So, I have the following proof question that I'm doing and I'm hoping that someone can look over the structure of my proof and give me comments.
Let $A \subset U$ and $B \subset U$. Then, prove that $A \subset B \iff (U-B) \subset (U-A)$
Proof
Let $A \subset U$ and $B \subset U$. Since the given proposition is a biconditional, we have to prove it in two directions.
We will, first, prove the following:
$A \subset B \implies (U-B) \subset (U-A)$
Let $x$ be an object such that $x \in U-B$. Then:
$x \in U-B \iff (x \in U) \land \lnot{(x \in B)}$
By the Law of Simplification:
$(x \in U) \land \lnot{(x \in B)} \implies \lnot{(x \in B)}$
$(x \in U) \land \lnot{(x \in B)} \implies (x \in U)$
Hence, it is true that $x \in U$ and $\lnot{(x \in B)}$. Since $A \subset B$, it follows that:
$A \subset B \iff \forall x \in U: \lnot{(x \in B)} \implies \lnot{(x \in A)}$
This shows that $\lnot{(x \in A)}$ is true. Hence, the following conjunction must be true as well:
$\lnot{(x \in A)} \land (x \in U) \iff (x \in U-A)$
That proves the forward conditional.
Now, we have to prove that:
$(U-B) \subset (U-A) \implies (A \subset B)$
Let $x$ be an object such that $x \in A$. Then, $x \in U$, since $A \subset U$. Hence, $\lnot{(x \in U-A)}$ is true. This implies that $\lnot{(x \in U-B)}$ is true as well as this is just the contrapositive of the hypothesis.
This just means that $x \in U-B$ must be false. Since we know that $x \in U$ is true, it follows that $\lnot{(x \in B)}$ must be false and, hence, $x \in B$ must be true. This proves that $A \subset B$.
By proving the conditionals in both directions, we have proven the original biconditional.
Seems like a rather long proof so I'm hoping to see a relatively shorter one suggested by someone. Thanks in advance.
For short proof:
$$A\not\subseteq B\iff \exists x\in U, s.t\ \ x\in A- B\iff x\in U, x\notin B \iff x\in U-B $$ $$A\not\subseteq B\iff\exists x\in U, s.t\ \ x\in A- B\iff x\not\in U-A$$
From both you obtain $$A\not\subseteq B\iff \exists x\in U, s.t\ \ x\in U-B, x\not\in U-A\iff U-B\not\subseteq U-A $$