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 December 20th, 2010, 11:47 PM #1 Newbie   Joined: Sep 2009 Posts: 20 Thanks: 0 logical form of "need not" Hi. I met with a problem: Set-up: G = (V, E) |V| = n delta(G) = minimum of degrees of vertices in G, i.e. the minimum degree of G Problem: "Let G be a graph of order n. If delta(G) >= (n-2)/2, then show that G need not be connected." I came up with a logical form of that problem as follows: Set-up: U(n) is the universe set of all graphs of order n. Conn is the universe set of all connected graphs. proposition p(x) is "delta(x) >= (n-2)/2." proposition q(y) is "y is connected." -q(y) is the negative of q(y). Logical form: All G belonging to U(n), p(G) -> q(G) or -q(G) If my logical form is right, the problem will be vacuous because q(G) or -q(G) is always true so the implication is also true. My logical form is really right one for the problem? December 22nd, 2010, 12:15 AM #2 Newbie   Joined: Sep 2009 Posts: 20 Thanks: 0 Re: logical form of "need not" I've resolved this problem. The "need not" means a negative of a implication. Logical form: All G belonging to U(n), -(p(G) -> q(G)) Tags form, logical, need not Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post johnmath Math Books 6 January 27th, 2013 04:13 PM SedaKhold Calculus 0 February 13th, 2012 12:45 PM The Chaz Calculus 1 August 5th, 2011 10:03 PM katie0127 Advanced Statistics 0 December 3rd, 2008 02:54 PM Ujjwal Number Theory 2 September 29th, 2008 08:06 AM

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