Different patterns of parentheses must give different functions; the parentheses can be unambiguously derived from the truth table. Let B be the boolean function. First note that p_n=T must always give B(p_1...)=True, so fix p_n = F. Now p_{n-1} must be either the head or the tail of an implication, so we have either .... -> p_{n-1} ) -> F or ...-> (p_{n-1} -> F) We are in the second case if and only if p_{n-1}=F always yields B(p_1...)=T, in which case we fix p_{n-1}=T and proceed recursively. Otherwise the possibilities are ... -> p_{n-2} ) -> p_{n-1} ) ->F or ... -> (p_{n-2} -> p_{n-1}) -> F We are in the second case if and only if (p_{n-2} -> p_{n-1})=F always yields B(p_1...)=T, and we can keep going in this way to get the entire parenthesization. On Sat, Dec 2, 2017 at 10:20 AM, Allan Wechsler <acwacw@gmail.com> wrote:
THERE we go. I knew somebody more energetic than me would step up. Thank you, Tom.
On Sat, Dec 2, 2017 at 7:52 AM, Tom Karzes <karzes@sonic.net> wrote:
I think so too. I checked up to 12 variables and found no duplicates.
Tom
Allan Wechsler writes:
If I were just a little more diligent I would write the approximately ten-line Haskell program to look for different application-sequences of the boolean implication-function to sequences of N boolean arguments with identical truth-tables. (My intuition is that there aren't any; that these are all distinct.)
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