[Math] In what sense is propositional logic “zeroth-order logic?”

first-order-logiclogicpredicate-logicpropositional-calculus

The Wikipedia page on the propositional calculus says that another name for it is "zeroth-order logic." This sort of makes sense to me, but appears to break down in some pretty crucial ways with logical connectives.

The idea is that the propositions of propositional logic can be thought of as predicates of "zeroth order." They take in no input and can be thought of basically as "constant $\{0,1\}$-valued functions." First-order predicates take these as input. Second-order predicates take in either first-order predicates or zero-order predicates as input. And so on.

This gets weird as it pertains to logical connectives, though. In propositional logic, the objects really are thought to be "constant functions" in that they are considered to have a true or false value, and we can apply logical connectives to them. But if we are considering these to be the things that first-order predicates take as arguments, how does this sense? In first-order PA, for instance, the objects are basically just numbers. But if these are supposed to be "zeroth-order predicates," what meaning does $(12 \lor 13) \land (14 \lor \lnot 15)$ have?

It is kind of strange to think of logical connectives as applying to all predicates of order > 0, unless you're working in "zeroth-order logic," at which point they apply there as well. Is there some neat way to make sense of precisely how things are related here?

Best Answer

This gets weird as it pertains to logical connectives, though. In propositional logic, the objects really are thought to be "constant functions" in that they are considered to have a true or false value, and we can apply logical connectives to them. But if we are considering these to be the things that first-order predicates take as arguments, how does this sense? In first-order $PA$, for instance, the zeroth-order predicates are basically just numbers. What meaning does $(12 \vee 13)\wedge(14\vee \neg 15)$ have?

The emphasized part is very misleading. We don't generally take our constants to be "zeroth-order" predicates, but rather $0$-ary functions (and even that is not forced). An $n$-ary predicate takes an $n$-tuple and returns a truth-value, whereas an $n$-ary function takes $n$-tuples and returns an object of the domain, so it's natural to identify $0$-ary functions with constants. But then the constants will not be "zeroth-order" predicates!

EDIT: You also seem confused by the definition of "zeroth-order" logic, first-order logic, etc. Basically, a logic is of the $n$th-order if no variable of order $n+1$ is quantified, where the order of the variables can be defined recursively by setting a (relational) type hierarchy as follows:

  • $o, i$ are types of order $0$ and $1$, respectively (they are, respectively, the types of the truth-values and of the individuals);
  • if $\tau_1, \dots, \tau_n (n \geq 1)$ are of any type except $o$, then $\langle \tau_1, \dots, \tau_n\rangle$ is a type of order $1 + max\{ord(\tau_1), \dots, ord(\tau_n)\}$. Intuitively, $\langle \tau_1, \dots, \tau_n\rangle$ is the type of $n$-ary relations with arguments from $\tau_1, \dots, \tau_n$. So, for instance, first-order $n$-ary relations are variables of type $\langle i, \dots, i\rangle$ (that's $n$ $i$'s) and order $2$.

So first-order logic is so-called because no variable of the second order is quantified. Similarly, a zeroth-order logic is so-called because no variable of order $1$ is quantified (i.e. we don't quantify over indviduals). Note further that relation variables take individuals, not truth-values, as inputs.

Related Question