(W)" E = 1). We have H(W|Y", E = 0) = 0, since knowing Yn and that there is no codeword error is equivalent to knowing the input codeword label W. Second, we have H(W)Y, E = 1) < log(2nR - 1) < log(2nR) = nR, since knowing Yn and that there is a codeword error, the number of mistaken codeword possibilities is 2nR - 1, with uniform probability q = 1/(2nR - 1); hence the entropy can be upper bounded by H' = - log(q) < nR. Finally, we have H(E) < 1, since E is a binary random variable. Substituting the two upper bounds into Eq. (L3) yields: since the knowledge of the input message source Xn and the codeword label W are equivalent. Equation (L5) is Fano's inequality. Property B The second tool required for the converse proof of the CCT is the property according to which the channel capacity per transmission is not increased by passing through the data several times. Note that this property is true if one assumes that the channel is memoryless, namely that there is no possible correlation between errors concerning successive bits or successive codewords. As I established earlier, the channel capacity for an n-bit codeword is nC, where C is the capacity of the binary channel, corresponding to a single bit transmission. The proposed new property can be restated as H(X";Y") <nC. The corresponding proof of Eq. (L6) is relatively straightforward. Indeed, we have, by definition Xn;Y")=H(Y")-H(Y"|X"). Converse proof of the channel coding theorem Let us now develop the second term in the right-hand side in Eq. (L7) as follows: H(Y"|Xn)=H(y1X")+H(y2ly1, = X") +...+ -H(yn/y1,y2,...,yn-1,X") The first equality stems from substituting the extended output source Yn = y1, Yn-1 Yn-1 = y2, Yn-2, etc., (here yk means the binary source of bit or rank k in the codeword), into the chain rule (Eq. (5.22)) as follows: = H(y1|X") + -H(Y"-1/y1, X") = H(y1|X") + H(y2Yn-21y1, X") = H(y1)X) + H(y2|y1, Xn) + H(Y"-2|y1, y2, X") = H(y1)X) + H(y2|y1, The last equality in Eq. (L8) stems from the fact that, assuming a memoryless chan- nel, all receiv