Proving $(U \cap V) \cap u(U \cap v) = (u \cap V)(U \cap v)$

Question

Let $U, V$ be subgroups of a group. Let $u \trianglelefteq U, v \trianglelefteq V$. I proved like this. Then by applying modular law, $$\mathrm{(LHS)} = (U \cap V) \cap u(U \cap v) = U \cap V \cap uv \\ \mathrm{(RHS)} = (u \cap V)(U \cap v) = U \cap (u \cap V)v = U \cap V \cap uv \\ \therefore \mathrm{(LHS)} = \mathrm{(RHS)}$$ However I think it's overkill. Is there a shorter proof?