Some sources include the symbols $\top$ and $\bot$ as elements of the alphabet for the formal language of propositional logic and call them nullary connectives, meaning they are logical connectives that take zero arguments. Though I've seen them being called logical constants as well.

I don't understand how I am supposed to think about them, how does a logical connective work that doesn't connect any statements?

Is $\top$ just always true and $\bot$ always false no matter what? If so, where's the distinction between those symbols (part of the formal language) and the (metalinguistic) concept of tautology/contradiction?

Why should one choose to include it in the language or not?

What especially trips me up is that wikipedia lists sets like $\{\rightarrow,\bot\}$ or $\{\nrightarrow,\top\}$ as functionally complete.

Sorry if the question is incoherent.

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    $\begingroup$ Well, they work exactly like the propostional symbols, except that for any interpration $\phi$, $\phi (\bot)=0$, and $\phi (\top)=1$. $\endgroup$ Feb 6 at 14:47
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    $\begingroup$ Also there is no big deal about $\{\rightarrow,\bot\}$(for instance) being functionally complete, you can get $\neg a$ as $a\to\bot$ and $a\land b$ as $(a\to (b\to\bot )) \to \bot$ since $\land,\neg$ is functionally complete you may conclude. $\endgroup$ Feb 6 at 14:55
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    $\begingroup$ Yes a constant truth-function $\endgroup$ Feb 6 at 16:02

1 Answer 1


First of all, $\top$ and $\bot$ are special constants whose valuations are always set to true and false respectively. In other words, we always have $\models \top$ and $\models \neg \bot$.

It is, however, convenient to consider them as connectives ($0$-ary connectives, hence) when one is dealing with sets of connectives. One never writes the braces, but you can imagine eg $\top$ as an operator $\top ()$.

As to why $X :=\{\rightarrow,\bot\}$ and $Y := \{\nrightarrow,\top\}$ are functionally complete sets of connectives, it is enough to show that you can express the operators $\wedge$ and $\neg$. I leave you as an exercise to show that:

  1. $\neg A \equiv A \rightarrow \bot$
  2. $A \wedge B \equiv \big( A \rightarrow (B \rightarrow \bot) \big) \rightarrow \bot$ (with 1. that makes $X$ functionally complete)


  1. $\neg A \equiv \top \nrightarrow A$
  2. $A \land B \equiv A \nrightarrow (\top \nrightarrow B)$ (with 3. that makes $Y$ functionally complete)
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    $\begingroup$ I did the exercise and tried to show functional completeness of those sets treating $\top$ and $\bot$ as propositional "variables" just like $A$ and $B$ except they're either just true or just false in the truth tables. However I got different results than you for 3. and 4. : $\neg A \equiv \top \nrightarrow A$ and $A \land B \equiv A \nrightarrow (\top \nrightarrow B)$ Is this correct? $\endgroup$ Feb 6 at 17:58
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    $\begingroup$ @LinusDieLinse There isn't a unique solution indeed and yes you are correct, well done. Oh and by the way I made a mistake on point 3. and 4! I'll correct my answer using your suggestion. :-) $\endgroup$ Feb 6 at 21:11
  • $\begingroup$ thank you! :) I understand $\top$ and $\bot$ as constants, but could you elaborate on how and why it is convenient to consider them as nullary connectives, and what a nullary connective is? $\endgroup$ Feb 8 at 19:36
  • $\begingroup$ @LinusDieLinse It is convenient because you can speak of a "set of connectives instead of a "set with connectives and eventually some constants". That is really all there is to it. $\endgroup$ Feb 11 at 5:16
  • $\begingroup$ @LinusDieLinse "What is a nullary connective?" Formally it is the unique function $\emptyset \mapsto \{a\}$ for some $a$. $\endgroup$ Feb 12 at 0:56

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