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The Locality property of SLE says the following. Let $D\subset \mathbb{H}$ be simply connected and agree with $\mathbb{H}$ in a neighborhood of $0$, and let $\psi :D\to \mathbb{H}$ be conformal with $\psi (0)=0$. The image under $\psi$ of an ${\mathrm{SLE}}_6$ in $\mathbb{H}$, stopped when it first leaves $D$, has the law of an ${\mathrm{SLE}}_6$ in $\mathbb{H}$, up to the corresponding stopping time and a change of half-plane-capacity parameterization.

Explore the two stopped curves in the opposite order. Begin with an ${\mathrm{SLE}}_6$ from $1$ to $\infty$, stopped on leaving $B(1,1)$, and then, in its unbounded complementary component, draw an ${\mathrm{SLE}}_6$ from $-1$ to $\infty$, stopped on leaving $B(-1,1)$. Before their respective stopping times, each curve is separated from the neighborhood in which the other hull changes the domain. The Locality property of SLE therefore says that mapping out the other stopped hull does not change either stopped marginal law.

The two exploration orders consequently define the same joint law for the pair of stopped hulls. Disintegrating this joint law with respect to the second curve shows that, conditional on ${\gamma }_2{|}_{[0,{\tau }_2]}$, the first curve is an ${\mathrm{SLE}}_6$ in the unbounded component offrom $-1$ to $\infty$, stopped when it leaves $B(-1,1)$, as required.

Choose a Möbius transformation of $\mathbb{H}$ that fixes $0$ and exchanges $x$ with $\infty$. By Conformal invariance of SLE, it transforms an ${\mathrm{SLE}}_6$ from $0$ to $x$ into one from $0$ to $\infty$. Until $x$ is disconnected from infinity, the discrepancy between the two target domains lies beyond the component visible from the growing tip. The Locality property of SLE therefore makes the two initial curve laws identical up to that disconnection time. Hence an ${\mathrm{SLE}}_6$ from $0$ to $x$, stopped at ${\tau }_x$, has the law of an ${\mathrm{SLE}}_6$ from $0$ to $\infty$ stopped when it disconnects $x$ from infinity.