I am new to the study of analysis and I decided to start with Terence's book in my endeavor. I want to show my "proof" of backwards induction since I have some difficulty in understanding this. I want to now if my proof is correct or have some error, because if have,$a $can't infer that. Any feedback is appreciated.
Let $n$ be a natural number, and let $P(m)$ be a property pertaining to the natural numbers such that whenever $P(m\text{++})$ is true, then $P(m)$ is true. Suppose that $P(n)$ is also true. Prove that $P(m)$ is true for all natural numbers $m ≤ n$; this is known as the principle of backwards induction. (Hint: apply induction to the variable $n$.)
First i want to show $P(m)$ is true $\forall$ $0\geq m$.
H1: $\forall m$ $P(m\text{++})\implies P(m)$
H2: $P(0)$
C: $P(m)$ is true $\forall$ $0\geq m$.
$0\geq m$ means $0=m+a$ for some natural number $a$, then $m=a=0$ for corollary 2.2.9. But $P(0)$ is true for H2, then the case $n=0$ is proved.
Suppose now that works for $n$ and prove $n\text{++}$. then:
H1: $\forall m$ $P(m\text{++})\implies P(m)$
H2: $P(n)\implies P(m)$ $\forall$ $n\geq m$
H3: $P(n\text{++})$.
In H1, for $m=n$ we have $P(n\text{++})\implies P(n)$ and for H2 we now $P(n)\implies P(m)$ (specifically for $n=m$), then $P(n\text{++})\implies P(m)$ for $n\text{++}>m$. We need to prove that works for $n\text{++}=m$ but for that $P(n\text{++})$ is true for H3. We conclude that $P(n\text{++})\implies P(m)$ $\forall$ $n\text{++}\geq m$.