On 21 January 2014 08:38, Bruno Marchal <[email protected]> wrote: > > Are the following laws? I don't put the last outer parenthesis for reason > of readability. > > p -> p >
This is a law because p -> q is equivalent to (~p V q) and (p V ~p) must be (true OR false), or (false OR true) which are both true > > (p & q) -> p > using (~p V q) gives (~(p & q) V q) ... using 0 and 1 for false and true ... (0,0), (0,1) and (1,0) give 1, (1,1) gives 1 ... so this is true. So it is a law. I think. > (p & q) -> q > Hmm. (~(p & q) V q) is ... the same as above. > > p -> (p V q) > (~p V (p V q)) must be true because of the p V ~p that's in there (as per the first one) > q -> (p V q) > Is the same...hm, these are all laws (apparently). I feel as though I'm probably missing something and getting this all wrong. Have I misunderstood something ? -- 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.
