Divisible numbers comprehension with multiple terms

Write a function that receives a list of numbers and a list of terms and returns only the elements that are divisible by ALL the terms. You must use two nested list comprehensions to solve it.

divisible_numbers([12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1], [2, 3]) == [12, 6]

Warning, this assignment is difficult. If you get stuck, try working it with a for-loop first and then start turning it into list comprehensions.

Test Cases

test empty list - Run Test 

def test_empty_list():
    assert divisible_numbers([], [5, 7]) == []

test many divisible numbers - Run Test 

def test_many_divisible_numbers():
    result = divisible_numbers([12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1], [2, 3])
    expected = [12, 6]

    assert result == expected

test no result - Run Test 

def test_no_result():
    assert divisible_numbers([2, 4, 8], [5, 7]) == []

test one divisible numbers - Run Test 

def test_one_divisible_numbers():
    result = divisible_numbers([16, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1], [2, 3, 4])
    assert  result == [12]

test both empty lists - Run Test 

def test_both_empty_lists():
    assert divisible_numbers([], []) == []

Solution 1 

def divisible_numbers(a_list, terms):
    return [x for x in a_list if len([1 for term in terms if x % term == 0]) == len(terms)]

Solution 2 

def divisible_numbers(a_list, a_list_of_terms):
    return [e for e in a_list if all([e % t == 0 for t in a_list_of_terms])]
def divisible_numbers(a_list, a_list_of_terms): pass
Python 3.6