Minimal V: Membership Operator

Check the principles of minimalist code in the [intro to the first challenge] (https://innokodakademija.com/challenge/2XLjgZhmACph76Pkr).

In the Code tab you will find a code that is missing a single character in order to pass the tests. However, your goal is to submit a function as minimalist as possible. Use the tips in the tips section below.

Write a function that returns the boolean True if the given two lists do not share any numbers, and False otherwise.

Tips

The operators in and not in test for membership. x in s evaluates to True if x is a member of s, and False otherwise.

For example, the code:

def startswithvowel(word):
    for vowel in "aeiou":
        if word[0] == vowel:
            return True
    return False

Can be simplified to:

def startswithvowel(word):
    return word[0] in "aeiou"

Bonus

Here are more examples:

12 in [1, 50, 12, 43, 7] ➞ True
1 in [12, 111111, "x"] ➞ False
[3, 4] in [1, 2, 3, 4, 5] ➞ False
3 in (True, 3, ["odd", "even"]) ➞ True
"odd" in (True, 3, ["odd", "even"]) ➞ False
"hello" in "hellomyfriend" ➞ True

"myfriend" not in "hello my friend" ➞ True
"bye" not in "bye my friend" ➞ False
2 not in {0: "even", 1: "odd"} ➞ True
1 not in {0: "even", 1: "odd"} ➞ False

Notes

• This is an open series: there isn't a definite list of features for the challenges. Please, do not hesitate to leave your suggestions in the Comments.
• Readability is indeed a subjective concept. Let's discuss it! Feel free to leave your opinion in the Comments.
• You can find all the exercises in this series [over here].