Q:

prove if A contain not(A intersect b) then A=U

I am trying to prove that if $(A\cap B)^{\complement}\subseteq A$ then $A=U$.
I used De-Morgan rule on $(A\cap B)^{\complement}  $  and got
    $A^{\complement} \cup B^{\complement} \subseteq  A$.
Now I am not sure how to continue from here 
If I am right I said that there are two solutions to this equations :
1) $A=A^{\complement}$
2) $A=B^{\complement}$
but I don't know how derive from here that $A=U$.
edit - U = the world 

A:

De-Morgan rule gives you $A^c \cup B^c \subseteq A$ (and not $A^c \cup B^c= A$ as you wrote). It follows that $A^c \subseteq A$ and hence $A^c \cap A^c \subseteq A \cap A^c$, that is $A^c \subseteq \emptyset$. Thus $A^c$ is empty and $A = U$.

