Differential Propositional Calculus • 6

Cactus Calculus

Table 6 outlines a syntax for propositional calculus based on two types of logical connectives, both of variable $k$ -ary scope.

A bracketed sequence of propositional expressions $\texttt{(} e_1 \texttt{,} e_2 \texttt{,} \ldots \texttt{,} e_{k-1} \texttt{,} e_k \texttt{)}$ is taken to mean exactly one of the propositions $e_1, e_2, \ldots, e_{k-1}, e_k$ is false, in other words, their minimal negation is true.
A concatenated sequence of propositional expressions $e_1 ~ e_2 ~ \ldots ~ e_{k-1} ~ e_k$ is taken to mean every one of the propositions $e_1, e_2, \ldots, e_{k-1}, e_k$ is true, in other words, their logical conjunction is true.

$\text{Table 6. Syntax and Semantics of a Calculus for Propositional Logic}$

All other propositional connectives may be obtained through combinations of the above two forms. As it happens, the concatenation form is dispensable in light of the bracket form but it is convenient to maintain it as an abbreviation for more complicated bracket expressions. While working with expressions solely in propositional calculus, it is easiest to use plain parentheses for bracket forms. In contexts where parentheses are needed for other purposes “teletype” parentheses $\texttt{(} \ldots \texttt{)}$ or barred parentheses $(\!| \ldots |\!)$ may be used for logical operators.

The briefest expression for logical truth is the empty word, denoted $\boldsymbol\varepsilon$ or $\boldsymbol\lambda$ in formal languages, where it forms the identity element for concatenation. It may be given visible expression in textual settings by means of the logically equivalent form $\texttt{((} ~ \texttt{))},$ or, especially if operating in an algebraic context, by a simple $1.$ Also when working in an algebraic mode, the plus sign ${+}$ may be used for exclusive disjunction. For example, we have the following paraphrases of algebraic expressions.

$\begin{matrix} x + y ~=~ \texttt{(} x \texttt{,} y \texttt{)} \\[6pt] x + y + z ~=~ \texttt{((} x \texttt{,} y \texttt{),} z \texttt{)} ~=~ \texttt{(} x \texttt{,(} y \texttt{,} z \texttt{))} \end{matrix}$

It is important to note the last expressions are not equivalent to the triple bracket $\texttt{(} x \texttt{,} y \texttt{,} z \texttt{)}.$

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