The proof by Gordon and Luecke (1989) is a tour de force of combinatorial 3-manifold topology.

Step 1: Reduction to surgery. Suppose $h: S^3\setminus K_1 \xrightarrow{\cong} S^3\setminus K_2$. The complement $S^3\setminus K_i$ determines the knot exterior $E_{K_i}$. The homeomorphism $h$ sends the boundary torus $\partial E_{K_1}$ to $\partial E_{K_2}$, but may map the meridian $\mu_1$ to a different slope $p\mu_2 + q\lambda_2$. If $q = 0$ (meridian goes to meridian), the homeomorphism extends to give $K_1 \cong K_2$. If $q \neq 0$, then $S^3_{p/q}(K_2) \cong S^3$, meaning non-trivial surgery on $K_2$ yields $S^3$.

Step 2: Show non-trivial surgery cannot yield $S^3$. Suppose $S^3_{p/q}(K) \cong S^3$ for some nontrivial knot $K$ and $p/q \neq \infty$. Consider the dual knot $K' \subset S^3_{p/q}(K) = S^3$ (the core of the surgery solid torus).

Step 3: Intersection graph analysis. Take a minimal-genus Seifert surface $F$ for $K$ (in $E_K$) and a meridian disk $D$ for the surgery solid torus. After put these surfaces in general position, their intersection consists of arcs and closed curves. The arcs on $F$ form a graph $G_F$ on $F$, and arcs on $D$ form a graph $G_D$ on $D$.

Step 4: Combinatorial argument. Analyzing the combinatorics of $G_F$ and $G_D$ using the parity of intersection numbers and the Scharlemann cycle argument: if the graphs satisfy certain combinatorial properties (related to the signs of intersection points), one derives that either the graphs contain a specific forbidden configuration or the surgery slope is forced.

Step 5: Contradiction. Through an intricate case analysis of the graph combinatorics, Gordon and Luecke show that the existence of these graphs (satisfying all topological constraints) leads to a contradiction unless $K$ is the unknot. The key ingredients are:

• The Scharlemann cycle lemma (certain cycles in intersection graphs force topological triviality).
• The reducing sphere argument (if a Scharlemann cycle exists, one constructs a reducing sphere for the surgered manifold, constraining its topology).
• A final counting argument showing that all possible graph configurations have been exhausted.

Therefore no non-trivial surgery on a nontrivial knot yields $S^3$, completing the proof. $\square$