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U aint gonna do this with Haskal anytime soon!

Name: W.T. Crackers 2016-09-02 0:44

Perl is king and here is the proof:
sub _ (&;@) {
my $f = shift;
my $x = sub {
my $x = shift;
sub { &$f(@_, &$x($x)) };
};
&$f(@_, &$x($x));
}

Name: Anonymous 2016-09-02 17:36

I think this is pretty much the same thing:
fixs :: [[a->b]->a->b] -> [a->b]
fixs fl = U (\pp -> map ($pp) fl)

U f = f g where g = f g

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