1.15 $\forall a, \: b \in \mathbb{F} \quad (-a)(-b)=ab$

Proof:
\begin{align*}
(-a)(-b) &= (-a)(-b)+0 &&\text{(by 1.3)} \\
&=(-a)(-b) + ((-(ab))+ab) &&\text{(by 1.4)} \\
&=((-a)(-b)+(-(ab)))+ab &&\text{(by 1.2)} \\
&=((-a)(-b)+(-1(ab)))+ab &&\text{(by 1.14)} \\
&=((-a)(-b)+((-1a)b))+ab &&\text{(by 1.6)} \\
&=((-a)(-b)+(-a)b)+ab &&\text{(by 1.14)} \\
&=(-a)(-b+b) + ab &&\text{(by 1.9)} \\
&=(-a)0+ab &&\text{(by 1.4)} \\
&=0+ab &&\text{(by 1.13)} \\
&=ab &&\text{(by 1.10)}
\end{align*}