Every such truth assignment gives a value of 1 (representing true) or 0 (representing false) to each variable. We can therefore think of a truth assignment $\tau $ as determining a four-bit integer ${x}_{\tau}$ depending on the values given to ${x}_{0},{x}_{1},{x}_{2}$ and ${x}_{3}$, and a four-bit integer ${x}_{\tau}$ depending on the values given to ${y}_{0},{y}_{1},{y}_{2}$ and ${y}_{3}$.

More precisely, with $\tau ({x}_{i})$ being the truth value assigned to ${x}_{i}$, we can define the integers ${x}_{\tau}={2}^{3}\tau ({x}_{3})+{2}^{2}\tau ({x}_{2})+{2}^{1}\tau ({x}_{1})+\tau ({x}_{0})$ and ${y}_{\tau}={2}^{3}\tau ({y}_{3})+{2}^{2}\tau ({y}_{2})+{2}^{1}\tau ({y}_{1})+\tau ({y}_{0})$

Write a formula that is satisfied by exactly those truth assignments $\tau $ for which ${x}_{\tau}>{y}_{\tau}$. Your formula may use any of the Boolean connectives introduced so far. Explain how you obtained your formula, and justify its correctness

Am I right in assuming that I would need to construct a formula for every single possibility where x is the greater bit?