   Propositions
  I'm doing an assignment which requires me to find the negation of statements. There are 12 statements and each "apparently" has a negation. I have to match up each statement with its negation and so far I've done it alright. I've got the problem below though...Is:Quote:∃m∈M,∃p∈P, S(p,m)The negation of:Quote:∀p∈P,∀m∈M, ~S(p,m)in other words...Is:Quote:There is at least one movie that some person has seen.The negation of:Quote:Nobody has seen any of the moviesI understand that to find the negation of (for example):Quote:∃m∈M,∀p∈P, ~S(p,m)you more or less swap existential for universal symbols (vice versa) and add or remove the compliment/not symbol:Quote:∀m∈M,∃p∈P, S(p,m)but I'm left with the problem above... where there are no negated statements left where the sets are in the same order.Cheers to anyone that can help.   
  Well we're not allowed to do your homework for you, but I think this is legal help:First realise that the binding order of the predicates allows you to bracket expressions from the right. Another way to look at this is pulling the first predicate out into a hypothesis (under some interpretation of the deduction theorem, and its converse). Once this is done, you can simplify the inner expression using laws you're more comfortable with.I hope you'd agree that the following statements are equivalent:1. There is a movie m and a person p for which p has seen m.2. There is a movie m for which (There is a person who has seen m). Here, m is a free variable in the bracketed expression. 3. There is a movie for which (Not all people haven't seen it).4. There is a movie for which not (All people haven't seen it).5. Not all movies are such that (All people haven't seen it).6. Not (For all movies m and people p : p hasn't seen m)And so the negation follows by removing the initial 'Not', exactly as you suspected. I haven't used anything other than associativity laws (i.e. rearranging brackets) and DeMorgan's law. Can you rewrite the argument in formal predicate calculus?Admiral   ;Quote:Original post by TheAdmiralWell we're not allowed to do your homework for you, but I think this is legal help:Admiral,Yes, your reply is fine. Thank you (!) for paying attention to and honoring my policies. I appreciate it, sincerely. I like your discussion, which I believe to be clarifying without doing the OP's homework!;  Thanks for your reply.. but before I start to read the rest of it, can I just say how degrading it is when you're accused of asking people to do your homework for you. I'm aware of this issue in forums so I can assure you that I am merely trying to understand the rules of negation - I didn't want anyone to do my homework no matter how badly I phrased it.Quote:but I'm left with the problem above... where there are no negated statements left where the sets are in the same order.;Quote:Original post by TheAdmiralWell we're not allowed to do your homework for you, but I think this is legal help:First realise that the binding order of the predicates allows you to bracket expressions from the right. Another way to look at this is pulling the first predicate out into a hypothesis (under some interpretation of the deduction theorem, and its converse). Once this is done, you can simplify the inner expression using laws you're more comfortable with.I hope you'd agree that the following statements are equivalent:1. There is a movie m and a person p for which p has seen m.2. There is a movie m for which (There is a person who has seen m). Here, m is a free variable in the bracketed expression. 3. There is a movie for which (Not all people haven't seen it).4. There is a movie for which not (All people haven't seen it).5. Not all movies are such that (All people haven't seen it).6. Not (For all movies m and people p : p hasn't seen m)And so the negation follows by removing the initial 'Not', exactly as you suspected. I haven't used anything other than associativity laws (i.e. rearranging brackets) and DeMorgan's law. Can you rewrite the argument in formal predicate calculus?AdmiralI'm not sure I understand what you mean... could you perhaps convert your examples to symbols? I'm just wondering if it's possible to re-order parts of the proposition to get it's negation like in my first example:Quote:Is:    Quote:    ∃m∈M,∃p∈P, S(p,m)The negation of:    Quote:    ∀p∈P,∀m∈M, ~S(p,m)in other words...Is:    There is at least one movie that some person has seen.The negation of:    Nobody has seen any of the moviesI don't want to know if that negation is correct, I just need to know if you can re-order parts of the proposition.Cheers.;  This is a somewhat mechanical way of doing this, always correctly without too much of thinking if you have the initial equation at hand.So, for instance, negating(∀ε ∈ R+)(∃nε ∈ N)(∀n ∈ N)(n ≥ nε ->  |xn − x| < ε)neg(∀ε ∈ R+)(∃nε ∈ N)(∀n ∈ N)(n ≥ nε -> |xn – x| < ε)= (∃ε ∈ R+)neg(∃nε ∈ N)(∀n ∈ N)(n ≥ nε -> |xn − x| < ε)= (∃ε ∈ R+)(∀nε ∈ N)neg(∀n ∈ N)(n ≥ nε -> |xn − x| < ε)= (∃ε ∈ R+)(∀nε ∈ N)(∃n ∈ N)neg(n ≥ nε -> |xn − x| < ε)= (∃ε ∈ R+)(∀nε ∈ N)(∃n ∈ N)(n ≥ nε ∧ neg(|xn − x| < ε))= (∃ε ∈ R+)(∀nε ∈ N)(∃n ∈ N)(n ≥ nε ∧ |xn − x| ≥ ε)You can just "transport" the negation inwards until you are done. Just use the usual rules provided in any book or course regarding negating quantifiers, implications, inequalities and so forth whilst moving inwards.[Edited by - Naurava kulkuri on September 8, 2007 3:56:48 AM]