On 21 January 2014 22:29, Bruno Marchal <marc...@ulb.ac.be> 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 everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
