Do I count too? Someone here has often accused me of trolling.
Chess will never be solved, here's why

Just a random person we both know. I talk to people sometimes but maybe you don't. It was out of the blue because it was ages since we talked or messaged.
The point is that both of you make the assumption that you aren't trolls.
As do you.
You were upsetting people for aged before our paths crossed.
I'm just the messenger.
Anyway, this is just yet another sidetracking project basically because you aren't very good at debating and you're trying to protect your own.
As were you. Our paths first crossed when you were beating somebody else over the head with your claimed IQ. Not surprising, really.
You are a messenger for a kingdom of ne'er-do-wells .

And the answer to my question is ... there is not a single person who is willing to back up @Optimissed's inexplicable claim that a model example of deductive reasoning is inductive.
I don't believe he is capable of seeing why that is.
Back on topic.
The game-theoretic value is the outcome when all participants play optimally.
Ultra-weakly solved means that the game-theoretic value of the initial position has been determined.
There is massive evidence that the game-theoretic value of the initial position of Chess is a draw.
Weakly solved means that for the initial position a strategy has been determined
to achieve the game-theoretic value against any opposition.
Strongly solved is being used for a game for which such a strategy has been determined
for all legal positions. For Chess this would mean a 32-men table base
and there are 10^44 legal positions, too much for present technology.
So the question is: will Chess be weakly solved?
So this means for Chess that for the initial position a strategy (i.e. one strategy, not all strategies) has been determined to achieve the draw for black against all opposition by white. i.e. white tries to win, black tries to draw, white fails, black succeeds, then Chess is weakly solved.
All participants play optimally, this means that white must oppose to the draw.
1 Nh3 opposes less than 1 Nf3. 1 a4 opposes less than 1 e4 or 1 d4.
1 c3 opposes less than 1 c4.
1 e4 e5 2 Ba6 is sure not to be optimal play.
1 e4 e5 2 Nf3 Nc6 3 Ng1 does not oppose: 3...Nb8 draws by repetition.
1 e4 e5 2 Nf3 Nc6 3 Nd4, 3 Nxe5, 3 Ng5, 3 Nh4 are sure not to be optimal play.
Good assistants, i.e. (ICCF) (grand)masters must contribute such knowledge and more.
The use of such knowledge is beneficial and allowed in weakly solving a game.
That leaves 10^17 relevant positions.
The latest cloud engines calculate a billion positions per second.
A year contains 31557600 seconds.
Thus 3 cloud engines (or 3000 desktops) can weakly solve chess in 5 years.
That is what GM Sveshnikov prophecised:
"Give me five years, good assistants and the latest computers
- I will bring all openings to technical endgames and "close" chess."
He was right.
The obstacle is the money: 3 million $ to hire 3 (ICCF) (grand)masters full time
and rent 3 cloud engines (or 3000 dekstops) non stop 24/7 during 5 years.

One who eschews and dismisses all sources of knowledge cannot really know much. That just follows, yes?

You must imagine that you and Elroch are sources of knowledge!
That does *not* follow. Better take the whole exam over, you might not get a good score .

I have a very fundamental question that I'm not sure has been answered:
Given any position in a game, does there exist, for the player about to move, an "optimal" move, such that if he makes that move, he will have a certain path to at least a draw no matter what the opponent does?
I am not convinced that such a move exists. Obviously, you can talk about percentages and data bases, but here we're looking for certainty, and I don't think we've established that it exists.

In "any position", there is one or more optimal moves which allows the forcing of the optimal result against any counterstrategy. In some positions this optimal result is a win; in others, a draw (and in some, a loss, which means all moves are optimal in the pure sense).

In "any position", there is one or more optimal moves which allows the forcing of the optimal result against any counterstrategy. In some positions this is a draw, in some positions a win (and in some a loss, which means all moves are optimal in the pure sense).
I'm not convinced. You're making an assertion that I don't agree with, but you're offering no rationale.

If we were talking tic tac toe or checkers, we'd all agree there is an "optimal" move that assures at least a path to a draw. How do we know if this is true in chess?
Also, I'm not sure how you can have one "or more" optimal moves.

Just to demonstrate that @Optimissed's projection about those who contribute to Wikipedia was not accurate:
An example peer-reviewed paper about the non-trivial ultra-weak solution of a game in the same general class as chess, checkers, go etc.
Another peer-reviewed paper listing many important games that have been weakly solved and discussing the prospects for solving others.
A milestone paper, explaining how checkers had finally been weakly solved.

In "any position", there is one or more optimal moves which allows the forcing of the optimal result against any counterstrategy. In some positions this is a draw, in some positions a win (and in some a loss, which means all moves are optimal in the pure sense).
I'm not convinced. You're making an assertion that I don't agree with, but you're offering no rationale.
I didn't, but this is a theorem from the theory of finite games, so you can be absolutely sure it is true.
The proof is not trivial but is quite easy.

