Exercise 4, § 1

This solution was written by laguna-s-2.1-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

We use induction on the number of signs in the assembly $A$.

Base Case: $A$ has one sign. Suppose first that $A$ is a term. A term of length 1 is a single letter. $AB$ then begins with that letter followed by the assembly $B$, which cannot be a term (as terms of length $\geq 2$ must start with $\tau$, a substantific sign, and $B \neq \varnothing$ would violate the first species condition) or a relation (since relations start with $\vee$, $\neg$, or a relational sign). If $A$ were a relation, it would have to be one of length 1, which does not occur since relations require initial $\vee$, $\neg$, or a relational sign (and these cannot be of length 1). Thus, $AB$ is neither a term nor a relation.

Inductive Step: Assume the result for all terms/relations $A$ with $n$ signs. Let $A$ have $n + 1$ signs.

1. If $A$ is a term: - $A$ starts with $\tau$ or a substantific sign, or is a single letter (length 1). - If $A = \tau_x(B)$, where $B$ is a relation of length $n$, then $AB = \tau_x(B) B$. To form a term, $\tau_x(B)B$ would require $\tau_x(C)$ where $C$ is a relation. The assembly following $\tau$, which incorporates $B$ followed by $B$, must be a relation. However, $B$ has $n$ signs as a relation. By the inductive hypothesis, appending $B$ to $B$ (i.e., forming $BB$) is not a relation, so $C$ is not a relation. Hence, $\tau_x(C)$ cannot form a term, and since $AB$ begins with $\tau$, it cannot be a relation. - If $A = s A_1A_2 \ldots A_m$, where $s$ is a substantific sign of weight $m$, then $AB = s A_1A_2 \ldots A_m B$ has $m + k$ terms (where $B$ has $k$ terms), exceeding $s$'s weight $m$, making $AB$ not a term.

2. If $A$ is a relation: - $A$ starts with $\vee$, $\neg$, or a relational sign. - If $A = \neg B$, $AB = \neg B B'$. By the inductive hypothesis, $B$ with $n$ signs (as a relation) cannot have $BB'$ form a relation. Thus, $\neg BB'$ fails to meet the relation formation condition (b). - If $A = \vee BC$, then $\vee BC B'$ must decompose into two relations. Since $BC$, formed by appending $B$ and $C$, invalidates their relation status by induction. - If $A = s A_1 \ldots A_m$, where $s$ is a relational sign of weight $m$, then $AB = s A_1 \ldots A_m B$ exceeds $s$'s weight $m$, violating CF4.

Thus, in all cases $AB$ cannot be a term or relation. $\blacksquare$