Not quite.

The actual examples in perldelta are:

@x->[2]
scalar(@x)->[2]
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yours,
Michael

Nuh-uh!

@$ref->[0] can be parsed as @{$ref->[0]}

That's not what's happening.

From perlref:
Because of being able to omit the curlies for the simple case of $$x, people often make the mistake of viewing the dereferencing symbols as proper operators, and wonder about their precedence.

Example code:

use strict;
my $r = [qw/foo bar/];
print $r->[0];    # 'foo'
print @$r->[0];   # 'foo', due to bug.
print @{$r->[0]}; # Can't use string ("foo") as an ARRAY ref...
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I bet that you've several times have heard "use references as variable names". That is a good way to describe references, and it works here too. Having $r = \@x, @$r->[0] will be the same as @x->[0]. So it is indeed the same issue as in perldelta.