[css-lists] `<reversed-counter-name> <integer>` seems useless

#6096 added the reversed() syntax to the counter-reset property.

However, unless I'm missing something, these are totally equivalent:

counter-reset: c 7;
counter-reset: reversed(c) 7;

The only difference is that the latter instantiates the counter with a reversed flag, but this flag only seems relevant in these cases:

• When omitting the starting value (it's 7 in the example)
• In the implicit list-item counter (it's c in the example)

So IMO we have two reasonable possibilities:

• Drop the useless <reversed-counter-name> <integer> syntax, and just keep <reversed-counter-name>.

  The reversed flag of the counter would then be set by <reversed-counter-name> and also when counter-reset instantiates a list-item counter in an <ol reversed>. The preshint for reversed HTML lists would be:

  ol[reversed] {
    counter-reset: reversed(list-item);
  }
  ol[reversed][start] {
    counter-reset: list-item calc(attr(start integer) + 1); /* with reversed flag */
  }
  

• Keep <reversed-counter-name> <integer>, but then negate increments not just to list-item but to all counters. That is, list-item counters would get counter-increment: list-item 1 by default, not -1 for reversed lists. And when incrementing a reversed counter, the specified amount would actually be decremented. So

  <style>
  #parent { counter-reset: reversed(c) 7; }
  #parent > .child { counter-increment: c 2 } /* Decrements since the counter is reversed */
  #parent > .child::before { content: counter(c) ". " }
  </style>
  <div id="parent">
    <div class="child">foo</div>
    <div class="child">bar</div>
  </div>
  

  would look like

  5. foo
  3. bar
  

The latter seems more useful for authors, and makes list-item less magic.

Read the CSS Lists draft sections on counter-reset and counter instantiation first, then compare the two alternatives described in this issue. Done means reaching and recording a standards decision, with the affected reversed-counter syntax and semantics updated consistently.