Let $A \le_m B$ denotes that language A is mapping reducible (also known as many-to-one reducible) to language B. Which one of the following is FALSE?

Properties of many-to-one (mapping) reducibility $A \le_m B$:

1. Decidability / Recursiveness:

   • If $B$ is recursive (decidable), then $A$ is recursive. (Statement (A) is TRUE)

   • Its contrapositive: If $A$ is undecidable, then $B$ is undecidable. (Statement (B) is TRUE)

2. Turing-Recognizability / Recursive Enumerability:

   • If $B$ is recursively enumerable (RE), then $A$ is recursively enumerable. (Statement (C) is TRUE)

   • Its contrapositive: If $A$ is not recursively enumerable, then $B$ is not recursively enumerable.

Statement (D) claims: *"If $A \le_m B$ and $B$ is not recursively enumerable then $A$ is not recursively enumerable."* This is FALSE because a trivial/regular language (which is RE) can easily be reduced to a non-RE language $B$. For example, choose $A = \emptyset$ and any fixed reject string $w_0 \notin B$, then $f(x) = w_0$ provides a valid reduction $A \le_m B$, yet $A$ is decidable and RE while $B$ is not RE.

Thus, option (D) is the FALSE statement.