And/Or elemental-wise for list of bools in python

You can use list comprehension

l3 = [(i and j) for i,j in zip(l1,l2)]
l4 = [(i or j) for i,j in zip(l1,l2)]

When you are doing

l3 = l1 and l2
l4 = l1 or l2

then it is getting reduced to

l3 = bool(l1) and l2
l4 = bool(l1) or l2

Now since l1 is non-empty list, so bool(l1) = True

Now, what I assume in the case of l3 = l1 and l2, l1 is evaluated as True so for short circuit evaluation, it returns l2.

In case of l4 = l1 or l2, again due to short circuit evaluation, l1 is returned since l1 is True.

So, you are getting the result like that. short circuit evaluation is nothing but--

1. In the case of A and B, if A evaluates to True go on evaluating B and return the result of B. If A evaluates to False, then no point of evaluating B. Return the result of A.

False and print('hello')
>>> False
True and print('hello')
>>> hello

2. In the case of A or B, if A evaluates to True, then there is no point in evaluating B. Just return the result of A. If A evaluates to False, then return the result of B.

True or (1/0)
>>> True
False or (1/0)
>>> ZeroDivisionError: division by zero

Note:

bool([]) = False
bool([1,2,'a']) = True
bool(['False']) = True

You can use the built-in all() and any() methods in a list comprehension:

l1 = [True, False, True]
l2 = [False, True, True]

l3 = [all(z) for z in zip(l1, l2)]
l4 = [any(z) for z in zip(l1, l2)]

print(l3)
print(l4)

Output

[False, False, True]
[True, True, True]

You can also use the built-in map() method:

l1 = [True, False, True]
l2 = [False, True, True]

l3 = map(all, zip(l1, l2))
l4 = map(any, zip(l1, l2))

print(list(l3))
print(list(l4))

Output:

[False, False, True]
[True, True, True]