[nilkn]

So you agree it’s ambiguous and one cannot formally deduce the meaning of the expression without making assumptions. One must assume that objects that don’t fit the definition of list can be substituted in for lists within some contexts (but not all).

Beyond that, dotted lists are introduced via the undefined example (a b . c), which requires one to go even further and make a second assumption (namely, assume the intention was to refer to (a . (b . c)), then assume the quoted transformation rule applies to certain non-lists allowing that to be rewritten as (a b . c)).

[SamReidHughes]

It is not the case that dotted lists don't fit the definition of list. What you point to as a definition wasn't a definition. Just examples. A definition would describe it as the minimal set of objects meeting those criteria. (A non-minimal set could include circular lists and/or pairs whose cdr is not a list.)

Since the document does use list in ways that encompass dotted lists and goes out of its way to define proper lists, we can infer that list includes dotted lists. Also, it's a long-running convention that the term "proper X" implies there are other kinds of X's.

[nilkn]

I'd disagree with you on that. If the document can't clearly define the terms it's using, it cannot function as a formal specification of a language. The writing as it stands now is far too ambiguous to serve such a role. You have no reasonable retort to me except to say that the definition wasn't even a definition, and after that there is indeed no more abstract definition you can even point to. You're left using undefined terms to defend your interpretation as the "correct" one. That's not sufficient to defend the document as a specification. In fact, that's just further criticism of it.

Like I said, I'm confident your interpretation is the intended one. But you cannot prove it as such, and that's where we disagree.

[SamReidHughes]

I never claimed it was a formal specification. The question was, is the meaning of (a b . c) defined, and the answer is, yes, because (b . c) is a list.

[nilkn]

And I say the answer is no, because (b . c) is an arbitrary pair, not a list. It would appear that we’re at an impasse.