diff --git a/latex/content.tex b/latex/content.tex
index 433621d..2003c8f 100644
--- a/latex/content.tex
+++ b/latex/content.tex
@@ -145,17 +145,19 @@ Let $ A_k = \{ U_1 + U_2 + \ldots + U_{k-1} < 1 \}$. Hence,
 	 & = P(A_{k-1}) - P(A_k) \qquad \text{ (as $ A_k \subset A_{k-1} $)} \\
 \end{align*}
 
-We will try to show that $ \forall k \geq 2 $, $ P(A_k) = \frac{1}{k!} $.
-To do so we will use induction to demonstrate that
+We will try to show that $ \forall k \geq 2 $, $ P(A_k) = \frac{1}{k!} $. To do
+so, we will use induction to prove the following proposition \eqref{eq:induction},
+$ \forall k \geq 2 $:
 
 \begin{equation}
 	\label{eq:induction}
-	P(U_1 + U_2 + \ldots + U_{k-1} < a) = \frac{a^k}{k!} \qquad \forall a \in [0, 1], \forall k \geq 2
+	\tag{$ \mathcal{H}_k $}
+	P(U_1 + U_2 + \ldots + U_{k-1} < a) = \frac{a^k}{k!} \qquad \forall a \in [0, 1],
 \end{equation}
 
 Let us denote $ S_k = U_1 + U_2 + \ldots + U_{k-1} \qquad \forall k \geq 2 $.
 
-\paragraph{Base cases} $ k = 2 $ : $ P(U_1 < a) = a $
+\paragraph{Base cases} $ k = 2 $ : $ P(U_1 < a) = a \neq \frac{a^2}{2}$ supposedly proving $ (\mathcal{H}_2) $.
 
 $ k = 2 $ : \[ P(U_1 + U_2 < a) = \iint_{\cal{D}} f_{U_1, U_2}(x, y) \cdot (x + y) dxdy \]
 
@@ -184,6 +186,43 @@ Hence,
 \end{align*}
 
 
+\paragraph{Induction step} For a fixed $ k > 2 $, we assume that $
+	(\mathcal{H}_{k-1}) $ is true. We will try to prove $ (\mathcal{H}_{k}) $.
+
+\[
+	P(S_{k-1} + U_{k-1} < a)
+	= \iint_{\cal{D}} f_{S_{k-1}, U_{k-1}}(x, y) \cdot (x + y) dxdy                 \\
+\]
+where $ \mathcal{D} = \{ (x, y) \in [0, 1]^2 \mid x + y < a \} $.
+As $ S_{k-1} $ and $ U_{k-1} $ are independent,
+\[
+	P(S_{k-1} + U_{k-1} < a)
+	= \iint_{\cal{D}} f_{S_{k-1}}(x) \cdot f_{U_{k-1}}(y) \cdot (x + y) dxdy \qquad \\
+\]
+
+$ (\mathcal{H}_{k-1}) $ gives us that $ \forall x \in [0, 1] $,
+$ F_{S_{k-1}}(x) = P(S_{k-1} < x) = \frac{x^{k-1}}{(k-1)!} $.
+
+By differentiating, we get that $ \forall x \in [0, 1] $,
+
+\[
+	f_{S_{k-1}}(x) = F'_{S_{k-1}}(x) = \frac{x^{k-2}}{(k-2)!}
+\]
+
+Furthermore, $ U_{k-1} $ is uniformly distributed on $ [0, 1] $, so
+$ f_{U_{k-1}}(y) = 1 $.
+
+\begin{align*}
+	\text{Hence, }
+	P(S_{k-1} + U_{k-1} < a)
+	 & =
+	 & = \frac{a^{k}}{k!}
+\end{align*}
+
+
+
+
+
 \section{Complexity and implementation optimization}
 
 The NFBP algorithm has a linear complexity $ O(n) $, as we only need to iterate