If ${A \subseteq B}$, then ${A \cup B = B}$

Question

I'm stuck with a family of set related proofs. They are in the form if A subset of B, then some kind of subset/equals property holds.

For example.

If ${A \subseteq B}$ then ${A \cup B = B}$

In this case we need to prove

${B \subseteq A \cup B}$

And

${A \cup B \subseteq B}$

Lets pick the first one to demonstrate where my problem is.

${B \subseteq A \cup B}$

Now it starts to become a bit rocky and where I'm looking for help. I want to build up the above statement based on definitions/axioms.

${B \subseteq B}$ reflexive property of a subset

${B \subseteq B \cup B}$ idempotency law

${\forall_x x \in B \to x \in B \cup B}$ def subset

${\forall_x x \in B \to x \in B \lor x \in B}$ def union

My problem: Since every element of A is in B, we should be able to replace ${x \in B}$ by ${x \in A}$

${\forall_x x \in B \to x \in A \lor x \in B}$ def subset.

${\forall_x x \in B \to x \in A \cup B}$ def union

${B \subseteq A \cup B}$ def subset

The question is if this approach is correct.

Answer 1

A simpler approach might be to note that $B \subset B \cup A$ by definition of $\cup$, and if $A \subset B$ then $A \cap B \subset B \cup B = B$. Hence the result.