= Solution
For events $A_i$ with a dependency <graph>, assume each event is independent of the <sigma-algebra> generated by its nonneighbours. If numbers $x_i\in[0,1)$ satisfy $P(A_i)\le x_i\prod_{j\in\Gamma(i)}(1-x_j)$, the general <Lovász local lemma> gives
$$
P\left(\bigcap_iA_i^c\right)\ge\prod_i(1-x_i)>0.
$$
The symmetric version says that if every event has probability at most $q$, the dependency <graph> has <maximum degree> at most $d$, and $eq(d+1)\le1$, then none occurs with positive probability. For $d\ge1$, set $x_i=1/(d+1)$. Since $(1-1/(d+1))^d\ge e^{-1}$, the general condition follows from the symmetric numerical hypothesis. The logarithmic inequality $\log(1+1/d)\le1/d$ proves that estimate. When $d=0$, the events are mutually independent and each has probability below one, so the conclusion is immediate. This is the intended implication from the general to the symmetric form; the source repeats the word symmetric in its final clause.
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