# Is there some n such that lambda calculus with only n variables is Turing-complete?

Typically in lambda calculus you have an infinite stock of variables. Could we get away with a finite set?

• If I understand what you mean by "get away with", yes, since only three variables are needed to encode a complete combinator basis such as SKI or BCKW. Conversely, Richard Statman proved that two variables are not enough, as explained in my answer here. Aug 29 at 13:21
• Good point. That was my intuition about it. Thanks Aug 29 at 14:09