On Page 11 of Second edition of Concrete Mathematics, it reads as  The induction step has two parts, depending on whether $l$ is even or odd. If $m &gt; 0$ and $2^m + l= 2n$, then $l$ is even and  $J(2^m +l) = 2J(2^{m-1} +l/2)-1 = 2(2l/2+1)-1 = 2l+1$  by (1.8)   Here I didn't get how $2J(2^{m-1} +l/2)-1 $ is reduced to $2(2l/2+1)-1$  Please Help.
I've been going through  and have a question on the inductive proof for the . Recurrent relation:  $J(1) = 1$ $J(2n) = 2J(n) - 1$ $J(2n+1) = 2J(n) + 1$  Closed form hypothesis:  $J(2^m + l) = 2l + 1$, for $m \ge 0$ and $0\le l&lt;2^m$  Inductive Proof (for the even case):  $J(1)=1$ $J(2^m+l)= 2J(2^{m-1}+l/2)-1=2(2l/2+1)-1=2l+1$  How do the authors get from $2J(2^{m-1}+l/2)-1$ to $2(2l/2+1)-1$?
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