Hypothesis Given1 : ~ Youngest Alice -> Youngest Bill.

The text says: Alice is the youngest unless Bill is

Which IMO should translate to:

Hypothesis Given1 : ~ Youngest Bill -> Youngest Alice.

I think the Lemma YoungBill (and the entire goal) cannot be proven without classical with this encoding anymore. It can be proved with double negative elimination.

Also, the firstorder tactic does not give an answer to this.

When cdsmith said “Goal Oldest Alice /\ Youngest Bill.” s/he was literally saying “Here’s a theorem I want to prove.” “Goal” essentially introduces anonymous theorems, so that’s where “Unnamed_thm” came from. S/he could also have said “Theorem OldestAliceAndYoungestBill Oldest Alice /\ Youngest Bill” and the prompt would have become “OldestAliceAndYoungestBill < ” instead. When you say you want to prove something to Coq, it goes into proof-editing mode and the prompt changes to reflect what you’re proving. This is the context in which you ask Coq to apply tactics in an attempt to simplify (away) the current (sub)goal. Coq has a lot of built-in tactics, and many, many more are available in libraries, but again, this is an extremely simple puzzle, so the built-in ones are enough. cdsmith is a newcomer to Coq, so s/he used the most basic tactics that all of the introductory materials teach you about (intro, split, apply, exact…) and some understanding of both classical and intuitionistic logic variants of Pierce’s law. What I did was to pull out one of Coq’s sledgehammer tactics, firstorder, which implements a very powerful first-order proof-search procedure, the basis of which is documented in Pierre Corbineau’s paper, First-order reasoning in the Calculus of Inductive Constructions. So the firstorder tactic does automatically what cdsmith did somewhat more manually, although it’s worth emphasizing that both cdsmith’s proof script and mine generate a term in the Calculus of Inductive Constructions that is quite a bit larger than either script. :-) You can see the term by asking Coq to “Print Unnamed_thm.” after it is defined.

This, by the way, is another difference between Prolog and Coq: typically in Prolog, you just want the result(s) of the resolution, and in fact one of the nice things is you get to ask “Who is oldest and who is youngest?” In Coq, you have to say what you want to prove (“I intend to prove that Alice is oldest and Bill is youngest”), but you don’t just get True/False, you also get the proof. When you’re just doing logic, this probably doesn’t matter much, but one of Coq’s most interesting applications is that you can develop software with it that is proven correct, and then literally extract the code from the proof, in any of Scheme, Haskell, or OCaml. This code is “correct by construction” with respect to its specification in Coq, and in Coq, specifications can be extremely strong, again thanks to the expressive power of the Calculus of Inductive Constructions. A nice example of this is this implementation of Finger Trees in Coq.

By the way, I hope you don’t believe I’m attempting to downplay the power of Prolog here—my point really is more about the simplicity of the puzzle. My copy of “PROLOG Programming for Artificial Intelligence,” 3rd edition, is about 6″ away from my right elbow, and I recently got ODBC working on my Mac OS X box, which means that I can play with FLORA-2’s Persistent Module package. FLORA-2, in turn, relies on the XSB Prolog system. I’m a huge fan.

I hope this helps clarify what Coq and firstorder are a bit!

But again, you should ignore most of that, because all it shows is that Coq does at least as much as Prolog. What Coq is really for is helping people prove things that can’t be automatically proven at all, e.g. the four-color map theorem, or that a compiler for a programming language correctly compiles all possible source programs into their semantically-identical respective target programs.

youngest(alice) :- not(oldest(alice)).
youngest(bill) :- not(oldest(bill)).
oldest(alice) :- not(youngest(carl)).

?- oldest(X),youngest(Y).
X = alice,
Y = bill.