Functionally complete set of a particular trivalued logic Consider the V-shaped partial order $\mathbf V$. Denote the three elements $\mathsf F > \bot < \mathsf T$. $\mathbb S_n$ is the set of n-ary connectives, that is, the set of monotonous functions $\mathbf V^n \to \mathbf V$. (They do not need to preserve $\bot$.)
Some of the connectives include:

*

*The "strict" $\dot\lor$, that agrees with the boolean $\lor$, and takes the value $\bot$ everywhere else.

*The "parallel" $\tilde\lor$, where $x \mathop{\tilde\lor} \mathsf T = \mathsf T \mathop{\tilde\lor} x = \mathsf T$, even if $x = \bot$.

*The "left-biased" $\stackrel \leftarrow \lor$, where $\bot \overleftarrow\lor \mathsf{T} = \bot$, but $\mathsf T \overleftarrow\lor \bot = \mathsf T$. This appears in many programming languages.

*The parallel majority function $\mathrm{maj}(x,y,z)$, on boolean inputs, it takes the value of the majority. This is the source of some important counterexamples.

*There cannot be a function $f(x)$ so that $f(\bot) = \mathsf{T}$, and otherwise $f(x) = \mathsf{F}$, because this is not monotone.

Is there a finite set of connectives such that all the other connectives can be expressed in terms of them? With some research it seems that Kleene's trivalued logic is close in spirit, but I haven't found any work that considers the monotone restriction. Some works on paraconsistent logic also has similar truth values, but they are sometimes too restrictive because of the need to eliminate the principle of explosion.
If there is no such finite set, an alternative is to allow some higher-order connectives, i.e. the least-fixpoint operation, or the strictification operation. Is there any relevant paper on this matter?
 A: Yes, here is one finite set of connectives which can build a formula (inspired by the disjunctive normal form) for any monotone function $f:\mathbf{V}^n \to \mathbf V$.

*

*The "nullary connectives" i.e. constants $\bot, \mathsf T, \mathsf F$, if these aren't otherwise included in the formal language.

*Unary prefix $\lnot$, with interpretation $\bot \mapsto \bot, \mathsf T \mapsto \mathsf F, \mathsf F \mapsto \mathsf T$

*Unary prefix $\tau$, with interpretation $\bot \mapsto \bot, \mathsf T \mapsto \mathsf T, \mathsf F \mapsto \bot$.

*Binary connective $\dot\land$, with the interpretation where $\bot \mathop{\dot\land} x \equiv x \mathop{\dot\land} \bot \equiv \bot$ for any $x \in \mathbf V$, and $\mathsf F \mathop{\dot\land} x \equiv x \mathop{\dot\land} \mathsf F \equiv \mathsf F$ for any $x \in \{\mathsf T,\mathsf F\}$, and $\mathsf T \mathop{\dot\land} \mathsf T \equiv \mathsf T$.

*Binary connective $\tilde\lor$, with the interpretation where $\mathsf T \mathop{\tilde\lor} x \equiv x \mathop{\tilde\lor} \mathsf T \equiv \mathsf T$ for any $x \in \mathbf{V}$, and $\bot \mathop{\tilde\lor} x \equiv x \mathop{\tilde\lor} \bot \equiv \bot$ for any $x \in \{\bot, \mathsf F\}$, and $\mathsf F \mathop{\tilde\lor} \mathsf F \equiv \mathsf F$.

*Binary connective $\star$, with the interpretation where $\bot \star x \equiv x$ for any $x \in \mathbf V$, and $x \star y \equiv x$ for any $x \in \{\mathsf T,\mathsf F\}$ and $y \in \mathbf V$.

