[(b)]{} If the domain $\dom(F)$ is quasi-regular, then we have the opposite inclusion $$\mbox{\rm qri}\big(\gph(F)\big)\supset\big\{(x, y)\in X\times Y\;\big|\;x\in\mbox{\rm qri}\big(\dom(F)\big),\;y\in\mbox{\rm qri}\big(F(x)\big)\big\}.$$

[**Proof.**]{} To verity the inclusion “$\subset$" in (a), pick $(\ox,\oy)\in\qri(\gph(F))$ and suppose on the contrary that $\ox\notin\qri(\dom(F))$. By Lemma \[proper separation\] we find $v^*\in X^*$ such that $$\la v^*,x\ra\le\la v^*,\ox\ra\;\mbox{ whenever }\;x\in\dom(F)$$ and also have $x_0\in\dom(F)$ for which the strict inequality $$\la v^*,x_0\ra<\la v^*,\ox\ra$$ is satisfied. Then for all $(x,y)\in\gph(F)$ we get $$\la(v^*,0),(x,y)\ra=\la v^*,x\ra\le\la v^*,\ox\ra=\la(v^*,0),(\ox,\oy)\ra,$$ and for each $y_0\in F(x_0)$ arrive at the conditions $$\la(v^*,0),(x_0,y_0)\ra=\la v^*,x_0\ra<\la v^*,\ox\ra=\la (v^*,0),(\ox,\oy)\ra.$$ This shows that the sets $\gph(F)$ and $\{(\ox,\oy)\}$ are properly separated, and hence $(\ox,\oy)\notin\qri\big(\gph(F)\big)$ by Lemma \[proper separation\]. The obtained contradiction tells us that $\ox\in \qri(\dom(F))$.

To proceed further with the proof of (a), let us verify that $\oy\in\mbox{\rm qri}(F(\ox))$. Fix any $y\in F(\ox)$ with $y\ne\oy$, and so $(\ox,y)\in\gph(F)$. The assumed quasi-regularity of $\gph(F)$ gives us $(\tilde{x},\tilde{y})\in\gph(F)$ and $t\in(0,1)$ such that $$(\ox,\oy)=t(\ox,y)+(1-t)(\tilde{x},\tilde{y}).$$ This yields $\tilde{x}=\ox$ and $\oy=ty+(1-t)\tilde{y}$ with $\tilde{y}\in F(\ox)$. It follows therefore that $\oy\in\mbox{\rm iri}(F(\ox))\subset\mbox{\rm qri}(F(\ox))$, which completes the proof of (a).

To verify now assertion (b) under the quasi-regularity of $\dom(F)$, fix $\ox\in\qri(\dom(F))$ and $\oy\in\qri\big(F(\ox)\big)$. Arguing by contradiction, suppose that $(\ox,\oy)\notin\qri(\gph(F))$. Then Lemma \[proper separation\] ensures the existence of $(u^*,v^*)\in 