How do I find out if at least one 2-itemsets is in a 3-itemsets list?

Question

Need to find out if an 3-itemset list is a superset of at least one 2-itemset. Every 3-itemset that has a frequent size-2 subset is already in your list. The list does not contain duplicated sets.

This is the last code I tried. The output should be a small list of sets if there are any subsets/supersets. With this code though I seem to be getting a larger list instead of a smaller list. Edited...

itemset2 =[{'', ''}, {'', ''}, {'', ''}, {'', ''}, {'', ''}, {'', ''},
 {'', ''}, {'', ''}, {'', ''}, {'', ''}, {'', ''}]


itemset3 =[{'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''},
 {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''},
 {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''},
 {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, 
{'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''},
 {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}]

stuff = itemset2

final = [set(y) for y in {frozenset(x) for x in stuff}]

final

nostuff=itemset3
ablanklist=[]
ablanklist2=[]

blanklist=set()
for things in nostuff:
    ablanklist.append(list(things))

for stuff in ablanklist:
    for items in final:
        if stuff[0] and stuff[1] in items:
            print(items)

#print(final)

Answer 1

The condition is a simple application of any. Given a trio element from three-itemset

if any(pair < trio for pair in two-itemset):

will tell you whether any pair is a subset of the given 3-element set.

Answer 2

def is_subset(big: set, little: set):
    return little - big == set()

set_a = set([1,2,3])
set_b = set([1,2])
is_subset(set_a, set_b)

edit: better to use the built in set_b.issubset(set_a)