The code in perlfaq4 sadly assumes no duplicates. But nonetheless hashes are the solution:

foreach @b { $hb{$_}++ }
foreach @a { $ha{$_}++ }

foreach keys %hb { 
  push @c,$_ if ( not exists $ha{$_} );
}
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This code should be easier to grok than the dense code in the perlfaq. Because %hb is a hash, "keys %hb" is identical to @b without the duplicates. In the loop a value gets pushed into @c if it is not in %ha and so also not in @a.

You might be able now to create a second loop that finds "all the elements in @a are all belong to @b".