K!! Posted March 29, 2008 Share Posted March 29, 2008 Simplify applyin' properties: [math](p\implies q)\implies[p\implies(\sim q\wedge p)].[/math] Link to comment Share on other sites More sharing options...
the tree Posted March 29, 2008 Share Posted March 29, 2008 We really aren't here to do your homework. At a quick glance it looks like you have a contradiction. Link to comment Share on other sites More sharing options...
thedarkshade Posted March 29, 2008 Share Posted March 29, 2008 Simplify applyin' properties: [math](p\implies q)\implies[p\implies(\sim q\wedge p)].[/math] Try to do this with the 'table thing', it's very easy and fun too. Link to comment Share on other sites More sharing options...
the tree Posted March 29, 2008 Share Posted March 29, 2008 Truth tables are fun?! For serious? You might possibly have a very different view of fun. Anyways, the task asked for a specific approach so that's the one he should use since this is clearly homework. (as a side note, my first glance was wrong, very wrong) Link to comment Share on other sites More sharing options...
thedarkshade Posted March 29, 2008 Share Posted March 29, 2008 Truth tables are fun?! For serious? You might possibly have a very different view of fun. Since when 'truth seeking' (even through tables) has not been fun? And it's fully logical, which in this case could apply to the title of the thread:-) Link to comment Share on other sites More sharing options...
YT2095 Posted March 29, 2008 Share Posted March 29, 2008 Since when 'truth seeking' (even through tables) has not been fun? And it's fully logical, which in this case could apply to the title of the thread:-) don`t waste your keyboard ink, he regularly Trolls threads to do with Logic, as evidenced here: http://www.scienceforums.net/forum/s...ad.php?t=27799 Don`t Feed The Troll! Link to comment Share on other sites More sharing options...
Aeternus Posted March 30, 2008 Share Posted March 30, 2008 Simplify applyin' properties: [math](p\implies q)\implies[p\implies(\sim q\wedge p)].[/math] One thing to remember is that [math]a \implies b \equiv a \wedge \neg b [/math], so then you can just apply that, demorgan's law and some of the other basic rules and derive a rather short equivalent formula. Link to comment Share on other sites More sharing options...
thedarkshade Posted March 30, 2008 Share Posted March 30, 2008 Well actually solving this is just a piece of cake, you just need to know what conjunction, disconjunction, equivalence and implication are. The rest is just procedure. Link to comment Share on other sites More sharing options...
the tree Posted March 31, 2008 Share Posted March 31, 2008 You don't need silly long words to solve anything, really. It's surprisingly easy, when you come to look at it though, quite a nice solution for a seemingly convoluted problem. Solving through truth tables just seems incredibly boring to me, it's like getting an ugly numerical solution when you could get a nice exact one (except that it's the process that is different, rather than the end result). Link to comment Share on other sites More sharing options...
Aeternus Posted March 31, 2008 Share Posted March 31, 2008 You don't need silly long words to solve anything, really. It's surprisingly easy, when you come to look at it though, quite a nice solution for a seemingly convoluted problem. Solving through truth tables just seems incredibly boring to me, it's like getting an ugly numerical solution when you could get a nice exact one (except that it's the process that is different, rather than the end result). There's also the issue that as you add more and more variables, basic truth tables become completely infeasible. Link to comment Share on other sites More sharing options...
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