## 2013年11月25日 星期一

### Second Principle of Mathematical Induction

$$\because$$ PMI $$\leftrightarrow$$ LNNP, we proove by suppose LNNP holds.

Let $$S = \{ n \in N : 1 \in S : \{ 1, 2, 3 \dots n\} \subseteq S \implies n+1 \in S \}$$ Suppose $$N - S \not = \emptyset$$ By LNNP $$\exists m \in N - S , m \not = 1$$ is the smallest member. Therefore $$m - 1 \in S$$ Furthermore $$m - 2, m - 3 \dots 1 \in S$$ $$\because 1, 2, 3 \dots m - 1 \in S \\ \implies m \in S$$ Thus we have a contradiction. Hence $$S = N$$