= Solution
A <logical right shift> by $k$ moves $a_{i+k}$ into position $i$ and fills the high positions with zero. Its <unsigned interpretation> is
$$
\operatorname{bin}(A\mathbin{\mathrm{LSR}}k)=\sum_{i=k}^{n-1}a_i2^{i-k}=:q.
$$
Meanwhile the original <unsigned interpretation> decomposes as
$$
a=2^kq+\underbrace{\sum_{i=0}^{k-1}a_i2^i}_{r},\qquad 0\leq r\leq2^k-1.
$$
By <integer division>, \b[$q=\lfloor a/2^k\rfloor$], exactly the unsigned quotient. Thus the source instruction puts the required quotient into `r1`. This is a proof from <binary representation>, rather than an assumption that every kind of right shift implements every kind of division. The displayed <bit> argument applies to $0\leq k\leq n$, with endpoint operations implemented as a copy or zero where appropriate.
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