# Proving two sets are equal using bi-directional set inclusion

Prove(A∩B)’=A’∪B’.
let x ∈ (A ∩ B)’
∴ (x ∈ A ∩ B)’
∴ (x ∈ A ∧ x ∈ B)’
∴ (x ∈ A)’ ∨ (x ∈ B)’
∴ x ∈ A’ ∨ x ∈ B’
∴ x ∈ A’ ∪ B’
∴ (A ∩ B)’ ⊆ A’ ∪ B’

The above is the solution provided. Sorry if it seems trivial but I don't understand how
∴ (x ∈ A ∧ x ∈ B)’
∴ (x ∈ A)’ ∨ (x ∈ B)’
and not
∴ (x ∈ A)’ ∧ (x ∈ B)’

-
If it is not true that ($2 + 2 = 4$ and I am the pope), then either $2 + 2 \neq 4$, or I am not the pope. But it is not the case that both $2 + 2 \neq 4$ and I am not the pope. en.wikipedia.org/wiki/De_Morgan%27s_laws – Rahul Oct 21 '10 at 18:29
All that needs to happen to make the statement (A and B) false is for at least one of the parts to be false. This is equivalent to (not A or not B) being true. – Paul VanKoughnett Oct 21 '10 at 18:38
For future reference, it would be clearer if you distinguished in your notation between set complement and logical negation. The standard symbol for the latter is prefix ¬, so the first 3 lines of your proof would read let x ∈ (A ∩ B)’ ∴ ¬(x ∈ A ∩ B) ∴ ¬(x ∈ A ∧ x ∈ B) – Stewart Nov 20 '10 at 18:25