Operator precedence in C for the statement z=++x||++y&&++z

Question

I was studying operator precedence and I am not able to understand how the value of x became 2 and that of y and z is

Answer 1

x=y=z=1
z=++x||++y&&++z

is equivalent to

x=y=z=1
z=((++x)||((++y)&&(++z)));

Since ++x returns 2, which is nonzero, the ++y && ++z branch is never executed, and thus the code is equivalent to:

x=y=z=1;
z=(++x)||(anything here);