How C execute z = ++x + ++x + ++x;

Disassemble C z = ++x + ++x + ++x;

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int x = 2;
    int z = 0;

    z = ++x + ++x + ++x;

    printf("z = %d\n",z);

    return 0;
}

Disassembled:
08048420 <main>:
 ...
 ...
 ...
 sub    $0x20,%esp           ;reserve 32 bytes of stack for var x and z
 movl   $0x2,0x1c(%esp)      ;int x = 2
 ...
 movl   $0x0,0x18(%esp)      ;int z = 0
 ...
 addl   $0x1,0x1c(%esp)      ;x = x + 1
 addl   $0x1,0x1c(%esp)      ;x = x + 1
 mov    0x1c(%esp),%eax      ;eax = x
 lea    (%eax,%eax,1),%edx   ;edx = eax + eax
 addl   $0x1,0x1c(%esp)      ;x = x + 1
 mov    0x1c(%esp),%eax      ;eax = x
 add    %edx,%eax            ;eax = eax + edx
 mov    %eax,0x18(%esp)      ;z = eax
 ...
 ...
 ...
 call   80482f0 <printf@plt> ;printf()
 mov    $0x0,%eax            ;return 0
 leave
 ret
 ...
 ...
 ...
 ...
 ...
 ...

notes: "..." = instruction that is not important


To simplify this, here the step-by-step instructions performed by the program:
------------------------------------------------------------------------------

Problem: z = ++x + ++x + ++x, find z = ?
Given: x = 2

Step 1:
        (++x + ++x) + ++x
        x = x + 1
        x = 2 + 1
        x = 3
        x = x + 1
        x = 3 + 1
        x = 4
        (4 + 4) + ++x
        8 + ++x

Step 2:
        (8 + ++x)
        x = x + 1
        x = 4 + 1
        x = 5
        (8 + 5)
        13

        z = 13