Second Principle of Mathematical Induction

前幾個禮拜我有堂課沒去上，在圖書館念書。那節課老師證了各種Principle的等價，我在放學後也有做一點，不過只有MI的部份。今天有題題目是要用PMI去證SPMI，我看同學是用LNNP來做，下面試著做做看，不確定是否正確。

\( \because \) 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 $$ S = N $$