in [@KMRT (2.9)]. Since this relative discriminant classifies symplectic involutions on $A$ up to conjugation, it suffices to show:

\[deg4symp.lem\] There exists an invertible element $y\in \operatorname{Sym}(Q',\tau')$ such that $$\Delta_\tau(\sigma)=(\operatorname{Nrp}_\tau(1\otimes y))\cup [A].$$

The involution $\sigma$ can be written as $\sigma=\operatorname{Int}(x)\circ \tau$ for some $x\in\operatorname{Sym}(A,\sigma)$, hence we already have $$\label{disc.eq} 
  \Delta_\tau(\sigma)=(\operatorname{Nrp}_\tau(x))\cup [A],$$ and we want to prove we can substitute for $x$ some element $1\otimes y$ with $y\in\operatorname{Sym}(Q',\tau')$. Since both $\sigma$ and $\tau$ become hyperbolic over $F_Q$, the relative discriminant $\Delta_\tau(\sigma)$ is killed by $F_Q$, hence $(\operatorname{Nrp}_\tau(x))\cup [Q']=(\lambda)\cup[Q]$ for some $\lambda\in
  F^\times$. By the common slot lemma [@A Lemma 1.7], we may even assume $$(\operatorname{Nrp}_\tau(x))\cup [Q']=(\lambda)\cup[Q']=(\lambda)\cup[Q],$$ from which we deduce $$\label{la.eq}
  (\lambda)\cup([Q]+[Q'])=0,\quad\text{and}\quad(\lambda\operatorname{Nrp}_\tau(x))\cup
  [Q']=0.$$ Hence the relative discriminant of $\sigma$ is $$\Delta_\tau(\sigma)=(\operatorname{Nrp}_\tau(x))\cup([Q]+[Q'])=(\lambda
  \operatorname{Nrp}_\tau(x))\cup([Q]+[Q'])=(\lambda \operatorname{Nrp}_\tau(x))\cup[Q].$$ Moreover, the quadratic space $(\operatorname{Sym}(A,\sigma),\operatorname{Nrp}_\tau)$ is an Albert quadratic space for the biquaternion algebra $A$ by [@KMRT (16.8)]. Hence, its Clifford invariant is $e_2(\operatorname{Nrp}_\tau)=[A]=[Q]+[Q']$. We deduce from (\[la.eq\]) that the quadratic form ${{\langle\!\langle{\lambda}\rangle\!\rangle}}\otimes\operatorname{Nrp}_\tau$ has trivial Arason invariant $e_3$, and hence is hyperbolic by the Arason–Pfister Hauptsatz. So there exists an element $x'\in\operatorname{Sym}(A,\sigma)$ such that $\lambda\operatorname{Nrp}_\tau(x)=\operatorname{Nrp}_\tau(x')$, and we have proven $$\label{Q'.eq} 
  \Delta_\tau(\sigma)=(\operatorname{Nrp}_\tau(x'))\cup [Q]
  \quad\text{and}\quad(\operatorname{Nrp}_\tau(x'))\cup [Q']=0$$ for some $x'\in\operatorname{Sym}(A,\sigma)$.

Let us now pick a pure quaternion $i'\in Q'$ such that 