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We will use:   to prove $p$ implies $q$, one should assume the hypothesis $p$ and prove the conclusion $q$.


Use induction to prove that if a simple connected graph $G$ has at least 3
vertices, and if each vertex is of degree 2, then $G$ is a cycle.

Proof by induction on $|V(G)|$

{\bf Base case: } Suppose $|V(G)| = 3$.  Then each vertex of $G$ is adjacent to the other two vertices of $G$.  Thus $G$ is the complete graph on 3 vertices.  Thus $G =K_3$.  Note $K_3$ is a cycle.

Let $S(n)$  be the statement that ``If $G$ is a simple connected graph  with  $n$
vertices, and if each vertex is of degree 2, then $G$ is a cycle.''

We need to prove that $S(n)$ implies $S(n+1)$.

Thus we assume the hypothesis, $S(n)$, and prove the conclusion, $S(n+1)$.
\eject
{\bf Induction hypothesis:}  If $G'$ is a simple connected graph with  $n$
vertices, and if each vertex is of degree 2, then $G'$ is a cycle.

{\bf Claim: } If $G$ is a simple connected graph with  $n+1$
vertices, and if each vertex is of degree 2, then $G$ is a cycle.

To prove that the claim is true, we again assume the hypothesis and prove the conclusion:

Suppose $G$ is a simple connected graph with  $n+1$
vertices, and each vertex is of degree 2.

Claim: $G$ is a cycle.

We need to use both the hypothesis that $G$ is a simple connected graph with  $n+1$
vertices, and each vertex is of degree 2,  as well as the induction hypothesis.  To use the induction hypothesis, we need to create a graph with $n$ vertices that satisfies the hypotheses of the induction hypothesis.

Let $v$ be a vertex of $G$.  Let $N(v) = \{x, y\}$

Let $G^* = G - v$.  Note $G^*$ does not satisfy the  hypotheses of the induction hypothesis ($G^*$  has 2 vertices with degree 1, but we need all vertices to have degree 2 to use the induction hypothesis).
So instead 

Let $G' = (V', E')$ where $V' = V(G) - v$ and $E' = E(G) \cup \{<x, y>\} - \{<v, x>, <y, v>\}$

Note $G'$ is a simple connected graph with  $n$
vertices, and each vertex is of degree 2.  Thus by the induction hypothesis, $G'$ is a cycle.

We now need to show $G$ is a cycle.  Sometimes it helps to be specific.

 $G'$ is a cycle means we can write  $G'$ as the cycle, $x, u_1, u_2, ..., u_{n-2}, y, x$.  

Then $G$ is the cycle 
 $v, x, u_1, u_2, ..., u_{n-2}, y, v$~~~~~~~








\eject
~
\eject

Note induction gives you lots of hypothesis to work with.  

Let $S(n)$ be the statement that $p(n)$ implies $q(n)$ 

where $p(n)$ is the hypothesis that depends on $n$ and $q(n)$ is the conclusion that depends on $n$ (for example if $G$ has $n$ vertices).


To use induction, you prove

(1) Base case:  $S(n_0)$ is true.

and that

(2)  $S(n)$ implies $S(n+1)$


(1) and (2) implies that  $S(n_0)$ is true which implies that  $S(n_0 +1)$ is true which implies that  $S(n_0 +2)$ is true which implies that 
 $S(n_0 +3)$ is true which implies that ....  \hfil \break

Thus  $S(m)$ is true for any integer $m \geq n_0$.

\eject

To prove $S(n)$ implies $S(n+1)$, you assume hypothesis and prove conclusion:

Induction hypothesis:  Suppose $S(n)$ is true.

Claim:  $S(n+1)$ is true

Note $S(n+1)$ is the statement that $p(n+1)$ implies $q(n+1)$.  Thus we need to prove:

Claim:  $p(n+1)$ implies $q(n+1)$

To prove $p(n+1)$ implies $q(n+1)$, you assume hypothesis and prove conclusion:


Suppose $p(n+1)$ holds.

Claim $q(n+1)$ holds.


Thus to prove the claim that  $q(n+1)$ is true, you can use both $p(n+1)$ and $S(n)$.  Thus you take an arbitrary graph, $G$, satisfying the hypothesis $p(n+1)$, modify this graph, creating a new graph, $G'$ so that you can use $S(n)$.  %In our example, we inducted on the number of vertices, so we removed a vertex (









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