Johnstone, Topos theory, 0.3, page 13, asserts that, given a Grothendieck pretopology $P$, if the equalizer condition on a presheaf $F$ is satisfied for a family of arrows $R=\{U_i\to U\}$, then it is satisfied for every family $S=\{W_j\to U\}\supset R$.
By equalizer condition I mean that $$F(U)\to \prod_R F(U_i)\rightrightarrows \prod_{R\times R}F(U_i\times_U U_j),$$ with the natural restriction maps, is an equalizer.
I can't see why this is true. First of all, should I assume that $R$ is a covering in the given pretopology? Otherwise it would sound very strange to me. For example, take a discrete space of two points $X=\{p,q\}$, the pretopology being given by the topological coverings, and take $R=\{p\to X\}, S=\{p\to X,q\to X\}$; then the equalizer condition for $R$ means that every local section on $p$ comes from a global section on $X$. So take $F(p):=\{0\}, F(q)=\mathbb Z, F(\varnothing)=\{0\},F(X)=\{0\}$. This is a presheaf (with the natural maps, identity or zero) and satisfies the equalizer condition for $U=X$ and the family $R$ just defined: in $$\{0\}\to \{0\} \rightrightarrows \{0\}$$ one can always take the global section $0$ on the left. However, the equalizer for $S$ is: $$\{0\}\to \{0\}\times \mathbb Z\rightrightarrows \{0\}\times \{0\}\times \{0\}\times \mathbb Z$$ where both arrows on the right are equal to $(0,1)\mapsto (0,0,0,1)$ (note that the "mixed" fiber products $p\times_X q$ and $q\times _X p$ are the empty set). So take $(0,1)\in \{0\}\times \mathbb Z$ in the central node of the last diagram. The maps on the right are the same, hence $F$ satisfies the equalizer condition for $S$ if and only if there exists $s\in F(X)=0$ s.t. $s|_{\{p\}}=0,s|_{\{q\}}=1$, but this is of course impossible.
Is my conterexample correct?
Thank you in advance.