User blog comment:King2218/Infinite Extensibility/@comment-25418284-20140314181820/@comment-24509095-20140316041504

Hmm. A bit. $$a \right-arrow b$$ is the same as $$a \{\{1\}\} b=\{a,b,1,2\}$$.

Then, $$a \uparrow_d^c b=\{a,b,c,d\}$$.

Close enough. :P