The sheaf condition can be checked on a basis with basis-refinable intersections

Assume $\mathcal{F}$ is a sheaf. Let $U={\bigcup }_i{B}_i$ be a basis cover and let $(s_i)$ satisfy the basis-overlap hypothesis. Fix $i,j$. The sets $C\in \mathcal{B}$ with $C\subseteq B_i\cap B_j$ cover $B_i\cap B_j$ by [L1] and the intersection hypothesis on $\mathcal{B}$. On each such $C$ the restrictions of $s_i$ and $s_j$ agree, so locality from [L2] gives $s_i{\mathrm{\mid }}_{B_i\cap B_j}=s_j{\mathrm{\mid }}_{B_i\cap B_j}$. Gluing in [L2] then produces a unique $s\in \mathcal{F}(U)$ with $s{\mathrm{\mid }}_{B_i}=s_i$.

Assume the displayed basis condition. Let $U={\bigcup }_{\alpha \in A}{U}_{\alpha }$ be an arbitrary open cover and let $t_{\alpha }\in \mathcal{F}(U_{\alpha })$ be compatible on overlaps. For each $x\in U$, choose $\alpha (x)$ with $x\in U_{\alpha (x)}$, and then choose $B_x\in \mathcal{B}$ with $x\in B_x\subseteq U_{\alpha (x)}$ by [L1]. Put $r_x:=t_{\alpha (x)}{\mathrm{\mid }}_{B_x}$.

By the assumed basis condition, there is a unique $s\in \mathcal{F}(U)$ with $s{\mathrm{\mid }}_{B_x}=r_x$ for every $x\in U$.

Let $C\in \mathcal{B}$ satisfy $C\subseteq B_x\cap B_y$. Then $C\subseteq U_{\alpha (x)}\cap U_{\alpha (y)}$, so compatibility of the original family gives $r_x{\mathrm{\mid }}_C=t_{\alpha (x)}{\mathrm{\mid }}_C=t_{\alpha (y)}{\mathrm{\mid }}_C=r_y{\mathrm{\mid }}_C$. Therefore the family $(r_x{)}_{x\in U}$ satisfies the displayed basis condition for the basis cover $U={\bigcup }_{x\in U}{B}_x$.

Fix $\alpha \in A$. For each $y\in U_{\alpha }$, choose $C_y\in \mathcal{B}$ with $y\in C_y\subseteq B_y\cap U_{\alpha }$ by [L1] and the basis-refinement hypothesis. Then the $C_y$ form a basis cover of $U_{\alpha }$. Since $C_y\subseteq U_{\alpha (y)}\cap U_{\alpha }$, step 2.1 gives $r_y{\mathrm{\mid }}_{C_y}=t_{\alpha (y)}{\mathrm{\mid }}_{C_y}=t_{\alpha }{\mathrm{\mid }}_{C_y}\mathrm{.}$ Because $s{\mathrm{\mid }}_{C_y}=r_y{\mathrm{\mid }}_{C_y}$ by step 1.3, the two sections $s{\mathrm{\mid }}_{U_{\alpha }}$ and $t_{\alpha }$ restrict to the same family on the basis cover $\{C_y{\}}_{y\in U_{\alpha }}$. By the uniqueness part of the assumed basis condition, $s{\mathrm{\mid }}_{U_{\alpha }}=t_{\alpha }$. Since this holds for every $\alpha$, the arbitrary compatible family glues uniquely, so [L2] implies that $\mathcal{F}$ is a sheaf.