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I think I'm familiar with most of the arguments for and against the submission of solutions arrived at with 'assistance'. I hope not to be identified in some future posting as the person single-handedly responsible for spreading disharmony on the subject, but the strict Plymouth Brethren approach has always struck me as, at least, debatable.

I don't have a co-solver and I assume that use of Chambers and the internet - to verify suppositions, confirm themes etc - is above reproach. But what if I use Mrs Bradford's book to identify synonyms that my reading of a clue fails to suggest? What if I consult the online Chambers Word Wizard or my very battered copies of Chambers Anagrams or Chambers Back-words when I can't think how a clue can possibly yield anything to fit the letters I already have? What if - though I never do - I use one of the online Playfair code-breakers? Am I thereby rendered a cheat, who should at once cease sending in my entries lest I ever be invited to a dinner I'd have no right to attend?

Does the asceticism of those who wish not to see the stats 'devalued' compel them to forswear all the above sources of help and inspiration? As Peter Simple used to write, I only ask because I want to know.

There are in fact usually several puzzles each year (and not just the numericals) that can't be solved outright with any online help, so the all-corrects aren't likely to include solvers who are scheming or fraudulent. They don't get a free dinner any more, either. And it's not the Nobel or the Pulitzer, is it?

Well I can hardly wait for Friday evening to see who is first aboard the Club Train, and which regulars are able to resist.
Even better would be to have one of them, ideally Shirley, explain just why it's so important to them to complete The Listener before it has even been published in the paper.

OK, chaps, we now have reached 72 posts, of which only 17 are actually relevant to the puzzle. Is it possible that this inexplicable outburst of inter-board rivalry has itself put people off coming here? Just get over yourselves, admit different people have different approaches. I am sure we all have far better things to do than arguing over crosswords - like helping the people who may come here for help.

Saoralba - in my opinion this is entirely down to the solver. For example, I think a solver who is perhaps relatively new to crosswords who has sweated away to complete a Listener using all the devices you refer to and perhaps even had the odd hint here and there more than merits posting off their hard won solution grid. I can't imagine anyone pursuing the elusive all correct just to go to the dinner - I think it is more that crossword solvers seem to like a challenge and all correct is a pretty tough one!

To Jim.
Catalysis, I accept. ΔF=ΔH-T.ΔS : The initial and final values for energy and entropy in a system are unchanged regardless of whether there is a catalyst. In atomic theory terms, the only question is when the cat actually dies.
For my part, no you're not a puppet, but maybe verbose.Hopefully we can co-exist from here on.

Meursault
delta G rather than delta F, surely? (Gibbs free energy) - and surely the cat can collapse the wave function itself?

Helmholtz free energy is F. Later interpretations of the Schrödinger paradox (eg Niels Bohr) weren't too concerned with the wave function.

Hi meursault
I know I'm being pedantic, but Helmholtz free energy (recommended symbol A rather than F) should use delta U instead of delta H.

Gibb's Free Energy: G = H - TS

Helmholtz Energy: A = U - TS

where A is Helmholtz free energy
U is internal energy of the system
T is absolute temperature
S is the entropy

I seem to have been catalyst for another debate, but this time one I welcome. There's no problem with being pedantic - we're all looking for the same thing, that is the truth.

In old money Helmholtz was also denoted as F.
Part of the reason for the two different denotations (Gibbs & Helmholtz) was that not every reaction remains at the same temperature and pressure (how true for the situation I was originally describing). But there is a simple relationship : ΔH = ΔU + P.ΔV