Strong induction. What is responsible for basis step?

Question

I have no problems with weak induction. There is a basis step and inductive step. But it seems that basis step is missing in the strong induction. It says that$$(\forall n[ \forall m<n, P(m)\implies P(n)]) \implies \forall n P(n))$$ If it is true, then we can proof that $\forall n \in \mathbb{N} (n<n)$ which can't be true. We have $P(m)$ which says that $1<1, 2<2,...,(n-1)<(n-1)$. From here we can construct implication by adding $n-m$ to the all inequalities $m<m$ s.t. $m<n$. So we have $n<n$.

Answer 1

Such a base case is automatically included in the strong induction hypothesis. Indeed, it applies to well-ordered sets; if $0$ is the minimal element of $X$, then by specializing the hypothesis "if $P(m)$ is true for all $m\in X$ with $m <_X n$, then $P(n)$ is also true" to the $n= 0$ case, we automatically obtain that $P(0)$ is true. (Note that a "for all $x$ which satisfies $P$,$\dotsc$" statement is vacuously true if there is no such $x$ that satisfies the property $P$.)

A subtle, you assume the consecuence $n<n$ from the strong hypothesis. But it is false, we kown that $n<n$ is false. So we have not proven the implication.