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December 20th, 2010, 10:47 PM   #1
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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?
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December 21st, 2010, 11:15 PM   #2
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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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