Verified answer

I believe the best way to formally prove that would be to treat being in sets as boolean values and use a truth table to prove it by exhaustion:

S: 0 0 1 1

T:: 0 1 0 1

A: 1 0 0 0

B: 1 0 0 0

(where A means (SUT)′ and B means S′∩T′ - needed to do this to get it to display right)

Since (SUT)′ is the same as S′∩T′ at all input values, they are equal

I would try to use truth tables to prove this...what is wrong with using them-they are a valid form of proof!

The proof is by observing that the following are equivalent:

x ∈ (S∪T)´

x ∉ S ∪ T

x ∉ S and x ∉ T

x ∈ S´ and x ∈ T´

x ∈ S´ ∩ T´