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Jethro_T

Discrete Math Problem

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Yes, this is homework but I'm not just dropping a bunch of questions and asking for answers. I'm looking to learn how to solve this type of problem so I can do the rest myself.

¬∀y : ((∃x : P(x,y)) ∨ (∀x : ¬Q(x,y)))

The goal is to rewrite the proposition so that the negations occur only within the predicates. I've been reading through my textbook but I still don't even know where to start with this one. Can someone help?


I'm not sure what the policy on homework is here but I know that people typically look down on people who ask homework related questions. Hopefully it's not a big deal, I'm willing to actually learn and I don't have this class again until Wednesday.

[Edited by - Jethro_T on October 1, 2010 11:32:57 PM]

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I'm drunk right now, try it tomorow. I think I tried converting to enlgisj. Btw, the professor always uses that ":" but my TA always omits it, so no, I didn't mean to use the implication operator. Basically, ignore it I guess. I'll try this stuff tomorow.

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For all values of y, ((∃x : P(x,y)) ∨ (∀x : ¬Q(x,y))) is false

There exists atleast one value for x such that P(x, y) is true

OR

for all values of x, ¬Q(x,y) is true

I don't really know how to take this any farther.

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>>¬∀y : ((∃x : P(x,y)) ∨ (∀x : ¬Q(x,y))

Translates to : NOT FOREACH y, there exist an x such that P(x,y) is true OR FOREACH x NOT Q(x,y) is true


Hint 1) !∀ = ∃
Hint 2) Use Demorgans law

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