A non-reduced positive-definite form admits an equivalent positive-definite form with smaller reduction measure

Since $f$ is positive definite, [F3] gives $a>0$ and $\mathrm{\Delta }:=b^2-4ac<0$.

If $c<a$ or if $c=a$ and $b<0$, let $g=f\mid \left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=(c,-b,a)$. This matrix lies in ${\mathrm{S}\mathrm{L}}_2(\mathbb{Z})$, so $g$ is properly equivalent to $f$; by [F2] and [F3] it is again positive definite because its leading coefficient is $c>0$ and its discriminant is still $\mathrm{\Delta }<0$. Its measure satisfies $\mu (g)=3c+a+\epsilon (g)<3a+c+\epsilon (f)=\mu (f)$ because $c<a$ gives a drop of at least $1$, and when $c=a$ with $b<0$ the boundary defect disappears so $\epsilon (g)=0<1=\epsilon (f)$.

Assume now that step 2.1 does not apply. Then $a\le c$ and, because $f$ is not reduced, one must have $b\mathrm{\notin }(-a,a]$. Choose the unique integer $k$ for which $b^{\mathrm{\prime }}:=b+2ak$ lies in $(-a,a]$, and let $g=f\mid \left(\begin{array}{cc}1& k\\ 0& 1\end{array}\right)=(a,b^{\mathrm{\prime }},c^{\mathrm{\prime }})$, where $c^{\mathrm{\prime }}=a{k}^2+bk+c$.

The new form $g$ is properly equivalent to $f$, so it has the same discriminant $\mathrm{\Delta }<0$ by [F2]; its leading coefficient is still $a>0$, so [F3] makes it positive definite. Also $4a(c^{\mathrm{\prime }}-c)=b^{\mathrm{\prime }2}-b^2$, hence $c^{\mathrm{\prime }}\le c$, with strict inequality when $\mathrm{\mid }{b}^{\mathrm{\prime }}\mathrm{\mid }<\mathrm{\mid }b\mathrm{\mid }$.

If $\mathrm{\mid }b\mathrm{\mid }>a$, then $\mathrm{\mid }{b}^{\mathrm{\prime }}\mathrm{\mid }<\mathrm{\mid }b\mathrm{\mid }$, so step 4.1 gives $c^{\mathrm{\prime }}<c$. In this case $\epsilon (f)=0$: if $a<c$ there is no boundary defect, while if $a=c$ then step 2.1 was excluded and therefore $b>0$.

If $\mathrm{\mid }b\mathrm{\mid }=a$, then step 2.1 is excluded, so $a<c$ and the only way $b\mathrm{\notin }(-a,a]$ can occur is $b=-a$. Then the chosen residue is $b^{\mathrm{\prime }}=a$, so $c^{\mathrm{\prime }}=c$ and the boundary defect disappears: $\epsilon (g)=0<1=\epsilon (f)$. Hence again $\mu (g)<\mu (f)$.

If $\mathrm{\mid }b\mathrm{\mid }>a$ and $c^{\mathrm{\prime }}>a$, then $-a<b^{\mathrm{\prime }}\le a<c^{\mathrm{\prime }}$, so $g$ already satisfies the reduced-form boundary sign conditions and $\epsilon (g)=0$. Therefore $\mu (g)=3a+c^{\mathrm{\prime }}<3a+c=\mu (f)\mathrm{.}$

If $\mathrm{\mid }b\mathrm{\mid }>a$ and $c^{\mathrm{\prime }}=a$, then step 5.1 gives $c>a$ because otherwise $c=a$ and $c^{\mathrm{\prime }}=c$ would force $\mathrm{\mid }{b}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }b\mathrm{\mid }$, contradicting $\mathrm{\mid }{b}^{\mathrm{\prime }}\mathrm{\mid }<\mathrm{\mid }b\mathrm{\mid }$. If $b^{\mathrm{\prime }}\ge 0$, then $\epsilon (g)=0$ and $\mu (g)=4a<3a+c=\mu (f)\mathrm{.}$ If instead $b^{\mathrm{\prime }}<0$, let $h=g\mid \left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=(a,-b^{\mathrm{\prime }},a)$. Then $h$ is properly equivalent to $f$, still positive definite, and $\epsilon (h)=0$, so again $\mu (h)=4a<3a+c=\mu (f)\mathrm{.}$

If $\mathrm{\mid }b\mathrm{\mid }>a$ and $c^{\mathrm{\prime }}<a$, let $h=g\mid \left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=(c^{\mathrm{\prime }},-b^{\mathrm{\prime }},a)$. Then $h$ is properly equivalent to $f$ and positive definite. Since $c^{\mathrm{\prime }}\le a-1$ and $\epsilon (h)\le 1$, $\mu (h)=3c^{\mathrm{\prime }}+a+\epsilon (h)\le 3(a-1)+a+1=4a-2<4a\le 3a+c=\mu (f)\mathrm{.}$

Step 2.1 covers the case $c<a$ or $c=a$ with $b<0$; steps 6.1, 6.2, and 6.3 cover the remaining case $\mathrm{\mid }b\mathrm{\mid }>a$; and step 5.2 covers the case $\mathrm{\mid }b\mathrm{\mid }=a$ with $b<0$. Therefore every non-reduced positive-definite form is properly equivalent to a positive-definite form of smaller reduction measure.