$i'^2=a'$ and $\tau'=\operatorname{Int}(i')\circ{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}$. Denoting by $Q^0$ the vector space of pure quaternions in $Q$, we have $$\operatorname{Sym}(A,\tau)=(1\otimes\operatorname{Sym}(Q',\tau'))\oplus(Q^0\otimes i').$$ Hence $x'$ can be written as $x'=x_0+1\otimes\xi_1+\xi_2\otimes i'$ for some $x_0\in F$, some pure quaternion $\xi_1\in
  Q'^0\cap\operatorname{Sym}(Q',\tau')$ and some $\xi_2\in Q^0$. The pure quaternion $\xi_1$ being $\tau'$-symmetric, it anticommutes with $i'$, and changing of quaternionic basis if necessary, we may assume that $\xi_1^2=b'x_1^2$ for some $x_1\in F$. Similarly, we may assume that $\xi_2^2=ax_2^2$ for some $x_2\in F$. Note that we allow $x_i=0$ for $i=0$, $1$, and $2$ since some term might be zero in the decomposition of $x'$. We then have $$\label{x'.eq} 
  \operatorname{Nrp}_\tau(x')=x_0^2-b'x_1^2-aa'x_2^2,$$ and the following lemma finishes the proof:

\[fin.lem\] There exists $z_0$, $z_1$, $y_0$, $y_1$ and $y_2$ such that $\operatorname{Nrp}_\tau(x')(z_0^2-az_1^2)=y_0^2-b'y_1^2+a'b'y_2^2\not=0$.

Indeed, if we let $y=y_0+y_1j'+y_2i'j'$, where $j'$ is a pure quaternion in $Q'$ such that $j'^2=b'$ and $i'j'=-j'i'$, we have $\operatorname{Nrp}_\tau(1\otimes y)=Nrd_{Q'}(y)=y_0^2-b'y_1^2+a'b'y_2^2$ (see [@KMRT (2.11)]). Hence, $$\begin{aligned}
\Delta_\tau(\sigma)& =(\operatorname{Nrp}_\tau(x'))\cup
[(a,b)_F]=(\operatorname{Nrp}_\tau(x')(z_0^2-az_1^2))\cup
[(a,b)_F]\\
& =(\operatorname{Nrd}_{Q'}(y))\cup[A]=(\operatorname{Nrp}_\tau(1\otimes y))\cup[A],
\quad\text{ with }y\in\operatorname{Sym}(Q',\tau').\end{aligned}$$

If ${{\left\langle{1,-b',a'b'}\right\rangle}}$ is isotropic, then it is universal, hence represents $\operatorname{Nrp}_\tau(x')$ and we are done. Otherwise, denote $\mu=\operatorname{Nrp}_\tau(x')$; by (\[x’.eq\]), we have $(\mu)\cup[(aa',b')_F]=0$, and combining with , we get $(\mu)\cup[(a,b')]=0$. In terms of quadratic forms, this means that the $3$-fold Pfister form $$\label{pform.eq}
  