*things*that you have around. The Untyped Lambda Calculus, however, allows programs that never terminiate. I guess the Y-combinator is a pretty good example here. I once learned that the Untyped Lambda Calculus can be embedded inside the Typed Lambda Calculus by having essentially one type,

*the everything type*. So I guess this makes sense, since there are no arrow types that can decrease monotonically...

Now I have more questions:

Does this mean the untyped lambda calculus

*is*Turing complete?

How does the encoding of the untyped lambda calculus into the lambda calculus work? I mean, don't functions always have to have function type? Or is the

*everything type*some kind of function type?

Yes, the un(i)typed lambda calculus is turing-complete.

ReplyDeleteIf you want a typed model of the untyped lambda calculus, the everything type, call it D, has to be isomorphic ("has to be effectively the same type as") to D → D. The everything type

isa function type, and its functions expect aneverythingand return aneverything.To actually describe what this set of things is carefully, you sometimes resort to tricky domain-theory tricks invented by Dana Scott.

Note that that typed lambda calculus that has D = D → D is decidedly

ReplyDeletenotthe simply-typed lambda calculus anymore, and of course it can't be, since it's turing complete and the STLC is not.

ReplyDeleteThe encoding of the untyped lambda calculus into the lambda calculus.I actually don't know what this means, and the way I interpret it it shouldn't be possible...

This is just me paraphrasing a comment made by Bob four years ago in 814 that a typed language is strictly more expressive than an untyped language because you can encode an untyped language in a typed one where every single value has the same type.

ReplyDeleteAs jcreed points out above, if you want to encode the untyped lambda calculus with a typed language, that language cannot be the simply typed lambda calculus, because it's not Turing complete.

I can give more details after we get to that lecture in 814 (which I am sitting in on, four years later). :-)

ReplyDeleteYes, that makes sense! Thanks.

ReplyDeleteAah, yes. You can encode the untyped lambda calculus into a lambda calculus with one recursive type "μx.x→x", but not the simply typed lambda calculus, per jcreed. I was just confused by the setup.

ReplyDeleteGrateful for sharingg this

ReplyDelete