) is the union of all the blocks in a parallel class. So from the property of resolvable designs the above set is equal to the set∩ t∈X \m Y t = ls = {is,js}, ∀s ∈ [z] (l1,l2,...,lz) = (a1,a2,...,az) {C 1,l1 ∪ C 2,l2 ∪ • • • ∪ C z,lz } ls = {is,js}, ∀s ∈ [2,z] {C 1,e1 ∪ C 2,l2 ∪ • • • ∪ C z,lz } ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ls = {is,js}, ∀s ∈ [2,z] (l2,l3,...,lz) = (a2,a3,...,az) {C 1,a1 ∪ C 2,l2 ∪ • • • ∪ C z,lz } ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ = C 1,e1 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ls = {is,js}, ∀s ∈ [2,z] (l2,l3,...,lz) = (a2,a3,...,az) {C 1,a1 ∪ C 2,l2 ∪ C 3,l3 ∪ • • • ∪ C z,lz } ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ls = {is,js},∀s ∈ [2,z] (l2,l3,...,lz) = (a2,a3,...,az)…”