SLE duality conjecture for boundary traces and conditional laws

Let $\kappa>4$ and set $\kappa'=16/\kappa$. In the configuration $(\mathbb{H},x,y,z,\infty)$ with $x<y<z$, let $(K_t)$ be the Loewner chain obtained by running an $\operatorname{SLE}_\kappa(\kappa/2-4,-\kappa/2)$ started from $(x,y,z)$ toward $\infty$ until it hits $z$, and then continuing as an $\operatorname{SLE}_\kappa(\kappa-4)$ in the remaining domain. Let $\delta$ be the Loewner trace of the right boundary of $K_\infty$. In a second configuration $(\mathbb{H},x',y',z',\infty)$ with $x'<y'<z'$, let $\gamma'$ be the trace of the chordal $\operatorname{SLE}_{\kappa'}(-\kappa'/2,\kappa'-2)$ started from $(z',x',y')$ toward $\infty$. SLE duality conjecture. The following statements hold: (i) the law of $\delta$ is that of $\gamma'$, with $x=x'$, $y=y'$, and $z=z'$. (ii) Conditionally on $\delta_{[0,u]}$, the law of $(\phi_{\delta_{[0,u]}}(K_t))_{t\geq 0}$ is, up to a time-change, that of a copy of $(K_t)$ started from $(x',y',z')=\phi_{\delta_{[0,u]}}(x,y,\delta_u)$. This conjecture identifies the right boundary of a non-simple $\operatorname{SLE}_\kappa$ construction with the corresponding simple dual $\operatorname{SLE}_{16/\kappa}$ and predicts the associated conditional conformal Markov law.