Final answer:
The statement that proving a language's complement as recognizable but not decidable implies the original language is not recognizable is false. Both a language and its complement must be unrecognizable for the language to be not recognizable. Recognizability and decidability are concepts grounded in computability theory, relevant to the Turing machine model.
Step-by-step explanation:
To determine whether a language is recognizably decidable involves an understanding of computability theory. When we talk about a language being recognizable, we refer to whether there exists some Turing machine that will accept any string in the language, but might not halt on strings not in the language. Decidability, on the other hand, refers to whether there is a Turing machine that will correctly accept strings in the language and reject strings not in the language and will always halt, no matter the input.
The statement in question suggests that to show a language is not recognizable, one might demonstrate that its complement is recognizable but not decidable. This statement is False. The reason for this falsehood is due to a fundamental theorem in computability theory: A language is recognizable if and only if its complement is unrecognizable. Conversely, a language is decidable (also known as recursive) if and only if both the language and its complement are recognizable.
If the complement of a language is recognizable but not decidable, this in fact implies that the original language itself is recognizable. Therefore, saying that a language is not recognizable because its complement is recognizable (whether decidable or not) is incorrect. To prove a language is not recognizable, one must show that no Turing machine exists that will accept all and only the strings in the language, potentially leaving the machine to run indefinitely without providing an answer for some strings.