On Fri, Dec 1, 2017 at 10:01 AM, James Propp <jamespropp@gmail.com> wrote:
I think Mike is thinking of my question in a different way than I intended.
If p, q, and r range over the set {0,1}, then (p-->q)-->r is one Boolean function on {0,1}^3 and p-->(q-->r) is another; they are different ternary Boolean functions because they have different truth-tables.
Do all five of the Boolean functions obtained by parenthesizing p-->q-->r-->s have distinct truth-tables?
What about the fourteen Boolean functions obtained by parenthesizing p-->q-->r-->s-->t?
Oh, I see, sorry. The truth value of the implication p_1 --> p_2 --> ... --> p_n is itself a function 2^n -> 2, and you'd like to compare different functions of that signature. I don't know, offhand. I need to think about it. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com