Let $\mathcal{I}(D)$ be the ideal of $\Omega^*(M)$ given by the dual description of the distribution.
A distribution $D$ is involutive if and only if $\mathcal{I}(D)$ is a differential ideal.
(See [Warner_1983] proposition 2.30)
Proof
Suppose $D$ is involutive, and let $\omega \in \mathcal{I}(D$ be a 1-form. Given $X, Y \in \Gamma(M,D)$, $d\omega (X,Y)=0$, by the infinitesimal Stokes' theorem. Therefore, by using proposition lemma of belongingness to ideal, $d\omega \in \mathcal{I}(D$. Since $\mathcal{I}(D)$ is generated by 1-forms, it is true for any of its elements.
Conversely, suppose $\mathcal{I}(D)$ is a differential ideal, and take $X, Y \in \Gamma(M,D)$. To see that $[X, Y] \in \Gamma(M,D)$ observe that $d\omega(X,Y)=0$ and so $\omega([X,Y]) =0$ for any $\omega \in \Gamma(M,D^*)$. Therefore $[X,Y] \in \Gamma(M,D^*$.$\blacksquare$
It can also be characterized in terms of $D^*$ (see dual description of the distribution): for every $\omega \in D^*$ it must be
$$ d\omega(X,Y)=0 $$for every $X,Y\in D$. See [Lychagin_2021] page 5/77.
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Author of the notes: Antonio J. Pan-Collantes
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