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Proposition $2.2.14$ (Strong principle of induction). Let $m_0$ be a natural number, and let $P(m)$ be a property pertaining to an arbitrary natural number $m$. Suppose that for each $m \geq m_0$, we have the following implication: if $P(m')$ is true for all natural numbers $m_0 \leq m' < m$, then $P(m)$ is also true. (In particular, this means that $P(m_0)$ is true since in this case, the hypothesis is vacuous .) Then we can conclude that $P(m)$ is true for all natural numbers $m\geq m_0.$

My Proof: Let $Q(n)$ be the property that $P(m)$ holds for all $m_0\leq m

Base Case: $Q(0)$ is vacuously true since there are not natural numbers less than $0$.

Inductive Step: Suppose $Q(n)$ holds true. Then this means that $P(m)$ is true for all $m_0\leq m

We have to show that $Q(n++)$ is true, where $n++$ denotes the successor of $n$. This is equivalent to proving that $P(n)$ holds because $P(m)$ is already true for all natural numbers $m_0\leq m

One way to fix this could be to let $Q(n)$ be the property that if $P(m)$ holds for all $m_0\leq m

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    Why the downvote? – 2017-01-19
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    I have read the answers to this question before, but none of them are satisfactory. They do not use weak induction to prove strong induction. And also they do not use Tao hint to construct a solution. – 2017-01-19
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    The base case is $Q(m_0)$, not $Q(0)$. Then to obtain that $Q(n+1)$ is true you just apply the second sentence of the hypothesis to prove that $P(n)$ is true, and $Q(n+1)$ follows trivially. – 2017-01-19
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    @Momo could you please sketch how we might obtain $Q(n+1)$. – 2017-01-19
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    I added some details in a separate answer. – 2017-01-19
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    @Momo If we want to prove using the principle of mathematical induction (Axiom 2.5), don't we need the base case of Q(0)? – 2018-10-18

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You've probably figured this out by now but here's my take:

We wanna show $Q(n)$ is true for $n\ge m_0$ given $Q(n)\Rightarrow P(n)$ by induction on $n$.

Let $n\le m_0$. Then $Q(n)$ is vacuously true. These are a bunch of base cases. Now suppose that $Q(n)$ is true for some $n\ge m_0$. Well $Q(n)\Rightarrow P(n)$ so $P(n)$ is true. This means that $P(m')$ is true for $m_0\le m' which is another way of saying $Q(n+1)$. So the induction is closed, and $Q(n)$ for every $n\ge m_0$.

I was confused similarly. The trick is that we need to suppose $Q(n)\Rightarrow P(n)$ because that's what strong induction is all about, and the weak induction to show $Q(n)$ uses that fact but is kinda orthogonal to it.

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$Q(m_0)$ is vacuously true.

Suppose Q(n) is true. This means that $P(n)$ is true for all $m_0\le m

From the implication in the hypothesis, which can be rewritten (with a slight change of notation) as:

"if $P(m)$ is true for all natural numbers $m_0 \leq m < n$, then $P(n)$ is also true."

it follows that $P(n)$ is true. As $P(m)$ is true for $m

$P(n)$ is true for all $m_0\le m

So $Q(n+1)$ is true.

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I think that a rewriting of the proposition will help in resolving the problem. What Tao actually meant was:

Proposition $2.2.14$ (Strong principle of induction). Let $m_0$ be a natural number, and let $P(m)$ be a property pertaining to an arbitrary natural number $m$ such that for each $m \geq m_0$, we have the following implication: if $P(m')$ is true for all natural numbers $m_0 \leq m' < m$, then $P(m)$ is also true. (In particular, this means that $P(m_0)$ is true since in this case, the hypothesis is vacuous .) Then we can conclude that $P(m)$ is true for all natural numbers $m\geq m_0.$

Thus the fact that $P(n)$ holds is automatically implied by the definition of $P(n)$. What that means is that whenever we say $P(n)$ holds for some $n\geq m_0$ then it is equivalent to saying that $P(m)$ holds for all natural numbers $m_0\leq m

This is what I think, but maybe I am wrong.