= Solution
The assertion is \b[false]. Suppose the <formal language> $L=\{ww:w\in\{0,1\}^*\}$ were a <regular language>, and let $p$ be its constant from the <pumping lemma for regular languages>. Consider
$$
t=0^p1\,0^p1\in L.
$$
In every decomposition $t=xyz$ with $|xy|\le p$ and $|y|\ge1$, the pumped factor is $y=0^d$ with $1\le d\le p$. Pumping down gives
$$
xz=0^{p-d}1\,0^p1.
$$
This <string> has exactly two ones and ends in a one. If it were $vv$, each copy of $v$ would contain exactly one one, and each copy would end in a one. The first copy would consequently be $0^{p-d}1$, while the second would be $0^p1$. They have different lengths because $d>0$, so they cannot be equal. This contradiction to the <pumping lemma for regular languages> proves
$$
\boxed{\{ww:w\in\{0,1\}^*\}\text{ is not regular}.}
$$
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