On 21 January 2014 22:29, Bruno Marchal <[email protected]> wrote: > > No, it is all good, Liz! > > What about: > > (p V q) -> p >
Using the same formula this is equivalent to(~(p V q) V p), which for (0,1) is 0, hence not a law. > and > > p -> (p & q) > And this is (~p V (p & q)) which is 0 for (1,0), hence also not a law :-) > > What about (still in CPL) the question: > > is (p & q) -> r equivalent with p -> (q -> r) > > is (~(p & q) V r) equal to (~p V (~q V r)) ? or is ~((p & q) & ~r) equal to ~(p & ~~(q & ~r)) i.e. is ~((p & q) & ~r) equal to ~(p & (q & ~r)) I'm going to take a punt and assume the order in which things are ANDed together doesn't matter, in which case the above comes out as equal (equivalent). Did I blow it? > Oh! You did not answer: > > ((COLD & WET) -> ICE) -> ((COLD -> ICE) V (WET -> ICE)) > Eek! That is even more difficult. Luckily you provided something that didn't involve so much typing... ((p & q) -> r) -> ((p -> r) V (q -> r)) Expanding furiously and trying not to make any mistakes... ~((p & q) & ~r) -> ( ~(p & ~r) V ~(q & ~r)) ~(~((p & q) & ~r) & ~(~(p & ~r) V ~(q & ~r))) Um! Assuming for a moment that's correct, we have 8 possible combinations of values for p,q,r r = 0 gives 1 (so that's half the values sorted) r = 1 also gives 1 (so that's the other half) So assuming I expanded it correctly, it's a law. I will try more later... -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
