# Proving Cut Vertices in Simple Graphs

• mooncrater

## Homework Statement

The given question is:
Show that a simple graph with at least two vertices has at least two vertices that are not cut vertices.

## Homework Equations

Cut vertices: The removal of cut vertices and all edges incident to them produces a subgraph more connected components than in the original graph.

## The Attempt at a Solution

What I did:
Assuming a ##S## represents that a graph is simple, and ##x## is the number of cut vertices in it. So we have to prove $$S→x>=2$$. We can also prove it through contradiction by proving the negation of the above statement to be always false( that is, a contradiction). So, we can write ##S→x>=2## as
~S√(x>=2) (where √ represents a disjunction)
Whose negation is
S^(x<2).
So, if we prove that having a simple graph having all vertices as cut-vertices or just one non-cut vertex is a contradiction, then does that prove our original statement? Do we still have to prove that 2 non-cut vertices are necessary?
Thanks
Moon

Last edited by a moderator:
Assuming a ##S## represents that a graph is simple, and ##x## is the number of cut vertices in it. So we have to prove $$S→x>=2$$.
No, that's not what is being asked. They are asking you to prove that, if ##n(S)\geq 2##,. where ##n(S)## is the number of vertices in ##S##, we have:

$$n(S)-x(S)\geq 2$$

$$x(S)\neq n(S)\wedge x(S)\neq n(S)-1$$
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