\begin{aligned} \forall n \in \N \quad &\forall l_{1}, l_{2} \in \text{QB} \\\\ \text{si } & \sum{l_{1}} = \sum{l_{2}} = n \\\\ \text{alors } & l_{1} = l_{2} \end{aligned}
Il faut montrer par récurrence \forall n \in \N :

• L’écriture de 0 est unique
• Si l’écriture de n - 1 est unique, alors l’écriture de n est unique
• Avec : \begin{aligned} \text{Soient } & \sum{l_{1}} = \sum{l_{2}} = n \\\\ \text{donc } & \sum{(\text{prev } l_{1})} = \sum{(\text{prev } l_{2})} = n - 1 \\\\ \text{donc } & \text{prev } l_{1} = \text{prev } l_{2} \\\\ \text{or } & \text{prev est \textbf{injective}} \\\\ \text{donc } & l_{1} = l_{2} \end{aligned}
  • \RA prev est injective.

Supposons : \text{prev } l_{1} = \text{prev } l_{2} \text{Mq} l_{1} = l_{2}
Cas :

• a. l_{1} = 1 :: t_{1}, l_{2} = 1 :: t_{2}
• b. l_{1} = h_{1} :: t_{1}, l_{2} = h_{2} :: t_{2} \quad h_{1} > 1, h2 > 1
• c. l_{1} = 1 :: t_{1}, l_{2} = h_{2} :: t_{2} \quad h_{1} > 1
• (d. l_{1} = 1 :: t_{1}, l_{2} = 1 :: t_{2} \quad h_{1} > 1) synonyme de c.
  Cas a :
  \left. \begin{array}{ll} \text{prev } l_{1} = t_{1} \\ \text{prev } l_{1} = t_{1} \end{array} \right \} \, \text{prev } l_{1} = \text{prev } l_{2} \text{ donc } t_{1} = t_{2} \text{ et } l_{1} = l_{2}
  Cas b :
  \left. \begin{array}{ll} \text{prev } l_{1} = \lfloor \frac{h_{1}}{2} \rfloor :: \lfloor \frac{h_{1}}{2} \rfloor :: t_{1} \\ \text{prev } l_{2} = \lfloor \frac{h_{2}}{2} \rfloor :: \lfloor \frac{h_{2}}{2} \rfloor :: t_{2} \end{array} \right \} \begin{array}{ll} \lfloor \frac{h_{1}}{2} \rfloor = \lfloor \frac{h_{2}}{2} \rfloor \, h_{1} \text{ et } h_{2} \text{ impairs } \RA h_{1} = h_{2} \\ t_{1} = t_{2} \end{array}
  Cas c :
  \begin{aligned} \text{prev } l_{1} = t_{1} \\\\ \text{prev } l_{2} = \lfloor \frac{h_{2}}{2} \rfloor :: \lfloor \frac{h_{2}}{2} \rfloor :: t_{2} \end{aligned}
  \text{prev } l_{1} n’a pas de doublon
  \text{prev } l_{2} a un doublon
  \RA contradiction (\text{prev } l_{1} = \text{prev } l_{2})