# The computational complexity of my answer in Question 1 is O(n^2)

def isSubsetArray(ArrayA, ArrayB):
    countSameSubset = 0                                                                             # 1 Step
    for i_index in range(len(ArrayB)):                                                              # n Step
        for j_index in range(len(ArrayA)):                                                              # n Step
            if (ArrayA[j_index].strip().upper() == ArrayB[i_index].strip().upper()):                    # 1 Step
                countSameSubset += 1                                                                    # 1 Step
                break                                                                                   # 1 Step

    if (countSameSubset == len(ArrayB)):                                                            # 1 Step
        return True                                                                                 # 1 Step
    else:
        return False                                                                                # 1 Step

def main():
    StringA 	= input("Input the 1st Array : ")                                                   # 1 Step
    StringB 	= input("Input the 2nd Array : ")                                                   # 1 Step
    isSubsetValue = isSubsetArray(StringA.split(','), StringB.split(','))                           # 1 Step
    print('isSubset([',StringB.upper(),'], [', StringA.upper(), ']) = ', str(isSubsetValue))        # 1 Step

if __name__ == '__main__':
	main()                                                                                          # 1 Step

# Because the comlexity time on line 5 is n Step and on line 6 is also n Step, so my computational complexity of my answer in Question 1 is O(n^2)