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I find it hard to find reductions that only work on one side: A is mapping reducible to B, but B is not mapping reducible to A. what could be an easy way to find such two languages A,B ? I tried Atm,Htm,EQtm,Etm,RegTm, and many others, and whenever I found that A is reducable to B, I also found that B is reducable to A.

thanks.

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    $\begingroup$ Use $\emptyset$ as one of your languages. $\endgroup$ – Yuval Filmus Feb 21 '18 at 9:39
  • $\begingroup$ Perhaps you could answer your own question now? $\endgroup$ – Yuval Filmus Feb 21 '18 at 14:39
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I now understand that there is a mapping reduction from the empty language to any other language except Sigma*, and there cannot be a reduction from a language to the empty language except for a reduction from the empty language to the empty language.

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