fix #791

lec_03a_computing_every_function.md

@@ -437,7 +437,7 @@ $$
 > ### {.proof data-ref="lookup-rec-lem"}
 If the most significant bit $i_{0}$  of $i$ is zero, then the index $i$ is in $\{0,\ldots,2^{k-1}-1\}$ and hence we can perform the lookup on the "first half" of $x$ and the result of  $LOOKUP_k(x,i)$ will be the same as $a=LOOKUP_{k-1}(x_0,\ldots,x_{2^{k-1}-1},i_1,\ldots,i_{k-1})$.
 On the other hand, if this most significant bit $i_{0}$  is equal to $1$, then the index is in $\{2^{k-1},\ldots,2^k-1\}$, in which case the result of $LOOKUP_k(x,i)$ is the same as $b=LOOKUP_{k-1}(x_{2^{k-1}},\ldots,x_{2^k-1},i_1,\ldots,i_{k-1})$.
-Thus we can compute $LOOKUP_k(x,i)$ by first computing $a$ and $b$ and then outputting $IF(i_{k-1},a,b)$.
+Thus we can compute $LOOKUP_k(x,i)$ by first computing $a$ and $b$ and then outputting $IF(i_0,b,a)$.
 
 
 __Proof of [lookup-thm](){.ref} from [lookup-rec-lem](){.ref}.__ Now that we have [lookup-rec-lem](){.ref},