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.
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.