Why are there multiple base cases in this strong induction? My understanding of needing a base case, in general, is that after proving the induction step, we can assert that the proposition is true for all values from the base case.
This question ($∀n ∈ Z,  n≥12$) $\implies$ ( $∃x, y ∈N$ such that $n= 4x + 5y$) uses 3 bases cases for its strong induction solution:

Base cases: $n = 12, 13, 14, 15$
Clearly, $12= 4(3) + 5(0)$, so $P(12)$ is true.
Also, $13= 4(2) + 5(1)$, so $P(13)$ is true.
And, $14= 4(1) + 5(2)$, so $P(14)$ is true.
And, $15= 4(0) + 5(3)$, so $P(15)$ is true

Meanwhile I only used base case, $n=12$. Their rationale for providing multiple cases is as follows

If we didn’t prove $P(13), P(14), P(15)$ as base cases, then the
inductive step to get $k+1 = 13, 14,15$ will fail, since proving
these assume that $P(r)$, for $ r < 12$, to be true, which we didn’t prove.
By establishing base cases $n = 12$ to $n = 15$, the inductive step can then work forward from $n ≥ 15$.


The first line starting from "If we..will fail" already contradicts with my understanding of the base case that I mentioned above as proving base case of $12$ should suffice to prove inductive steps for $k+1=13,14,15$.
The part where they said "since proving..we didn't prove" doesn't even make sense to me.Why are they talking about $ r < 12$.
And the last sentence says "work forward from n ≥ 15" confuses me, shouldn't we want it to work from $n ≥ 12$.
My main question is essentially why we do need multiple base cases here? The sub-questions are the 3 paragraphs talking about the solution's rationale for providing multiple cases.
Induction Step(included after a comment by fleabloods):

Note: If the sub-questions should be separately posted, do let me know

Your understanding of base case is incorrect. The base cases are simply those cases needed to provide a sufficient foundation for the induction step. 
Here the induction step to prove that $P(k+1)$ is true relies on knowing that $P(k-3)$ is true: if there are $x,y\in\Bbb N$ such that $4x+5y=k-3$, then clearly $$4(x+1)+5y=k+1\;,$$ where $x+1,y\in\Bbb N$, so $P(k+1)$ is true. If we’ve verified only that $P(12)$ is true, that argument won’t tell us that $P(13)$ is true: in that case $k=12$, so $k-3=9$, and we’d need to know that $P(9)$ is true in order to use that argument. But as your source says, we didn’t verify that $P(9)$ is true, so that argument simply isn’t valid.
As it happens, $P(9)$ is true, so if we’d checked it as a base case, we could have used the induction step to see that $P(13)$ is true. $P(10)$ is also true, so we could have verified it as part of the base case and used the induction step to see that $P(14)$ is true. But $P(11)$ is not true, so there is no possible way to use the induction step to derive $P(15)$. It turns out that $11$ is the last failure: $12$ is the smallest integer such that $P(n)$ is true for it and every larger integer.
The induction works only after we know that $P(n)$ is true for four consecutive integers $n$: if $P(n)$, $P(n+1)$, $P(n+2)$, and $P(n+3)$ are true, the truth of $P(n)$ allows us to use the induction step to prove that $P(n+4)$ is true, the truth of $P(n+1)$ allows us to use the induction step to prove that $P(n+5)$ is true, and so on. But we need those four consecutive integers: if $P(n)$ and $P(n+1)$ were true, but $P(n+2)$ were not, the induction step would give us the truth of $P(n+4)$ and $P(n+5)$, but not the truth of $P(n+6)$, because for that we’d need to know that $P(n+2)$ was true.
Thus, the induction argument in this example really does use all four parts of the base case, $P(12)$, $P(13)$, $P(14)$, and $P(15)$: without them, the argument falls apart at one of the next four cases.
