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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.

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    $\begingroup$ $\forall x\in B,x\in B$ is true, then $\forall x\in B,(x\in A)\vee (x\in B)$ just by simple logic $\endgroup$ – ElfHog Feb 2 '18 at 6:38
  • $\begingroup$ $A\subseteq B$ means $\forall x\in A,x\in B$ so $\forall x\in A\cup B,(x\in A)\vee(x\in B)\implies (x\in B)\vee(x\in B)\Longleftrightarrow (x\in (B\cup B))\implies x\in B$ $\endgroup$ – ElfHog Feb 2 '18 at 6:41
  • $\begingroup$ I try to imitate lines of proof from you (to be honest I haven't learnt about logics and set theory so correct me if I have errors in lines. Thankss $\endgroup$ – ElfHog Feb 2 '18 at 6:42
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    $\begingroup$ In "english"(Sometimes it's easier to write it like this then go to the proof): If an element is in B, then of course it is also in the union of A and B. In the other direction, if an element is in the union of A and B, then it must either be in A or B. By the subset condition, anything in A is also in B, so that element must also be in B. $\endgroup$ – DaveNine Feb 2 '18 at 6:56
  • $\begingroup$ I have serious reservations about my automatic replacement of ${x \in B}$ by ${x \in A}$. $\endgroup$ – pveentjer Feb 2 '18 at 6:56
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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.

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