If $P(n_0)$ is false then the implication $P(n_0) \implies P(n_0+1)$ is true (vacuously, as you've pointed out), however, we cannot deduce the truth or otherwise of $P(n_0+1)$ in this case.
The hardest part in a proof by induction is proving $P(n) \implies P(n+1).$ If you've proved this, then all you have to do is find a suitable $n_0$ such that $P(n_0)$ is true. If $P(n_0)$ is false, but $P(n) \implies P(n+1)$, then pick a different value for $n_0$ (assuming that one exists). If you can find just one such $n_0$ (and if you've proved that $ \forall n, P(n) \implies P(n+1)$, then $P(n)$ is true for all $n \geqslant n_0$, by the principle of mathematical induction.
In this case, pick a different value of $n_0$ until $P(n_0)$ is true.
Regarding your last sentence, if $P(n_0)$ is true and $P(n) \implies P(n+1)$, then, by the property of implication, $P(n_0+1)$ is true, as is $P(n_0+2), P(n_0+3), ...$, so you'd never have $P(n_1)$ being false (for any $n_1 \geqslant n_0$)!
First of all, there is an implicit assumption in your Weak Principle of Induction that $P(n)$ is a predicate defined on the natural numbers (integers $n \ge 0$). Likely if you carefully read the context where you found this, it will actually say somewhere that they are dealing with predicates defined on integers greater than or equal to the base case. So when it says "for all $n$" it actually means "for all integers $n \ge 0$".
Second, note that your Weak Principle explicitly states "$P(0)$ is true" as one of the hypotheses. So it is not a case that "$P(0)$ is vacuously true". That $P(0)$ is true is something you have to show explicitly for your predicate before you can apply this "Weak Principle".
As Gerry Myerson has pointed out, the equivalence is just a matter of re-indexing. If all you knew was the weak principle, and you have a predicate $P(n)$ defined on all integers $n \ge x$ where $x$ is some arbitrary integer. And if you can prove both that $P(x)$ is true, and that for all $n \ge x, P(n) \implies P(n+1)$, then you can show $P(n)$ is true for all $n \ge x$ as follows:
- Define the predicate $Q(m) = P(m + x), m\ge 0$
- $Q(0) = P(x)$. Since $P(x)$ is true, so is $Q(0)$.
- $Q(m) \equiv P(m + x) \implies P((m + x) + 1) \equiv Q(m + 1)$, so $Q(m) \implies Q(m+1)$ for all $m \ge 0$.
- By the weak principle, $Q(m)$ is true for all $m\ge 0$.
- That is, $P(m + x)$ is true for all $m \ge 0$.
- Setting $n = m + x, P(n)$ is true for all $n \ge x$.
There is a stronger version of induction. That version is probably where you got the idea of $P(0)$ being "vacuously true". That version (for natural numbers) is
If $P(n)$ is defined for $n \in \Bbb N$, and if $\forall n\in \Bbb N, (\forall m < n, P(m)) \implies P(n)$, then $\forall n \in \Bbb N, P(n)$.
This is stronger than the versions you have been discussing, since there are predicates $P$ for which just knowing $P(n-1)$ is not enough to show $P(n)$, but for which knowing $P(m)$ for every $m < n$ is enough to show $P(n)$. Such predicates can be proven always true from this version of induction, but not using the versions you gave.
You may note that there is no explicit mention of a base case in this version. But this version still has a base case. It is just that the induction case implies the base case, so it doesn't need to be mentioned separately. Since there is no $m < 0$ in $\Bbb N$, the statement $\forall m < 0, P(m)$ is "vacuously" true. And therefore if the induction case has been shown true, $P(0)$ must also be true.
Note that it is "$\forall m < 0, P(m)$" that is "vacuously true" (true because there is no $m < 0$ rather than any property of $P$), not $P(0)$, for which the phrase "vacuously true" has no definition. $P(0)$ is just a consequence of the vacuous truth.
Best Answer
Trivial source of examples: if $\sum_{k=0}^na_k=b_n$ for all $n\ge0$, $\sum_{k=0}^na_k=b_n+c$ meets your requirement for any constant $c\ne0$.
What you're missing is that we really do need the base case. The principle of induction says if two things are true (base case & inductive step) the result follows.