Thread: prove without set algebra

prove without set algebra:

A \ (A ∩B)=(A \ B)

this is my solution:

consider A \ (A ∩B)

let x ∈ A and x ∉ A ∩
--> x ∈ B'
--> x ∈ A\B

is this correct?

First, you showed only one inclusion: $A\setminus (A\cap B)\subseteq(A\setminus B)$. Second, could you explain more why $x\in A$ and $x\notin A\cap B$ imply that $x\notin B$? This problem is elemenratu enough so that every step has to be spelled explicitly.