# Can a formula with only unique variables be an intuitionistic tautology?

Consider intuitionistic propositional formulae, using only the connective "$$\rightarrow$$" and absurdity . Can there exist a formula such that it is a theorem/tautology and every pair of variables contained in it, is a pair of different variables?

• What about e.g. $\bot\to x$? May 27 at 18:13
• You are correct, I did not think about that. Thank you, I will remove absurdity from the question. May 27 at 18:32
• Do you mean that no variables are repeated? I.e. any variable that occurs in the formula only occurs once? May 27 at 18:36

You can prove this using the topological semantics for intuitionistic logic.

Our chosen topological space is $$\mathbb{R}$$ equipped with the standard topology. The open subsets of $$\mathbb{R}$$ are the truth values and $$\mathbb{R}$$ itself is the sole designated truth value. If you do this, you can construct explicit truth values for all of the variables in a well-formed formula that make the formula have a non-designated truth value.

Next I'll define a notion of a covariant position and a contravariant position.

In the well-formed formula, $$a \to b$$, $$a$$ is a contravariant position and $$b$$ is a covariant position.

In the well-formed formula $$(a \to b) \to c$$, $$b$$ is a contravariant position and $$a$$ and $$c$$ are covariant positions.

A position is covariant if and only if it is the antecedent of an even number of conditionals.

Consider a well-formed formula $$\varphi$$ with no repeated variable symbols. Let $$A$$ be the variables that appear covariantly and $$B$$ be the variables that appear contravariantly. Note that $$A$$ and $$B$$ are disjoint and their union is all the variables of $$\varphi$$.

Consider a variable $$x$$ that appears in $$\varphi$$.

If $$x$$ is in $$A$$, then give $$x$$ the truth value $$\varnothing$$.

If $$x$$ is in $$B$$, then give $$x$$ the truth value $$\mathbb{R}$$.

$$\varphi$$ will thus have the truth value $$\varnothing$$.

No such formula is a tautology. Since the arguments to the outermost $$\implies$$ are independent, there is always a model where the antecedent is true and the consequent is false.

Let's do an example. Consider the sentence $$(a\implies b) \implies (c \implies (d \implies e)).$$ Because they share no variables, we can easily make $$a\implies b$$ true while $$c \implies (d\implies e)$$ is false. Consider any assignment with $$a$$ and $$e$$ false, and $$c$$ and $$d$$ true. Such an assignment gives the formula the form $$T \implies F$$, so it can't be a tautology.

• This is really an inductive argument, right? To find an assignment making $\varphi\implies \psi$ false, you need to find an assignment making $\psi$ false (which is possible by induction). It's interesting to note from unwinding this inductive argument that you can always set the right-most variable to false and all others to true. May 27 at 19:02
• That's right, to prove it formally, you have to induct on the height of the sentence's syntax tree. May 27 at 19:20