I try to make some definition of [0] along back, but I stopped at nonassociativity of [0] and I expect that formula,,, a [0] a [0] ...[0] a {b+1 times} = a [0] (a [0] a [0] ... [0] a {b times}) is not correct.
How I can see your definition of [0] make correct results. It is very interesting that the [0] is not associative and ,,, a [0] a [0] ...[0] a {b+1 times} = a [0] (a [0] a [0] ... [0] a {b times}) is correct
I realize that the formula can be corect, bat other bracket construction(with identical number of operands) needn't to be equal betwen each other ,,, and than ,,, if some bracket construction isn't equal to another,the operation cannot be associative.
Good question is ,whether are here some other formulas with bracket construction ,which are equal betwen each other?
it could by interesting to find out behaviour of this special nonasociativity (I think ,this special behaviour is cauced by combination {comutativity vs. nonassociativity of [0]})
How I can see your definition of [0] make correct results. It is very interesting that the [0] is not associative and ,,, a [0] a [0] ...[0] a {b+1 times} = a [0] (a [0] a [0] ... [0] a {b times}) is correct
I realize that the formula can be corect, bat other bracket construction(with identical number of operands) needn't to be equal betwen each other ,,, and than ,,, if some bracket construction isn't equal to another,the operation cannot be associative.
Good question is ,whether are here some other formulas with bracket construction ,which are equal betwen each other?
it could by interesting to find out behaviour of this special nonasociativity (I think ,this special behaviour is cauced by combination {comutativity vs. nonassociativity of [0]})