GATE CS 2014 Set 2 — Question 26
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Theory of Computation → Turing Machines & Computability → Recursive & RE Languages
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Question
Let denotes that language A is mapping reducible (also known as many-to-one reducible) to language B. Which one of the following is FALSE?
Correct answer
(D) If A ₘ B and B is not recursively enumerable then A is not recursively enumerable.
Solution
Let be a computable function with (that is, ). To decide or recognise , compute and ask the question about .
- (A) True. If is recursive (decidable), run the decider for on ; this decides .
- (B) True. This is the contrapositive of (A): if were decidable, would be decidable. So if is undecidable, must be undecidable.
- (C) True. If is recursively enumerable, run the recogniser for on ; it accepts exactly when , so is RE.
- (D) False. The valid contrapositive of (C) is "if is not RE then is not RE"; (D) states the converse direction. Counterexample: let (decidable, hence RE) and let be any non-RE language, with and . The computable map , for gives , yet is RE while is not.
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