Tuple Intersection

Write a Python function `tuple_intersection` that takes two lists of tuples and returns the intersection of these tuples, considering the tuples as unordered pairs. This means that `(a, b)` is considered the same as `(b, a)`.

#### Function Signature
```python [main.nopy]
def tuple_intersection(list1: list, list2: list) -> set:
    pass
```

#### Input
- `list1`: A list of tuples, where each tuple contains exactly two elements.
- `list2`: Another list of tuples, where each tuple contains exactly two elements.

#### Output
- A set of tuples representing the intersection of the two lists, considering the tuples as unordered pairs.

#### Example Usage
```python [main.nopy]
list1 = [(3, 4), (5, 6), (9, 10), (4, 5)]
list2 = [(5, 4), (3, 4), (6, 5), (9, 11)]
print(tuple_intersection(list1, list2))  # Output: {(4, 5), (3, 4), (5, 6)}

list1 = [(4, 1), (7, 4), (11, 13), (17, 14)]
list2 = [(1, 4), (7, 4), (16, 12), (10, 13)]
print(tuple_intersection(list1, list2))  # Output: {(4, 7), (1, 4)}
```

#### Constraints
- Each tuple in the input lists will contain exactly two elements.
- The elements in the tuples are hashable.
- The order of elements in the tuples does not matter for comparison.