Write a function that checks if all the letters in the second string are present in the first one at least once, regardless of how many times they appear:

```
["ab", "aaa"]    =>  true
["trances", "nectar"]    =>  true
["compadres", "DRAPES"]  =>  true
["parses", "parsecs"]    =>  false
```

Function should not be case sensitive, as indicated in example #2. Note: both strings are presented as a **single argument** in the form of an array.

Source: codewars | Difficulty: EASY
Tags: String algorithms

Starter code: def letter_check(arr):
	

Solutions:
def letter_check(arr):
	return set(arr[1].lower()) <= set(arr[0].lower())


def letter_check(arr):
	(a, b) = (set(s.lower()) for s in arr)
	return b <= a


def letter_check(arr):
	for i in arr[1]:
		if i.lower() not in arr[0].lower():
			return False
	return True


def letter_check(arr):
	(a, b) = [set(i.lower()) for i in arr]
	return not b - a


def letter_check(arr):
	return all((c in arr[0].lower() for c in arr[1].lower()))


def letter_check(lst):
	return set(lst[0].lower()) >= set(lst[1].lower())


def letter_check(arr):
	arr = list([x.lower() for x in arr])
	for i in set(list(arr[1])):
		if i not in arr[0]:
			return False
	return True


def letter_check(arr):
	return set.issubset(*(set(s.lower()) for s in arr[::-1]))