find the intersection of characters given a list of strings

inter.py

# Fencepost loop using dictionaries.
def inters_dict(lst):
    dct = {ch: True for ch in lst[0]}
    for s in lst[1:]:
        dct = {ch: True for ch in s if ch in dct}
    return dct.keys()

# I suspect Pythonistas would find the use of (ch, True) as key-value
# pairs to be jarring, so here it is using sets---essentially, the
# "True" part is implied by virtue of being in the set.
def inters_set(lst):
    st = set(ch for ch in lst[0])
    for s in lst[1:]:
        st = set(ch for ch in s if ch in st)
    return [i for i in st]

# This is the "cool" one-line solution I was thinking of, but couldn't
# come up with on the spot.  It essentially creates a set (like in
# inters_set) for each string in the list, then takes the intersection
# of all of them using Python's built-in set.intersection.  The star (*)
# unpacks the list of sets so it's like calling set.intersection(s_1,
# s_2, ..., s_n) (but since we don't want to write it like that, we can
# instead write set.intersection(*sets), where sets = [s_1, s_2, ...,
# s_n]).
def inters_set_cool(lst):
    return list(set.intersection(*[set(ch for ch in i) for i in lst]))

test_matrix = [
    ["abbbbcd","abbbbbbb","ca", "bcdae", "efghaab"],
    ["abc","ab","bcba", "bcdae", "efghaab"],
    ["abc","ab","bca", "bcdae", "efghab","z"],
]

# This seems to be the "right" way to do "main" in Python.
if __name__ == '__main__':
    for lst in test_matrix:
        # User can verify all three come out to be the same.
        print(inters_dict(lst), inters_set(lst), inters_set_cool(lst))