= Solution
Put $M=2^n$ for an $n$-bit <processor register>. The <carry flag> for addition satisfies \b[$\operatorname{bin}(x)+\operatorname{bin}(y)=\operatorname{bin}(z)+MC$]: the result retains the low $n$ <bits> and $C=1$ records an <arithmetic carry> beyond them.
For the <ARM architecture family> subtraction convention, \b[$\operatorname{bin}(x)-\operatorname{bin}(y)=\operatorname{bin}(z)-M(1-C)$]. Thus $C=1$ means \i[no borrow], equivalently the unsigned minuend is at least the subtrahend. This flag is different from signed overflow: an <arithmetic carry> decision is not a test of the signed result.
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