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 sav08:deciding_quantifier-free_fol [2008/04/23 06:49]vkuncak sav08:deciding_quantifier-free_fol [2015/04/21 17:30] (current) Both sides previous revision Previous revision 2009/05/05 22:49 vkuncak 2008/04/23 06:49 vkuncak 2008/04/17 13:41 vkuncak 2008/04/17 13:38 vkuncak 2008/04/17 13:38 vkuncak 2008/04/17 13:33 vkuncak 2008/04/17 13:30 vkuncak 2008/04/17 13:23 vkuncak 2008/04/17 13:03 vkuncak 2008/04/17 12:56 vkuncak 2008/04/17 12:55 vkuncak 2008/04/17 12:52 vkuncak 2008/04/17 12:50 vkuncak 2008/04/17 12:49 vkuncak 2008/04/17 12:48 vkuncak 2008/04/17 12:47 vkuncak 2008/04/16 23:37 vkuncak 2008/04/16 23:37 vkuncak 2008/04/16 23:34 vkuncak created Next revision Previous revision 2009/05/05 22:49 vkuncak 2008/04/23 06:49 vkuncak 2008/04/17 13:41 vkuncak 2008/04/17 13:38 vkuncak 2008/04/17 13:38 vkuncak 2008/04/17 13:33 vkuncak 2008/04/17 13:30 vkuncak 2008/04/17 13:23 vkuncak 2008/04/17 13:03 vkuncak 2008/04/17 12:56 vkuncak 2008/04/17 12:55 vkuncak 2008/04/17 12:52 vkuncak 2008/04/17 12:50 vkuncak 2008/04/17 12:49 vkuncak 2008/04/17 12:48 vkuncak 2008/04/17 12:47 vkuncak 2008/04/16 23:37 vkuncak 2008/04/16 23:37 vkuncak 2008/04/16 23:34 vkuncak created Line 8: Line 8: Example: is the following formula satisfiable:​ Example: is the following formula satisfiable:​ - \[ + \begin{equation*} a=f(a) \land a \neq b a=f(a) \land a \neq b - \] + \end{equation*} is the following is the following - \[ + \begin{equation*} - ​(f(a)=b \lor (f(f(f(a))) = a \land f(a) \neq f(b))) \land f(f(f(f(a)) \neq b \land b=f(b) + ​(f(a)=b \lor (f(f(f(a))) = a \land f(a) = f(b))) \land f(f(f(f(a)) \neq b \land b=f(b) - \] + \end{equation*}