Why $x=5; $x+++$x++; equals with 11 in PHP?

According to the opcodes it should be 12. Am I getting it wrong?

number of ops:  8
compiled vars:  !0 = $x
line    #* E I O op                  fetch      ext   return  operands
-------------------------------------------------------------------------
  3     0  E >   EXT_STMT                                                 
        1        ASSIGN                                         !0, 5
  5     2        EXT_STMT                                                 
        3        POST_INC                               ~2      !0
        4        POST_INC                               ~3      !0
        5        ADD                                    ~4      ~2, ~3
        6        ECHO                                           ~4
  7     7      > RETURN                                         1

branch: #  0; line:     3-    7; sop:     0; eop:     7; out1:  -2
path #1: 0,

Edit

Also ($x++)+($x++); returns the same result (11). Actually this was the main reason for the question and opcode investigation.

It took me a few reads, but $x=5; $x++ + $x++; works like this:

In the case of a $x++, it first 'gets used', then increased:

• $x=5; — Set $x to 5
• $x++ + ... — Use, then increment
  • Place $x onto formulae stack (which is 5)
  • Increment(++) ($x is now 6, stack=[5])
• ... + $x++; — Use, then increment
  • Add $x onto stack (stack=[5,6], so 5+6 -> $x=11)
  • Adding is done, that outcome is 11
  • Increment $x(++) (which is isn't used further, but $x is now 7)

Actually, in this specific example, if you would echo $x; it would output 7. You never reassign the value back to $x, so $x=7 (you incremented it twice);