In "any position", there is one or more optimal moves which allows the forcing of the optimal result against any counterstrategy. In some positions this is a draw, in some positions a win (and in some a loss, which means all moves are optimal in the pure sense).
I'm not convinced. You're making an assertion that I don't agree with, but you're offering no rationale.
I didn't, but this is a theorem from the theory of finite games, so you can be absolutely sure it is true.
The proof is not trivial but is quite easy.
OK, I'm a bit out "over my skis" here, but what is the definition of a "finite game"?

In "any position", there is one or more optimal moves which allows the forcing of the optimal result against any counterstrategy. In some positions this is a draw, in some positions a win (and in some a loss, which means all moves are optimal in the pure sense).
I'm not convinced. You're making an assertion that I don't agree with, but you're offering no rationale.
I didn't, but this is a theorem from the theory of finite games, so you can be absolutely sure it is true.
The proof is not trivial but is quite easy.
OK, I'm a bit out "over my skis" here, but what is the definition of a "finite game"?
A finite two player game is where there are two players, they move alternately, there are a finite number of alternatives at each move and every game ends in a finite number of moves. An example of a game that fails to meet the definition would be noughts and crosses on an infinite plane. Chess only meets the definition if you assume draws are forced by the 50 move rule or a repetition, rather than needing to be claimed.
[Also, I don't think my adjective "easy" was really appropriate. Rather it is a concise proof!].

Do you mean that you admit you're not sources of knowledge? Does that mean that you don't know enough to be able to make a judgement??
Exam?? Dementia?
Lol, when you don't like your test results you just take the test over and over until you do...I would have expected you to pick up on that.

I have not much idea of what you're rabbitting on about unless it's inspired by jealousy regarding something or other.
Kennel!!
I was not addressing the general "you", but the specific you.
If you still don't get it, I don't know what to tell you, it's your own past anecdote I'm referring to.

In "any position", there is one or more optimal moves which allows the forcing of the optimal result against any counterstrategy. In some positions this is a draw, in some positions a win (and in some a loss, which means all moves are optimal in the pure sense).
I'm not convinced. You're making an assertion that I don't agree with, but you're offering no rationale.
I didn't, but this is a theorem from the theory of finite games, so you can be absolutely sure it is true.
The proof is not trivial but is quite easy.
OK, I'm a bit out "over my skis" here, but what is the definition of a "finite game"?
A finite two player game is where there are two players, they move alternately, there are a finite number of alternatives at each move and every game ends in a finite number of moves. An example of a game that fails to meet the definition would be noughts and crosses on an infinite plane. Chess only meets the definition if you assume draws are forced by the 50 move rule or a repetition, rather than needing to be claimed.
[Also, I don't think my adjective "easy" was really appropriate. Rather it is a concise proof!].
What if there were two players and they each had two moves alternately?
I am not sure what you mean. If the rules of a game permit one to go on forever, it is excluded by definition.
If you meant something else, do say.

In "any position", there is one or more optimal moves which allows the forcing of the optimal result against any counterstrategy. In some positions this is a draw, in some positions a win (and in some a loss, which means all moves are optimal in the pure sense).
I'm not convinced. You're making an assertion that I don't agree with, but you're offering no rationale.
I didn't, but this is a theorem from the theory of finite games, so you can be absolutely sure it is true.
The proof is not trivial but is quite easy.
OK, I'm a bit out "over my skis" here, but what is the definition of a "finite game"?
He's trying to blind us with theory. A finite game is one that is not infinite. Thet is, it ends at some point and does not have an infinity of permutations or positions. Chess might as well be infinite, for all the possibility there is in tracing all the possible games.
LOL. The difference is very important when you want to prove facts rigorously.
<<I didn't, but this is a theorem from the theory of finite games, so you can be absolutely sure it is true.>>
I stopped accepting Elroch's judgement many years ago.
While a reasonable amount of skepticism is always appropriate, with all due humility you would very rarely go wrong by believing what I say. That is because I have not lost a lifetime's habit for being precise and correct.
Your quote above is an example. Anyone who checks the body of knowledge will confirm it, via my link or any other good source.

You seem to place great faith in peer revue. Whereas, of course, it's necessary to have checks and balances in acedemia, a peer revue can only pick up on obvious errors since someone's peer is unlikely to be a specialist in the field under consideration. So peer revue isn't and cannot be a guarantee of correctness.
Also you're spouting the same old stuff about weak and strong and I've already explained that it's perfectly obvious that you do not understand the application of Games Theory sufficiently to have any authority on the matter.
What is "perfectly obvious" to an ignorant windbag doesn't matter as much as you think it does.
With all due humility, I have a perfect understanding of the degree to which the relevant part of game theory is applicable to chess. It is as clear to me as the meaningfulness and value of the three definitions of types of solutions of games which all the cognizant people in these forums accept, which are well-established in the academic literature, and which you falsely asserted were made up for wikipedia!
With more appropriate respect for others and less unfounded arrogance, your understanding would improve greatly. [That's not a prediction].
Just a random person we both know. I talk to people sometimes but maybe you don't. It was out of the blue because it was ages since we talked or messaged.
The point is that both of you make the assumption that you aren't trolls.
As do you.