Question: Show that for every decidable set A there is a decidable set B such that B is not polynomial-time reducible to A. Then show that

Show that for every decidable set A there is a decidable set B such that B is not polynomial-time reducible to A. Then show that this implies that there is no set that is complete for the class of Turing-decidable sets under polynomial reductions.

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Databases Questions!