Final answer:
The statement is true because if both a language L and its complement are partially decidable, we can run two Turing machines in parallel to decide whether a string belongs to L or not, hence making L decidable.
Step-by-step explanation:
If a language L is partially decidable (also known as semi-decidable or recognizable) and its complement is also partially decidable, then L is decidable. This statement is true.
A language is partially decidable if there is a Turing machine that accepts all strings in the language and either rejects or runs forever on strings not in the language. When both a language and its complement are partially decidable, we have two Turing machines: one for the language and one for its complement. If an input string is in the language, the first Turing machine will eventually accept it. If an input string is not in the language, it is in the complement of the language, and the second Turing machine will eventually accept it.
Thus, to decide whether a string is in language L or not, one could run both Turing machines in parallel (using dovetailing or another method to interleave their steps) and eventually one of the two machines will accept the input. Hence, the language L is decidable because there is a procedure (the parallel execution of both Turing machines) that will always make a decision (accept or reject) for every possible input string.