arrays and pointer arithmetic ~ clarification needed

Because the precedence of the operator [] is higher than the operator *, the following expression:

int x = *(a + i)[j];

is equal to:

int* p = (a + i)[j];int  x = *p;

which is also equal to:

int* p = ((a + i) + j);int  x = *p;

which in this case is equal to:

int  (*p0)[3]  = (a + i);int*  p        = (p0 + j);int   x = *p;

meaning that both i and j eventually shift the first index making p to point to element a[2][0], value of which is 7

And what has precedence of [] and * operators to do with the evaluation of this expression? Simple test by using () to make sure that * will be evaluated first will suffice here. Meaning that this:

int y = (*(a + i))[j];

is equal to:

int y = *(a[i] + j);

which is nothing but simple:

int y = a[i][j];

Lets say a+i is b

                        +--------------a+0---> {1, 2, 3,                         |              a+1--->  4, 5, 6,                         |              a+2--->  7, 8, 9};                        |*(a + i)[j] =       *(*(b+j)) = *(*(b+1)+1)                               = *(*(b+2))                              = *(*(a+2))                              = **(a+2)                               = a[2][0]                              = 7

int (*p)[3]; // Is pointer to array of 3 int s

When p=a same scenario , similar to using b