TITLE: How complicated is the set of tautologies?
QUESTION [6 upvotes]: Consider the set $\mathcal T$ of all tautologies in the propositional calculus in which the only operators allowed are $\to$ and $\neg$, and involving only the two variables $x$ and $y$.

How complicated is $\mathcal T$? Is it a context-free language? 

Does the answer depend on the fixed number of variables? I guess not... If one removes the limitation of the number of variables (and somehow adapts the notions of context-freeness to play nice with this...), does the answer change? I guess it does...
This question is a natural outcome of Doug's recent question here.

REPLY [3 votes]: Generalization of Yuval's answer:
For any finite set $A$ and operations $o_i \colon A^2 \to A$ the language of expressions using elements of $A$ and infix operators $o_i$ that evaluate to fixed $a \in A$ is context-free. Here $A = \{0,1\}^n \to \{0,1\}$.
If you regard the expressions as trees not words, it will be a regular tree language (recognizable with finite tree automaton).
I remember an exercise in automata theory: the set of Boolean formulas (as words) using 0,1,$\vee$,$\wedge$,$\neg$,(,) evaluating to 1 is regular, provided that arbitrary answer is allowed when input has mismatched brackets.
As a consequence of this there should exist a regular language $L$ such that the set of tautologies is $L \cap R$ where $R$ is the context-free language of all well-formed formulas (tautologies or not).