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[QUOTE="Old Wolf, post: 2466794"]
The phrase "partial ordering" is not used by the Standard. Can you
explain what you mean, and provide supporting quotes?

The following quotes are relevant:

C99 6.5.16#3 (Assignment operator)
    The side effect of updating the stored value of the left operand
    shall occur between the previous and the next sequence point.

C99 6.5.17#2
    The left operand of a comma operator is evaluated as a void
    expression; there is a sequence point after its evaluation. Then
    the right operand is evaluated; the result has its type and value

C99 5.1.2.3#2
    At certain specified points in the execution sequence called
    sequence points, all side effects of previous evaluations shall
    be complete and no side effects of subsequent evaluations
    shall have taken place.

Old Wolf's lemma:
    The right operand of the assignment operator must be evaluated
    before the assignment operator's side-effect can take place.

We are evaluating the expression (x = 5, 11).
6.5.17#2 says that there is a sequence point after (x=5) is
evaluated and before "11" is evaluated.
5.1.2.3#2 and 6.5.16#3 both say that the side-effect of (x=5)
must be complete before this sequence point.
Old Wolf's lemma says that storing the result of (x=5, 11)
in x must occur after (x=5,11) is evaluated.

Therefore, according to 6.5.16#3 and 5.1.2.3#2, at the comma
operator sequence point, x=5's side effect must be complete
and x=11's side effect must not be complete.

The only way you can deny this argument is to deny my
lemma; so you are claiming that in the situation:

x = (expression)

the side-effect of updating x can occur before (expression) is
evaluated.

Agree?

This is absurd. For example:

x = 1;
  x = (printf("x is %d\n", x), 11);

Do you still call this undefined and can it print 11?

How about
  int f() { printf("x is %d\n", x); return 11; }
  ...
  x = 1;
  x = f();

How far can you draw your bow?
[/QUOTE]

Verification