Let curve $C_1$ be defined by ${\mathbf{R}}_1\left(s\right)$, for which the TNB-frame consists of the vectors $\left\{{\mathbf{T}}_1\,{\mathbf{N}}_1\,{\mathbf{B}}_1\right\}$.

Let curve $C_2$ be defined by ${\mathbf{R}}_2\left(s\right)$, for which the TNB-frame consists of the vectors $\left\{{\mathbf{T}}_2\,{\mathbf{N}}_2\,{\mathbf{B}}_2\right\}$.

By hypothesis, both curves have $\mathrm{\κ}\left(s\right)$ as curvature and $\mathrm{\τ}\left(s\right)$ as torsion.

By translation and rotation, align the curves so that the TNB-frames at $s\=0$ coincide. Define

$Q\={\mathbf{T}}_1·{\mathbf{T}}_2\+{\mathbf{N}}_1·{\mathbf{N}}_2\+{\mathbf{B}}_1·{\mathbf{B}}_2$ 

and by the calculations in Table 2.7.10(a), $Q\prime \left(s\right)\=0$, so $Q\=c$, constant. This constant can be determined by evaluating $Q$ at $s\=0$ where ${\mathbf{T}}_1\={\mathbf{T}}_2\=\mathbf{T}\left(0\right)\,{\mathbf{N}}_1\={\mathbf{N}}_2\=\mathbf{N}\left(0\right)\,{\mathbf{B}}_1\={\mathbf{B}}_2\=\mathbf{B}\left(0\right)$. At this point (since $\mathbf{T}\left(0\right)\,\mathbf{N}\left(0\right)\,\mathbf{B}\left(0\right)$, are all unit vectors)

$Q\left(0\right)\=\mathbf{T}\left(0\right)·\mathbf{T}\left(0\right)\+\mathbf{N}\left(0\right)·\mathbf{N}\left(0\right)\+\mathbf{B}\left(0\right)·\mathbf{B}\left(0\right)\=1\+1\+1\+3$ 

so $c\=3$. It is now possible to conclude that ${\mathbf{T}}_1\={\mathbf{T}}_2\,{\mathbf{N}}_1\={\mathbf{N}}_2\,{\mathbf{B}}_1\={\mathbf{B}}_2$ for $s\ge 0$. The reasoning is as follows.

${\mathbf{T}}_1·{\mathbf{T}}_2\=∥{\mathbf{T}}_1∥$ $∥{\mathbf{T}}_2∥$ $\cos \left({\mathrm{\θ}}_1\right)$ = $1\cdot 1\cdot \cos \left({\mathrm{\θ}}_1\right)$

where ${\mathrm{\θ}}_1$ is the angle between ${\mathbf{T}}_1$ and ${\mathbf{T}}_2$. Applying the same thinking to ${\mathbf{N}}_1·{\mathbf{N}}_2$ and ${\mathbf{B}}_1·{\mathbf{B}}_2$ leads to

$Q\=\cos \left({\mathrm{\θ}}_1\right)\+\cos \left({\mathrm{\θ}}_2\right)\+\cos \left({\mathrm{\θ}}_3\right)\=3$ 

The only way this can be true is if ${\mathrm{\θ}}_1\={\mathrm{\θ}}_2\={\mathrm{\θ}}_3\=0$ so that each cosine has the value 1.

The essential step has been to obtain ${\mathbf{T}}_1\left(s\right)\={\mathbf{T}}_2\left(s\right)$ so that ${\mathbf{R}}_1^{\mathbf{\/}}\left(s\right)\={\mathbf{R}}_2^{\mathbf{\/}}\left(s\right)$ and ${\mathbf{R}}_1\left(s\right)\={\mathbf{R}}_2\left(s\right)\+\mathbf{A}$. But at $s\=0$, ${\mathbf{R}}_1\left(0\right)\={\mathbf{R}}_2\left(0\right)$, so $\mathbf{A}\=\mathbf{0}$ and ${\mathbf{R}}_1\left(s\right)\={\mathbf{R}}_2\left(s\right)$.