Proof that the subset relation is reflexive and transitive I'm teaching myself set theory, and I'm not sure how detailed I should be when asked to prove things.
Here is my proof that $A\subseteq A$ (the subset relation is reflexive):
$A \subseteq B$ iff $((x \in A) \implies (x \in B))$
$A \subseteq A$ iff $((x \in A) \implies (x \in A))$
$(x \in A) \implies (x \in A)$ is always true, as something implying itself must be true (is there a formal way to write this?)
Hence $ A \subseteq A$ is always true.
Is this proof formal enough, and does it contain the right amount of detail?
 A: Your proof is right.
Two comments :
(i) Instead of $A⊆B \implies ((x∈A) \implies (x∈B))$ I prefer :

$A⊆B$ iff $((x∈A) \implies (x∈B))$

because the RHS is the definition of set inclusion.
(ii) $x \in A \implies x \in A$ is an instance of the "logical law" :

$\mathcal A \implies \mathcal A$

which is a tautology of propositional logic.
The rules of logic allow us to substitute for the propositional letter $\mathcal A$ a formula whatever, and the result is still true, provided that we replace all the occurrences of $\mathcal A$ with the same formula.
A: User wrote:

$(x \in A) \implies (x \in A)$ is always true, as something implying itself must be true (is there a formal way to write this?)

This is certainly more than rigorous enough for most math courses, even at the highest level. 
However, to make it a truly formal proof, depending on the axioms and rules you are allowed, you would need something along the lines of:


*

*$x\in A$ (Premise)

*$x\in A\implies x\in A$ (Conclusion)

*$\forall a:[a\in A\implies a\in A]$ (Universal Generalization)
On each numbered line, you would have to cite the specific axiom or rule used to obtain that result. In my example, I refer to axioms of logic called Premise, Conclusion and Universal Generalization (names and usage will vary). Generally, in a formal proof, it's one line, one rule. Fortunately, most courses will not require this level of detail.
A: I'm practicing my set proving skills. Maybe somebody could benefit from it, or point out an error in the solution.
The proof for 
${A \subseteq A}$
${\forall_x, [(x \in A) \to (x \in A)]}$ def subset
${\forall_x, [\lnot(x \in A) \lor (x \in A)]}$ implication conversion law
${\forall_x, T}$ law of excluded middle
So the statement is reduced to a tautology, so therefor it must be true.
