$$gcd[(gcd(x,y),y)] = gcd(x,y)$$
Intuitively, I realize that GCD has these same properties, but I assume that this is not enough. normally showing that one operation is the same as another, we must prove that these functions have the same price in all possible sets of arguments. Please show me a way to prove these facts formally.
And how can I prove ab|cd if a|c,b|c and gcd(a,b)=d?
Put $d = \gcd (x,y)$. Then $\gcd (d,y)\leq d$. Since $d$ is a common divisor of $y$ and itself, we conclude $d = \gcd(d,y)$.