In the first lesson of a course in logic we defined as a logical formula the following:

  • propositional variables $p_0,p_1,...$

  • if $\phi$ is a formula then $\lnot\phi$ is a formula too

  • if $\phi$ and $\psi$ are formulas then $(\phi \lor \psi)$ , $(\phi \land \psi)$ , $(\phi \rightarrow \psi)$ , $(\phi \leftrightarrow \psi)$ are formulas

and nothing else is a formula.

Now by using only one propositional variable $p$,

and $0$ times any of the symbols $\lnot, \lor, \land, \rightarrow, \leftrightarrow$ we can create only $1$ logical formula

and $1$ time any of the symbols $\lnot, \lor, \land, \rightarrow, \leftrightarrow$ we can create $5$ logical formulas: $\lnot p$ and $(p*p)$ ,where $*$ can stand for any of the symbols $\lor, \land, \rightarrow, \leftrightarrow$ (I write it this way to write less)

and $2$ times any of the symbols $\lnot, \lor, \land, \rightarrow, \leftrightarrow$ we can create the following logical formulas: $\lnot\lnot p$, $(\lnot p * p)$, $(p*\lnot p)$, $\lnot (p*p)$, $(p*(p*p))$, $((p*p)*p)$ which are $1+4+4+4+4^2+4^2=45$

My question is how many logical formulas can we create using only one variable and $n$ times any of the symbols $\lnot, \lor, \land, \rightarrow, \leftrightarrow$? I can calculate it(I think) for very small $n$ as above but for larger $n$ the thing gets more complicated.


Every formula with $n$ symbols is an incomplete binary tree. It has:

  • exactly $n$ internal nodes (incl. the root),

  • each internal node having $1$ or $2$ children,

    • any internal node having $2$ children being decorated with any of $4$ symbols ($\lor, \land, \rightarrow, \leftrightarrow$),

    • any internal node having $1$ child being decorated with $\lnot$,

  • all the leaves decorated with the variable $p$.

The related problem where every internal node has $2$ children (i.e. complete binary tree), and there are no decorations, has a neat solution, but generalizing it to counting incomplete binary trees probably makes it very hairy, and adding decorations will make it even hairier.

Anyway here's an alternative: Let $f(n)$ be the number you seek.

  • If the root has $1$ child (the formula has a leading $\lnot$), then there are $f(n-1)$ ways to complete the tree.

  • If the root has $2$ children, i.e. $2$ subtrees, they can have $a, b$ internal nodes respectively as long as $a+b = n-1$, so there are $f(a)$ ways to make the left subtree and $f(b)$ ways to make the right subtree. And of course, need to multiply by $4$ for the $4$ binary op symbols.


$$f(n) = f(n-1) + 4 \times \sum_{a=0}^{n-1} f(a) f(n-1-a)$$

I am not sure how to turn this into something better (either closed-form, or at least some summation/product).

UPDATE: Just found this excellent CS.SE answer which suggests there is a way to count these using generating functions. Sadly, I'm out of my depth.

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