Set Inclusion Definition I would like to know if the following expression is correct:
\begin{equation}
A \subseteq B \quad \text{ iff } \quad (A \subset B) \lor (A = B).\tag{1}
\end{equation}
I know for sure that the following is correct, because of some books:
$$
A \subset B \quad \text{ iff } \quad (A \subseteq B) \land (A \neq B).\tag{2}
$$
If of any help, my attempt to disprove the first (1) using the second (2) is as follows:
\begin{align*} 
A \subset B \quad &\text{ iff } \quad (A \subseteq B) \land (A \neq B)\\
(A \subset B) \lor (A = B) \quad &\text{ iff } \quad \bigl((A \subseteq B) \land (A \neq B) \bigr)\lor (A = B)\\
(A \subset B) \lor (A = B) \quad &\text{ iff } \quad (A \subseteq B) \lor (A = B)\\
\end{align*}
After this I am stuck. Any help, of any kind, would be appreciated.
(Note: I know I'm being sloppy with the notation.)
 A: below a proof of the --> direction 

The logic of the proof is as follows. 
The GOAL is to prove : X --> ( Y OR Z). So, we will assume X is true and try to derive : (Y OR Z) under this assumption. 
Now, propositional logic tells us that  (Y OR Z) is equivalent to : ~Y --> Z. 
So, in order to get (Y OR Z) at the end, we will proceed indirectly and derive first : ~Y --> Z. 
So, inside our main conditional proof, we will assume : ~Y  and try to derive Z. 
If we manage, we will have ( ~Y --> Z ) and therefore ( Y OR Z) 
Having done that, we will close our subordinate conditional proof and finish the main one by concluding : X --> ( Y OR Z). 
A: I was not sure, but it looks like it is indeed true.
Provided:


*

*def. of proper subset: $\quad A \subset B \quad \text{iff} \quad (A \subseteq B) \land (A \neq B)$.

*absorption law: $\quad A \quad \text{iff} \quad A \lor (A \land B)$. 

*set equivalence: $\quad A = B \quad \text{iff} \quad (A \subseteq B) \land (B \subseteq A)$.


We need to prove $\quad A \subseteq B \quad \text{iff} \quad (A \subset B) \lor (A = B)$.
Direct Proof:
\begin{array}{c & l}
    A \subseteq B & \\
   (A \subseteq B) \;\lor\; (\,(A \subseteq B) \land (B \subseteq A)\,) & \qquad \text{absorption}\\
   (A \subseteq B) \;\lor\; (A = B) & \qquad \text{set equivalence} \\
   (\,(A \subseteq B) \lor (A = B)\,) \;\land\; \text{True} & \\
   (\,(A \subseteq B) \lor (A = B)\,) \;\land\; (\,(A \neq B) \lor (A=B)\,) & \\
   (\,(A \subseteq B) \land (A \neq B)\,) \;\lor\; (A=B) & \qquad \text{distributive prop.}\\
   (A \subset B) \;\lor\; (A=B) & \qquad \text{def. of proper subset}.
\end{array}
where every row in the previous proof follows the next by an equivalence relation. Hence the reasoning can follow backwards. 
