= Solution
Conversely, suppose the <exterior derivative> of every $(1,0)$-form has no $(0,2)$ part. For sections $U,V$ of $T^{0,1}M$, the same <exterior derivative> formula gives
$$
0=d\alpha(U,V)=-\alpha([U,V])\qquad\text{for every }(1,0)\text{-form }\alpha.
$$
The common annihilator of all $(1,0)$-covectors is $T^{0,1}M$. Local $(1,0)$-coframes therefore show that $[U,V]$ belongs to $T^{0,1}M$. Conjugating gives bracket closure of $T^{1,0}M$. Thus \b[the two criteria are equivalent]. These are the <bracket and differential-form criteria for integrability>, and establish integrability in the sense used here.
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