Are all defined bracketings equivalent for a partial associative operator?

Hi, I cannot find an answer to this question, but it seems intuitively true. Say an operator is partial and associative in the following sense:

a * (b * c) = (a * b) * c if both sides are defined

That is, not the stronger constraint that "either both sides are defined and equal or both sides are undefined". In this case, is it true that all possible defined bracketings of a1 * a2 *.... * aN are equivalent? If so, what would be a proof argument? The standard induction-based proof used for total operators does not seem to work as some bracketings may not be defined:

http://unafold.math.rpi.edu/Teaching...Binary_Ops.pdf

Re: Are all defined bracketings equivalent for a partial associative operator?

Quote:

Originally Posted by

**mraza** In this case, is it true that all possible defined bracketings of a1 * a2 *.... * aN are equivalent?

What do you mean by saying that the bracketings are equivalent? That expressions with all possible bracketings are defined simultaneously and are equal when defined? It could be that a * (b * c) is defined while (a * b) * c is not; then the fact that * is associative says nothing for these a, b and c. Or are you saying that they are equivalent in the sense that only those expressions with all possible bracketings that are defined are equal?

Re: Are all defined bracketings equivalent for a partial associative operator?

Yes I asked if "all DEFINED bracketings" are equivalent, that is, there may be many bracketings of a1*....*aN of which some are defined and some are not. I want to know whether all the defined bracketings will be equal under this definition of associativity. Or if not, a counterexample will be helpful.