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¿Question #5186?
[Deleted user]: [deleted question]

kurthl33t: r(p-q)/e^(x-q) > 1, therefore r is TRUE for all real numbers
TunaSled: Asshole. Asshole I say!
MisterCranky: I am going to ask DHEK what he thinks, and get back to you.
sirjeyhmis: Is "q" really a constant?
DerDa: p=(?q?Q: B(q)); q=(?q?C={c?Q: X(c)}: ¬B(q)); r=(C?Ø); p?q ? (?q?Q: B(q))?(?q?C: ¬B(q))?(C?Q) ? (?q?Q: B(q))?(?q?C: ¬B(q))?(?q?C: q?Q) ? (?q?C: B(q))?(?q?C: ¬B(q))?(?q?C: q?Q) ? C={q?Q: B(q)?¬B(q)} ? C=Ø ? ¬C?Ø ? r ?
Wabete: yikes and away
ricardofo: It is already answered above
Sideburn: ~r returns the complex conjugate of the right hand side of r. With r representing "Bill Cleary has asked a GeekQuestion", ~r would represent "Bill Cleary has NOT asked a GeekQuestion". Given that (p ^ q) = TRUE, all GeekQuestions blow and no GeekQuestions by Bill Cleary blow. Bill Cleary CLEARLY CANNOT have asked a GeekQuestion, therefore ~r is TRUE
bazik123: p^q => bill cleary's=0 => ~r
BUDDA: p and q are in conflict so r must be false
bigbadsteve: that one just about reduced me to tears
DHEK: I'm not even going to bother with this... Gee... The things you think up!
screwnut75: ummm....next question please.
naruto55: ............hu?
aleo09: thanks for asking
dougpartridge: I'm just as happy not to, thanks
Joel Baxter: Your mother.
richlowe32: Ow. New years day and my head hurts more now. Thanks!
dkwd: Who is this Bill Cleary person anyways? And why does he feel like he needs to have to try to plant the seeds of his non-blow-ness? Is he concerned?
Cut & Space: Huh?? So when's the Superbowl again?
brainrob: My brain hurts now ...
mrspank: So much for remembering geometry theorems.
vetinari7878: Well, if p and q are both true, then r is obviously false. Ad absurdum, if r was true then p or q had to be false.
WatchmanX2000: No.
POvidiusNaso: Since 1+(x^-r)=.0555 and therefore r=712, and 712 is a true number, ~r is also true, being the 4th letter in the game "Puerto Rico."
hacksword: It's turtles all the way down.
scriptorum: I'll have to dig out my logic textbook for this answer ... hold on, looking for it ... once second ... ah yes, here it is. You're an idiot.
rawdawg: your goofy
LemonyFresh: No.
confuse_a_cat: Let Ga = a is a GeekQuestion, Ba = a blows, Pa = a is of Bill Cleary. A symbolic representation of p is (x)(Gx -> Bx), and a representation of q is ~(there exists an x)(Gx ^ Px ^ Bx). Similarly, we represent r with (there exists an x)(Px ^ Gx). Now, assume r. We'll name the instance that is a GeekQuestion of Bill Cleary a, thus Pa ^ Ga. From Pa ^ Ga we derive Ga. From Ga and (x)(Gx -> Bx) we derive Ba. From Pa ^ Ga and Ba we derive Pa ^ Ga ^ Ba, from which we derive (there exists an x)(Px ^ Gx ^ Bx). Using reductio ad absurdum, we derive ~r. Question: does this answer make me nerdy, or would it make me geeky, instead?
Megadev: uhhhh..yeah
CraftyShafty: Me like pictures.
jasta6: get a life!
djlg: Um, well its obvious isn't it? Given p and q, BC can't have asked a question. ie, ~r=TRUE
Loquutus: ?
puffinge: That sure is a complicated way to say I suck.
Lord_Prussian: um, yah
chaddyboy_2000: Yippee!
skelebone: Goddammit Cranky, lay off the fucking math questions.
CDRodeffer: Mind your own p's and q's, Josh.
Spielfreak: Oh please, didn't your parents ever tell you to mind your P's and Q's...
jttm: Umm... basic math theory would state that the inverse is true actually.