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

You didn't include what the induction step was!
We can't evaluate their argument if we don't know what the induction step was.
But from their commentary I suspect the induction step is something like:
If it is true for $n$ and for all values between $12$ and $n$ then it should be true for $n-3$  (assuming $n-3$ is between $12$ and $n$);  then $n-3 = 4x + 5y$ for some $x,y$.  So $n+1 = 4(x+1) + 5$.
But what's wrong with that?
Let's try it on $12$.  Let's try to go from $12$ to $13$.  It is true for $12$ and for all values between $12$ and $12$.  So it should be true for $12-3=9$ (assuming $12-3=9$ is between $12$ and $12$) and ..... ....!!!!skreech!!!!!.....
NO!  $9$ is not between $12$ and $12$ so we can't use the induction step.
Okay we'll can figure out a special case for $13$.  So we now have that is it true for all $13$ and all numbers between $12$ and $13$.  
Let's try to use the induction step to go from $13$ to $14$.  It is true for $13$ and for all values between $12$ and $13$.  So it ought to be true for $13-3=10$ so long as $10$ is between.... Oh G#@%!  
So we do a special case for $14$.  
But, lord, how long can this go on?!  How many special cases will we have to do before we can do the induction step and never worry about special cases?
Well, we can do our induction step if we can assume that $n-3$ is between $12$ and $n$.  But for $n-3$ to be between $12$ and $n$ we must have $n$ at least $15$.  So we can't use our induction step untill we have gotten to $15$
So we need special base cases for $12,13,14$ and $15$.  
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Note:  We do have $12 -3  =9$ and we can do $9 = 1*4 + 1*5$ so $13 = (1+1)*4 + 1*5 = 2*4 + 5$.  But we have to have shown $9$ was possible first.
Likewis we could have gotten $14$ from $10$.  But we can't have gotten $15$ from $11$ because $11 = 4x + 5y$ is impossible.  Note: $15 = 0*4 + 3*5$ is the only way to do this and there are not $4$s so we can't have gotten $15$ from $11$ but just adding only more $4$.
