The square knot does not have Property 2R

Let $Q\subset S^3$ be the square knot, and let Property $2R$ mean that whenever $Q$ is a component of a $2$-component link whose integral surgery yields

$$\#_{2}(S^1\times S^2),$$

a sequence of handle slides converts the link into a $0$-framed unlink. Square-knot Property 2R conjecture. The square knot does not have Property $2R$. This proposed counterexample is motivated by a family of links whose surgery yields $\#_2(S^1\times S^2)$ and by the apparent difficulty of trivializing the associated presentations; the source presents it as a question motivated by evidence, not as an established result.