But can ($var || 2) ever be false ?

No way I can see. So yes, the OPs code should be reducible to just $var; though that will produce a warning: Useless use of private variable in void context ....

And $var || 2; gets Useless use of a constant in void context ...

Replace the 2 with 1: $var || 1; to take advantage of a special case that produces no warning and deparses as:

C:\Users\HomeAdmin>perl -MO=Deparse -Mstrict -wle"my $var = 1; $var ||
+ 1;"
BEGIN { $^W = 1; }
BEGIN { $/ = "\n"; $\ = "\n"; }
use strict 'refs';
my $var = 1;
'???' unless $var;
-e syntax OK
[download]

Or:

1 if $var;
[download]

which achieves the same thing.

Or, if he's happy there are no ties involved, ditch the whole darn thing.