Ok, let's start from the beginning. Every integer can be decomposed in its prime factors. For example:
$$\cases{12=2^2\times3^1=2\cdot2\cdot3 \\ 90=2^1\times3^2\times5^1=2\cdot3\cdot3\cdot5 \\ 2772=2^3\times3^2\times7^1\times11^1=2\cdot2\cdot2\cdot3\cdot3\cdot7\cdot11}$$
Those are prime factors. And this decomposition is unique. The only possible divisors of a number are combinations of its prime factors (and $1$). Take for example $90$. Except for $1$, it's divisors are combinations of $2$, $3$, $3$ and $5$. Let's list them so you can see what I mean:
$$\cases{2 \\ 3 \\ 5 \\ 2\cdot3=6 \\ 2\cdot5=10 \\ 3\cdot5=15 \\ 2\cdot3\cdot3=18 \\ 2\cdot3\cdot5=30 \\ 3\cdot3\cdot5=45 \\ 2\cdot3\cdot3\cdot5=90}$$
Now what happens if we are looking for a common divisor of two numbers ? Well for each of them, its divisors are combinations of its prime factors, and his only (because the decomposition is unique), so if you want a divisor of both, you can only use the prime factors that both have.
Take for example the two numbers I used : 12 and 2772. Their common prime factors are $2$,$2$ and $3$. Except for $1$ (as always), all the divisors of both are thus the combinations of these three common prime factors. Let's list them once more:
$$\cases{2 \\ 3 \\ 2\cdot2=4 \\ 2\cdot3=6 \\ 2\cdot2\cdot3=12}$$
You see the greatest of these common divisors is the product of all common prime factors (Or $1$ if they don't have any in common).
This explains the first line of your proof:
- The GCD of 2 numbers is the product of their common prime factors
The second line is just the notation you use to qualify these common prime factors:
- $F=\{f_0,\ldots,f_n\}$ is the set of common prime factors of both $a$ and $b$
Again, picture our two numbers $12$ and $2772$ as your $a$ and $b$. The set $F$ is actually: $F=\{2,2,3\}=\{f_0,f_1,f_2\}$
As we said, the gcd is the product of those common factors so: $d=2\cdot2\cdot3=f_0\times f_1\times f_2$. And this is where your third line comes from:
- $d=(f_0\times\ldots\times f_n)$
Remember this is for the gcd, but all other divisors also have to be the product of some of the common prime factors among $2$, $2$ and $3$ (check the second list again if you have a doubt).
So if you take $F'$ as a subset of $F$ which is basically a word that says you take none, some or all elements from $F$ and is written $F'\subset F$ then there is such a subset $F'=\{f'_0,\ldots,f'_n\}$ that verifies $c=f'_0\times\ldots\times f'_n$
And you get your next line:
- If $c|a$ and $c|b$ then $F′\subset F$ and $c=(f′_0\times\ldots\times f′_n)$
I'm not a fan of this notation by the way, it conveys the false sense that there is always $n$ elements in the subset which is wrong, there is actually atmost $n$ elements ...
I would've written it this way instead:
- If $c|a$ and $c|b$ then $F′\subset F$ and $c=(f′_0\times\ldots\times f′_m)$, $0\leq m\leq n$
I don't know what you meant by "Delta" in your next line so I'll just skip it for now.
However, you know that $d$ is the product of all prime factors from $F$ and $c$ the product of some of those prime factors (those in $F'$). So $c$ has to divide $d$. Plus, you know that it divides it by the remaining factors (the factors that are in $F$ and not in $F'$). So I guess that's what you meant by "Delta":
And now for the last line, well I answered it in the last paragraph. However I didn't mention a special case: If $c=d$ then there is no factor left in $D$ they're all in $F'$ so you have to say that $c=d$ divides $d$ by $1$. This is why you get your last line:
- Therefore $c|d$ by a factor of $\max(1,(d_0\times\ldots\times d_n))$
Once again they use $n$ as a subscript, but it should be something else, $k$ for example. (with $k=n-m-1$)
