wu :: forums
« wu :: forums - logic »

Welcome, Guest. Please Login or Register.
May 2nd, 2024, 12:53am

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   general
   wanted
(Moderators: SMQ, ThudnBlunder, towr, william wu, Grimbal, Eigenray, Icarus)
   logic
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: logic  (Read 3449 times)
hparty
Newbie
*





   


Gender: male
Posts: 15
logic  
« on: Sep 13th, 2010, 11:24am »
Quote Quote Modify Modify

it is known that the connective "or : v" can be defined using  => (implies) only as followes
(P v Q )<=> ((P => Q )=> Q )  
 
how we can show that we cannot express all the truth functions using only the set  {=>, v} or { ~, <=>}.
 
/where  ~ :not , <=> :equivalence/
IP Logged
towr
wu::riddles Moderator
Uberpuzzler
*****



Some people are average, some are just mean.

   


Gender: male
Posts: 13730
Re: logic  
« Reply #1 on: Sep 13th, 2010, 3:15pm »
Quote Quote Modify Modify

You could probably use exhaustion of the possible sets of truth values.
 
So to start,
 
P Q A:PvQ B:P=>Q
1 1   1      1  
1 0   1      0
0 1   1      1
0 0   0      1

 
Then try to get a new set of truth values with combinations of P,Q,A,B, etc
And at some point you've either covered all 24=16 possible sets, or it's impossible.
 
Or you could note that at least in the case of => (and the redundant addition of v) you can't get negation, because if P=Q=1, no application of =>'s will ever get you 0.
« Last Edit: Sep 13th, 2010, 3:15pm by towr » IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board