December 20th, 2010, 10:47 PM  #1 
Newbie Joined: Sep 2009 Posts: 20 Thanks: 0  logical form of "need not"
Hi. I met with a problem: Setup: 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) >= (n2)/2, then show that G need not be connected." I came up with a logical form of that problem as follows: Setup: 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) >= (n2)/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 21st, 2010, 11:15 PM  #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)) 

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