Confusion about proof that every submodule of a semisimple module contains a simple module

Question

Theorem 8.3 of these commutative algebra notes by Pete L. Clark characterizes semisimple modules.

Theorem 8.3. For an $R$-module $M$, TFAE:

1. $M$ is a direct sum of simple submodules.
2. Every submodule of $M$ is a direct summand.
3. $M$ is a sum of simple submodules.

The proof of (2)$\implies$(1) involves a claim whose proof I do not understand. Here's the argument.

I don't understand the equality sign. In general there's a canonical containment $$(a\vee b)\wedge (a\vee c)\geq a\vee(b\wedge c)$$ induced by universal properties of meets and joins. (In the proof $b\wedge c=0$ so the RHS is just $a$.) Being an equality is an instance of the codistributivity axiom which as a property of a poset is equivalent to distributivity. Why is there equality here?

Answer 1

Let $x\in (D\oplus F)\cap (D\oplus G)$, $x=d+f, d\in D, f\in F, x=d'+g, d'\in D, g\in G$, it implies that $d-d'=g-f$. $d-d'\in D, g-f\in F\oplus G$, since $(F\oplus G)\cap D=0$, we deduce that $d-d'=0$ and $g-f=0$. This implies that $g=f$ since $F\cap G=0$, we deduce that $f=g=0$.