[pron]

> you're wrong to say that LC has no notion of complexity

I didn't say that it has no notion of complexity; I said it "does not have a useful formulation of complexity", as reduction step count are not very useful in measuring algorithmic complexity, at least not the measures of complexity most algorithms are concerned with.

> It's foolish to think of this as equivalent to LC, though.

Oh, I don't think that at all, which is why I specifically said that some people make the mistake of confusing LC reductions with classical substitutions (equality). They may then think that computation can be equational (false), rather than say it may sometimes be useful to think of computation in equational terms, but that's an abstraction -- namely, a useful lie -- that has a cost, i.e. it is "leaky" (true).

[tel]

Fair enough.