= Solution
Under the <null hypothesis> of consistency, $\widehat\delta_D-\widehat\delta_I$ has approximately zero <expectation> and <variance> $v_D+v_I$, since the displayed trials are <independent>. Thus a two-sided <Wald test> uses
$$
\boxed{Z=\frac{\widehat\delta_D-\widehat\delta_I}{\sqrt{v_D+v_I}}\ \dot\sim\ N(0,1),\qquad Z^2\ \dot\sim\ \chi^2_1.}
$$
The observed difference is $-0.07$, and $Z=-0.07/\sqrt{0.29}\simeq-0.130$. Its two-sided <p-value> is approximately $0.897$, so there is no detectable discrepancy here. Failure to reject is not proof of consistency: this test has limited <statistical power>, and <transitivity in network meta-analysis> also needs substantive assessment. If evidence sources overlap, replace $v_D+v_I$ by $v_D+v_I-2\operatorname{Cov}(\widehat\delta_D,\widehat\delta_I)$.
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