Reduced assignment operator?

[Audrey Tang]

Consider these cases:
    [=] $x, $y, $z;
    [+=] $a, $b, $c;
S03 is currently inconsistent.  It first says these are not supported:

The final metaoperator in Perl 6 is the reduction operator.  Any
infix operator (except for non-associating operators and assignment
operators) can be surrounded by square brackets in term position to

But then implies it is on the defaulting table below:

    [=]()       # undef    (same for all assignment operators)

I don't see an obvious problem in supporting them as a syntactic  
expansion.
But would it be useful? And is there some hidden corners that I missed?

Thanks,
Audrey