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+\subsection*{Exercise 03}
+
+\subsubsection*{1.15 Proposition (a)}
+\begin{flalign*}
+                        xy &= xz                                      &\\
+\implies \frac{1}{x}{x}{y} &= \frac{1}{x}{x}{z} \text{ (by 1.12(M5))} &\\
+\implies       1 \cdot {y} &= 1 \cdot {z}       \text{ (by 1.12(M4))} &\\
+\implies               {y} &= {z} 
+\end{flalign*}
+
+\subsubsection*{1.15 Proposition (b)}
+\begin{flalign*}
+         x \cdot 1 &= x \text{ (by 1.12(M4))}               &\\
+\implies x \cdot y &= x \text{ (use $y$ to substitute $1$)} &\\
+\implies         y &= 1
+\end{flalign*}
+
+\subsubsection*{1.15 Proposition (c)}
+\begin{flalign*}
+         x \cdot \frac{1}{x} &= 1 \text{ (by 1.12(M5))}                         &\\
+\implies           x \cdot y &= 1 \text{ (use $y$ to substitute $\frac{1}{x}$)} &\\
+\implies                   y &= \frac{1}{x}
+\end{flalign*}
+
+\subsubsection*{1.15 Proposition (d)}
+\begin{flalign*}
+\frac{1}{\frac{1}{x}} &= \frac{1 \cdot x}{\frac{1}{x} \cdot x} 
+                       = \frac{x}{1} 
+                       = x \text{ (by 1.12(M5))} &
+\end{flalign*}
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