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+\subsection*{Exercise 03 (Gapry)}
+
+\subsubsection*{1.15 Proposition (a)}
+As we know $x \neq 0$
+\begin{flalign*}
+                 xy &= xz                                       &\\
+  \frac{1}{x}{x}{y} &= \frac{1}{x}{x}{z} \text{ (by 1.12 (M5))} &\\
+  {x}\frac{1}{x}{y} &= {x}\frac{1}{x}{z} \text{ (by 1.12 (M2))} &\\
+        1 \cdot {y} &= 1 \cdot {z}       \text{ (by 1.12 (M5))} &\\
+        {y} \cdot 1 &= {z} \cdot 1       \text{ (by 1.12 (M2))} &\\
+                {y} &= {z}               \text{ (by 1.12 (M4))}
+\end{flalign*}
+
+\subsubsection*{1.15 Proposition (b)}
+As we know $x \neq 0$
+\begin{flalign*}
+                 xy &= x                                     &\\
+  \frac{1}{x}{x}{y} &= \frac{1}{x}{x} \text{ (by 1.12 (M5))} &\\
+  {x}\frac{1}{x}{y} &= {x}\frac{1}{x} \text{ (by 1.12 (M2))} &\\
+        1 \cdot {y} &= 1              \text{ (by 1.12 (M5))} &\\
+        {y} \cdot 1 &= 1              \text{ (by 1.12 (M2))} &\\
+                {y} &= 1              \text{ (by 1.12 (M4))}
+\end{flalign*}
+
+\subsubsection*{1.15 Proposition (c)}
+As we know $x \neq 0$
+\begin{flalign*}
+                 xy &= 1                                          &\\
+  \frac{1}{x}{x}{y} &= \frac{1}{x} \cdot 1 \text{ (by 1.12 (M5))} &\\
+  {x}\frac{1}{x}{y} &= \frac{1}{x} \cdot 1 \text{ (by 1.12 (M2))} &\\
+        1 \cdot {y} &= \frac{1}{x} \cdot 1 \text{ (by 1.12 (M5))} &\\
+        {y} \cdot 1 &= \frac{1}{x} \cdot 1 \text{ (by 1.12 (M2))} &\\
+                {y} &= \frac{1}{x}         \text{ (by 1.12 (M4))}
+\end{flalign*}
+
+\subsubsection*{1.15 Proposition (d)}
+We know that $x \in F$ and $x \neq 0$. According to 1.12, there must exist an inverse element $\frac{1}{x} \in F$ such that $x \cdot \frac{1}{x} = 1$. Similarly, since $\frac{1}{x} \in F$, there must exist an inverse element $\frac{1}{\frac{1}{x}} \in F$ such that $\frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = 1$.
+\begin{flalign*}
+          \frac{1}{x} \cdot \frac{1}{\frac{1}{x}} &= 1         \text{               } &\\
+  x \cdot \frac{1}{x} \cdot \frac{1}{\frac{1}{x}} &= x \cdot 1 \text{               } &\\
+                    1 \cdot \frac{1}{\frac{1}{x}} &= x \cdot 1 \text{ (by 1.12 (M5))} &\\
+                    \frac{1}{\frac{1}{x}} \cdot 1 &= x \cdot 1 \text{ (by 1.12 (M2))} &\\
+                            \frac{1}{\frac{1}{x}} &= x         \text{ (by 1.12 (M4))}
+\end{flalign*}