Final answer:
Binary strings recursively defined to contain more O's than 1's can be proven using mathematical induction and iteration.
Step-by-step explanation:
The binary strings that contain more O's than 1's can be recursively defined as follows:
Base case: The empty string is in the set, as it has no 1's or O's.
Recursive step: If a string is in the set, then appending an O to the string will result in a new string that still has more O's than 1's, as every string in the set already satisfies this condition.
(a) To prove that every string in the set has more O's than 1's, we can use mathematical induction. The base case has been established, and for the recursive step, we assume that a string in the set has more O's than 1's. Appending an O to this string will increase the count of O's by 1 and leave the number of 1's unchanged, so the resulting string still has more O's than 1's.
(b) To prove that every string with more O's than 1's is in the set, we can use a similar approach. If a string has more O's than 1's, we can remove one O to create a new string. This new string has the same number of 1's as the previous string, but one fewer O. By repeating this process, we can eventually reach the empty string, which is in the set by the base case. Therefore, every string with more O's than 1's can be obtained by applying this process to a string in the set, and hence is in the set.