diff --git a/logique-programmable/exercices/main.tex b/logique-programmable/exercices/main.tex
index 84022ce..7694b41 100644
--- a/logique-programmable/exercices/main.tex
+++ b/logique-programmable/exercices/main.tex
@@ -5,21 +5,114 @@
 \date{Dernière compilation~: \today{} à \currenttime}
 
 \usepackage{../../cours}
+\usepackage{enumitem}
 
 \begin{document}
 
 \maketitle
 
 \section{Simplification des fonctions logiques}
+
     \subsection{Exercice 1}
+
+        \begin{multicols}{4}
+
+            \begin{enumerate}[label=\alph*)]
+
+                \item $a \cdot b \cdot c \cdot d = 1$
+
+                \item $a + b + c + d = 0$
+
+                \item $a + b + c + d = 1$
+
+                \item $a \cdot b \cdot c \cdot d = 0$
+
+            \end{enumerate}
+
+        \end{multicols}
+
     \subsection{Exercice 2}
+
+        \begin{enumerate}[label=\alph*)]
+
+            \item \begin{align*}
+                    F(a, b, c) &= a\bar{b} + abc + a\bar{c} \\
+                    F(0,1,1) &= 0 \\
+                    F(1,1,0) &= 1 \\
+                    F(1,0,0) &= 1 \\
+                \end{align*}
+
+            \item \begin{equation*}
+                    F(a, b) = \bar{a}\bar{b} + b = \bar{a} + b
+                \end{equation*}
+
+                \begin{center}
+                \begin{tabular}{cc|c|c|c|c}
+                    \toprule
+                    $a$ & $b$ & $\bar{a}\bar{b}$ & $\bar{a}\bar{b} + b$ & $\bar{a}$ & $\bar{a} + b$ \\
+                    \midrule
+                    0 & 0 & 1 & 1 & 1 & 1 \\
+                    0 & 1 & 0 & 1 & 0 & 1 \\
+                    1 & 0 & 0 & 0 & 0 & 0 \\
+                    1 & 1 & 0 & 1 & 0 & 1 \\
+                    \midrule
+                      &   &   & $\uparrow$ & = & $\uparrow$ \\
+                    \bottomrule
+                \end{tabular}
+                \end{center}
+
+            \item \begin{align*}
+                    F(a, b, c) &= a\bar{b} + ab + ac \\
+                               &= a (\bar{b} + b + c) \\
+                               &= a (1 + c) \\
+                               &= a \cdot 1 = a
+                \end{align*}
+
+            \item \begin{align*}
+                    F(a, b, c) &= \bar{a}\bar{b}c + a\bar{b}\bar{c} + a\bar{b}c + ab\bar{c} + abc \\
+                               &= \bar{a}\bar{b}c + a\bar{b}\bar{c} + ab\bar{c} + ac(b + \bar{b}) \\
+                               &= \bar{a}\bar{b}c + a\bar{b}\bar{c} + ab\bar{c} + ac \\
+                               &= \bar{a}\bar{b}c + a\bar{c} + ac \\
+                               &= \bar{a}\bar{b}c + a \\
+                \end{align*}
+
+        \end{enumerate}
+
     \subsection{Exercice 3}
+
+        \begin{enumerate}[label=\alph*)]
+
+            \item \begin{align*}
+                    F_1 &= (a + b) \cdot (\bar{a} + \bar{b}) \\
+                    \overline{F_1} &= \overline{(a + b) \cdot (\bar{a} + \bar{b})} \\
+                                   &= \overline{(a + b)} + \overline{(\bar{a} + \bar{b})} \\
+                                   &= \bar{a}\bar{b} + ab \\
+                                   &= \overline{a \oplus b} \\
+                                   &= \bar{a} \oplus b = a \oplus \bar{b} \\
+                \end{align*}
+
+            \item \begin{align*}
+                    F_2 &= a (c + d) + (\bar{a} + c) \cdot (\bar{b} + c + d) \\
+                    \overline{F_2} &= \overline{a (c + d) + (\bar{a} + c) \cdot (\bar{b} + c + d)} \\
+                                   &= \overline{a (c + d)} \cdot \overline{(\bar{a} + c)} + \overline{(\bar{b} + c + d)} \\
+                                   &= \overline{ac + ad} \cdot a\bar{c} + b\bar{c}\bar{d} \\
+                                   &= \overline{ac} \cdot \overline{ad} \cdot a\bar{c} + b\bar{c}\bar{d} \\
+                                   &= (\bar{a} + \bar{c}) \cdot (\bar{a} + \bar{d}) \cdot a\bar{c} + b\bar{c}\bar{d} \\
+                                   % TODO: finish
+                \end{align*}
+
+        \end{enumerate}
+
     \subsection{Exercice 4}
+
     \subsection{Exercice 5}
+
     \subsection{Exercice 6}
+
     \subsection{Exercice 7}
 
 \section{Circuits de logique combinatoire}
+
     \subsection{Exercice 1}
 
         \includegraphics[width=0.6\linewidth]{./img/2.1.png}
@@ -72,22 +165,22 @@
 
         a/ Donner la table de vérité de la fonction $H$.
 
-        \begin{tabularx}{0.7\linewidth}{X|Y|YYY|Y}
+        \begin{tabularx}{0.7\linewidth}{X|YYY|Y|Y}
             \toprule
-                   & D & A & B & C & H \\
+                   & A & B & C & D & H \\
             \midrule
             0      & 0 & 0 & 0 & 0 & 0 \\
             1      & 0 & 0 & 0 & 1 & 0 \\
             2      & 0 & 0 & 1 & 0 & 0 \\
-            3      & 0 & 0 & 1 & 1 & 0 \\
+            3      & 0 & 0 & 1 & 1 & 1 \\
             4      & 0 & 1 & 0 & 0 & 0 \\
-            5      & 0 & 1 & 0 & 1 & 0 \\
+            5      & 0 & 1 & 0 & 1 & 1 \\
             6      & 0 & 1 & 1 & 0 & 1 \\
             7      & 0 & 1 & 1 & 1 & 0 \\
             \midrule
             8 (0)  & 1 & 0 & 0 & 0 & 0 \\
             9 (1)  & 1 & 0 & 0 & 1 & 0 \\
-            10 (2) & 1 & 0 & 1 & 0 & 1 \\
+            10 (2) & 1 & 0 & 1 & 0 & 0 \\
             11 (3) & 1 & 0 & 1 & 1 & 1 \\
             12 (4) & 1 & 1 & 0 & 0 & 0 \\
             13 (5) & 1 & 1 & 0 & 1 & 1 \\
@@ -98,9 +191,10 @@
 
         b/ Exprimer $H(A,B,C,D)$ sous la forme disjonctive.
 
-        \begin{equation*}
-            H(A,B,C,D) = AB\bar{C}\bar{D} + \bar{A}B\bar{C}D + \bar{A}BCD + A\bar{B}CD + AB\bar{C}D
-        \end{equation*}
+        \begin{align*}
+            H(A,B,C,D) &= \bar{A}\bar{B}CD + \bar{A}B\bar{C}D + \bar{A}BC\bar{D} + A\bar{B}CD + AB\bar{C}D + ABC\bar{D} \\
+                       &= \sum(3, 5, 6, 11, 13, 14)
+        \end{align*}
 
     \subsection{Exercice 4}