Step 1: Decompose the complement. Project $K$ to the plane $\mathbb{R}^2 \subset \mathbb{R}^3$ to get the diagram $D$. The complement $S^3\setminus K$ is decomposed into pieces using the projection. Think of $S^3 = \mathbb{R}^3 \cup \{\infty\}$ and the knot as lying near the plane $z = 0$, with over-crossings slightly above and under-crossings slightly below.

Step 2: Generators. For each arc $x_i$ of the diagram, define a loop (based at a point far above the diagram plane) that descends, encircles the arc $x_i$ once in the positive direction (right-hand rule with the knot orientation), and returns to the basepoint. These loops represent elements of $\pi_1(S^3\setminus K)$, and we denote them also by $x_i$.

Step 3: The complement as a union. Decompose $S^3\setminus K$ into $n$ pieces $U_1,\ldots,U_n$, one for each arc: $U_k$ is the complement of $K$ in a "slab" region around arc $x_k$, plus the upper half-space. Each $U_k$ is simply connected (contractible to a point above the diagram). The intersections $U_k \cap U_{k+1}$ at crossings have fundamental group $\mathbb{Z}$ (generated by the loop around the over-strand).

Step 4: Apply Seifert-van Kampen. Build up the complement by successively attaching pieces. Start with $V_1 = U_1$. At each step, $V_k = V_{k-1} \cup U_k$ and we apply the Seifert-van Kampen theorem.

Away from crossings, attaching an arc introduces a new generator $x_k$ (for the next arc) and the relation $x_k = x_{k-1}$ (since adjacent arcs on the same strand are homotopic). At a crossing $c_k$ where arc $x_i$ crosses over and the under-strand transitions from $x_j$ to $x_\ell$:

Step 5: Crossing relations. Consider a small ball $B$ around the crossing. In $B\setminus K$, the fundamental group is generated by meridians of the four arcs meeting at the crossing, with one relation. The loop $x_\ell$ (meridian of the outgoing under-arc) is homotopic to a loop that: goes along $x_i$ (over-arc meridian), then $x_j$ (incoming under-arc meridian), then back along $x_i^{-1}$. This gives:

For a positive crossing: $x_\ell = x_i x_j x_i^{-1}$.

For a negative crossing: $x_\ell = x_i^{-1}x_j x_i$.

Step 6: Redundancy. The $n$ crossings give $n$ relations, but one is redundant. This follows from the fact that the Euler characteristic of the 2-complex associated to the presentation must match: the boundary of a regular neighborhood of the diagram gives $\chi = 0$ (it is a torus), so $n - n + 1 = 0$ implies one relation is dependent. Alternatively, the product of all relations (suitably conjugated and ordered around the diagram) is trivially the identity, so any one relation follows from the others.

Therefore $\pi_1(S^3\setminus K) = \langle x_1,\ldots,x_n \mid r_1,\ldots,r_{n-1}\rangle$. $\square$