It's simple to show the value function for each connective is monotone, and that $\dot\land$ and $\tilde\lor$ are symmetric and associative.
Let's say the formal language $\mathbf L$ has variables $\mathbf X = \{x_1,x_2,\ldots\}$. For each $\alpha = (\alpha_1,\ldots,\alpha_n) \in \mathbf V^n$, define the formula
$$ \begin{align*}
G_\alpha &= \left(\bigwedge_{\{i \mid \alpha_i = \mathsf T\}}^\bullet \tau\, x_i\right) \mathbin{\dot\land} \left(\bigwedge_{\{i \mid \alpha_i = \mathsf F\}}^\bullet \tau\, \lnot\, x_i \right)
\end{align*} $$
Then given our monotone function $f : \mathbf{V}^n \to \mathbf V$, define formulas
$$ \begin{align*}
A &= \bigvee_{\alpha \in f^{-1}(\{\mathsf T\})}^\thicksim (G_\alpha) \\
B &= \bigvee_{\alpha \in f^{-1}(\{\mathsf F\})}^\thicksim (G_\alpha) \\
\varphi &= (A) \star \lnot(B)
\end{align*}$$
Here each $\overset{\bullet}{\bigwedge}$ symbol and each $\overset{\thicksim}{\bigvee}$ symbol over a set of $n\geq 1$ elements means a formula with the corresponding binary connective appearing $n-1$ times, in the usual recursive way. Since the interpretation of all these connectives is symmetric and associative, order of the set elements in the formula doesn't affect the formula's interpretations. For a well-defined result, we can index each $\overset{\bullet}{\bigwedge}$ by increasing $i$, and index each $\overset{\thicksim}{\bigvee}$ by any full ordering of $\mathbf{V}^n$. Any $\overset{\bullet}{\bigwedge}$ symbol indexed over the empty set is the formula $\mathsf T$. Any $\overset{\thicksim}{\bigvee}$ symbol indexed over the empty set is $\bot$.
Let $\mu : \mathbf X \to \mathbf V$ be any variable assignment. Together with the connective interpretations above, $\mu$ determines an interpretation $I_\mu: \mathbf L \to \mathbf V$ on the well-formed formulas. Define $\beta = (\mu(x_1), \ldots, \mu(x_n)) \in \mathbf{V}^n$.
Given $\alpha = (\alpha_1,\ldots,\alpha_n) \in \mathbf{V}^n$, if $\alpha \leq \beta$, then for each $i \in \{1,\ldots,n\}$, we have $\alpha_i \leq \mu(x_i)$. So
$$\begin{align*}
\alpha_i=\mathsf T &\Rightarrow \mu(x_i) = \mathsf T \Rightarrow I_\mu(\tau\, x_i)=\mathsf T \\
\alpha_i=\mathsf F &\Rightarrow \mu(x_i) = \mathsf F \Rightarrow I_\mu(\tau\, \lnot\, x_i)=\mathsf T
\end{align*} $$
So all $\tau$ expressions in $G_\alpha$ are interpreted as $\mathsf T$, and again any $\overset{\bullet}{\bigwedge}$ over an empty set in $G_\alpha$ is also interpreted as $\mathsf T$. Therefore we also have $I_\mu(G_\alpha) = \mathsf T$.
If $\alpha \nleq \beta$, then for some $i \in \{1,\ldots,n\}$,
$$\alpha_i \nleq \mu(x_i) \Rightarrow \alpha_i \notin \{\bot, \mu(x_i)\}$$
Either $\alpha_i=\mathsf T$ and $\mu_i(x_i) \neq \mathsf T$ and so $I_\mu(\delta_T x_i)=\bot$, or else $\alpha_i=\mathsf F$ and $\mu_i(x_i) \neq \mathsf F$ and so $I_\mu(\delta_F x_i)=\bot$. Either way, this implies $I_\mu(G_\alpha) = \bot$.
So the interpretation of $G_\alpha$ is in general
$$ I_\mu(G_\alpha) = \begin{cases}
\mathsf T & \alpha \leq \beta \\
\bot & \alpha \nleq \beta
\end{cases} $$
Next consider the value of $f(\beta)$.
If $f(\beta) = \bot$, then for any $\alpha \in f^{-1}(\{\mathsf T\}) \cup f^{-1}(\{\mathsf F\})$, we know $f(\alpha) > f(\beta)$. Since $f$ is monotone, $\alpha \nleq \beta$. So for each $G_\alpha$ appearing in $A$ or $B$, $I_\mu(G_\alpha) = \bot$. Therefore $I_\mu(A) = I_\mu(B) = \bot$ (even if $f^{-1}(\{\mathsf T\})$ and/or $f^{-1}(\mathsf F\})$ is empty), and $I_\mu(\varphi) = \bot$ also.
If $f(\beta) = \mathsf T$, then $G_\beta$ appears in formula $A$. Since $\beta \leq \beta$, $I_\mu(G_\beta) = \mathsf T$. $A$ is a number of formulas including at least one interpreted as $\mathsf T$ joined with $\dot \vee$, so $I_\mu(A) = \mathsf T$. Since $\mathsf T \star y \equiv \mathsf T$ for any $y \in \mathbf V$, $I_\mu(\varphi) = \mathsf T$.
If $f(\beta) = \mathsf F$, then $G_\beta$ appears in formula $B$. As in the $\mathsf T$ case, this implies $I_\mu(B) = \mathsf T$. For any $\alpha \in f^{-1}(\{\mathsf T\})$, we have $f(\alpha) = \mathsf T \nleq \mathsf F = f(\beta)$. By monotonicity of $f$, $\alpha \nleq \beta$, so $I_\mu(G_\alpha) = \bot$. This is true of every $G_\alpha$ appearing in $A$, so $I_\mu(A) = \bot$ (even if $f^{-1}(\{\mathsf T\})$ is empty). Therefore $I_\mu(\varphi) = I_\mu(\bot \star \lnot \mathsf T) = \mathsf F$.
In all cases, then, $I_\mu(\varphi) = f(\mu(x_1),\ldots,\mu(x_n))$. Since this is true of every variable assignment $\mu$, formula $\varphi$ represents function $f$.
A: We claim that three connectives $\{\mathsf T, \bot, \tilde\uparrow\}$ is enough. First, using the parallel Sheffer's stroke, it is easy to construct parallel versions of conjunction and disjunction. Negation is $\neg p = p \mathop{\tilde\uparrow} p$. We proceed using a sort of disjunctive normal form.
For every input that evaluates to true, we add a disjunctive branch that is the conjunction of all the variables that are true, and the negation of all the variables that are false. This will not "overspill" by continuity. So now the expression agrees with the desired connective on the truthy inputs. Also, it will not return $\bot$ when the actual result should be $\mathsf F$. So we only need to consider the inputs that evaluate to $\mathsf F$ when it should be $\bot$.
For each one of these, we add a disjunctive branch that is conjunction of $x_i \mathop{\tilde\land} \bot$ when the input $x_i$ is true, and $x_i \mathop{\tilde\lor}\bot$ when the input is false, and $x_i\mathop{\tilde\lor}\neg x_i$ when the input is $\bot$. This is $\bot$ at the input, and takes on as many true values as allowed by continuity. Taking the disjunction completes the construction.
This is a rough sketch, and the details can be adapted from the other answer by @aschepler.
