Exercise 1, § 4

This solution was written by nemotron-3-ultra-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

Let $A$ and $B$ be relations in $\mathscr{T}$, and let $x$ be a letter which does not appear in $A$. By definition of implication, the relation $(A \Rightarrow B)$ is the assembly $(\neg A) \vee B$. Since $x$ does not appear in $A$, it does not appear in $\neg A$. The relation $\neg A$ is a relation in $\mathscr{T}$ by CF2 (§1, no. 4). Applying criterion C33 (no. 3) with $R = \neg A$ and $S = B$, we obtain that the equivalence