AB5 implies AB4

Let $(m_i:B_i\rightarrowtail A_i{)}_{i\in I}$ be a small family of monomorphisms, and let $m:{\coprod }_{i\in I}{B}_i\to {\coprod }_{i\in I}{A}_i$ be the induced coproduct map, which exists by the AB3 part of [L1]. Let $k:K\rightarrowtail {\coprod }_i{B}_i$ be its kernel. For each finite subset $F\subseteq I$, let $S_F\le {\coprod }_i{B}_i$ be the finite partial sum of the summands with indices in $F$. Because finite coproducts in an abelian category are biproducts, the restriction of $m$ to $S_F$ is a finite direct sum of monomorphisms and is therefore monic. Hence $K\wedge S_F=0$ for every finite $F$.

The family $(S_F)$ is directed and has join ${\coprod }_i{B}_i$. Applying the AB5 identity [L1] with the fixed subobject $K$ gives $K=\left({\bigvee }_F{S}_F\right)\wedge K={\bigvee }_F(S_F\wedge K)=0.$ So the kernel of $m$ is zero, and therefore $m$ is monic. By [L2], this is exactly AB4.