Let \(A \leq _m B\) denotes that language A is mapping reducible (also known…

Definition: A ≤_m B means there is a computable function f such that for every input x, x ∈ A iff f(x) ∈ B.

Use this definition to examine each claim:

• If A ≤_m B and B is recursive then A is recursive: True. Compute f(x) and run the decider for B on f(x). Because f is computable and the decider halts on all inputs, this procedure decides membership in A.

• If A ≤_m B and A is undecidable then B is undecidable: True. If B were decidable, composing its decider with the computable reduction f would decide A, contradicting that A is undecidable. Thus B must be undecidable.

• If A ≤_m B and B is recursively enumerable then A is recursively enumerable: True. Given a recognizer for B, on input x compute f(x) and run the recognizer on f(x). If f(x) is accepted exactly when x ∈ A, this yields a recognizer for A, so A is r.e.

• If A ≤_m B and B is not recursively enumerable then A is not recursively enumerable: False. Provide a counterexample. Let A be the empty language (which is recursively enumerable). Choose any language B that is not r.e. and pick a computable function f that maps every input to a fixed string w not in B. Then for every x, f(x) ∉ B, so x ∈ A iff f(x) ∈ B holds (both sides false). Thus A ≤_m B, but A is r.e. while B is not r.e. Therefore the claim is false.

Conclusion: The incorrect claim is the one asserting that non-recursively-enumerability of B implies non-recursively-enumerability of A.