Proof: $\sum_{i=1}^pE_i \doteq \bigoplus_{i=1}^p E_i \leftrightarrow \forall i\in \{1,...,p\}(E_i \cap \sum_{t \in \{1,...,p\}-\{i\}}E_t=\{0\})$
- Proof of $\bf\leftarrow$
Let $e_i\in E_i$ such that $$e_1+e_2+\cdots+e_n=0$$ so forall $i$ we have $$e_i=-\sum_{j\ne i} e_j\in E_i \cap \sum_{j \in \{1,...,n\}-\{i\}}E_j=\{0\}$$
- Proof of $\bf\rightarrow$ (by contraposition)
If there's $i$ such that $$E_i \cap \sum_{j \in \{1,...,n\}-\{i\}}E_j\ne\{0\}$$ so let $x\ne0$ in this intersection hence $$x=e_i=\sum_{j\ne i} e_j$$ so $$e_i-\sum_{j\ne i} e_j=0$$ but $$e_i\ne0$$ and this means that $E_1+E_2+...+E_p$ isn't a direct sum.QED.