In each part, decide if the statement is true. if it is true, prove it. if it is not true, give an explicit counterexample and demonstrate that your counterexample really is one.

(a) for any sets a, b and c, a ∩ b ⊆ c if and only if either a ⊆ c or b ⊆ c.

(b) for any sets a, b and c, a ⊆ b ∩ c if and only if both a ⊆ b and a ⊆ c

Hello!

ok, remember that "if and only if" implies that you need to prove the statement in both ways, this is represented with the ⇔ usually.

a) For any sets A, B and C, A ∩ B ⊆ C if and only if either A ⊆ C or B ⊆ C.

In this type of problems, i  find very useful start looking for some counterexample.

In this case, suppose that A = {1,2,3,4,5} , B = {3,4,5,6,7} and C = {3,4,5,6}

then is easy to see that A ⊄ C and B ⊄C.

And A∩B = {3,4,5}

then A∩B ⊂ C

then the statement is false (because one of the ways is false, remember that this is an "if and only if" statement)

b) For any sets A, B and C, A ⊆ B ∩ C if and only if both A ⊆ B and A ⊆ C.

the first way is true; because if A ⊆ B ∩ C. then all the elements of A are in the intersection of B and C (which are common elements for B and C) and then all the elements of A are in the set B and in the set C, and this means that A ⊆ B and A ⊆ C.

But let's see the other way now, suppose that A ⊆ B and A ⊆ C, now we want to know if A ⊆ B ∩ C.

if A ⊆ B and A ⊆ C, means that all the elements of A are in B, and all the elements of A are in C, then all the elements of A are common elements between B and C, this means that B ∩ C is at least equal to A (at least, because we know that all the elements of A are common elements between B and C, but there could be more common elements that don belong to A)

then A ⊆ B ∩ C.

the reporter's right

step-by-step explanation:

1/4 = .25 / 25%

x = 12

y = 1

for the points!