Is this a good way for proving $A - B \subseteq A$?

Question

I'm asked to prove $A - B \subseteq A$ and would like to verify if my proof is well-written, correct, and see how I can improve it.

Proof:

For $A - B \subseteq A$, we would need that $$\forall x(x\mid x \in (A-B) \to x \in A).\tag{$i$}$$

Since $A - B = \{ x \mid x \in A\, {\rm and }\, x \notin B\}$, we know that $A - B \subseteq A.$

If we return to $(i)$, we are left with the implication that $\forall x(x \in A \to x \in A)$, which is true.

Therefore, $A - B \subseteq A $.

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Is this a good way for proving this? What other elementary method could I use to analyze this problem, since I'm still staring to learn this subject?

Answer 1

Here's an alternative proof . . .

We have $x\in A\setminus B$ iff both $x\in A$ and $x\notin B$. But, surely, if both $x\in A$ and $x\notin B$, then, in particular, $x\in A$. Thus $A\setminus B\subseteq A$.

Answer 2

Well, a proof using quantifiers explicitly requires more knowledge about first-order predicate logic.

Informally you can get rid off the all-quantifier by saying ''$x$ is an arbitrary element''.

Then $x\in A\setminus B$ means that $x\in A\wedge x\not\in B$. Here you can infer that $x\in A$, since the implication $P\wedge Q\Rightarrow P$ is a tautology (and for the mathematicians here one needs modus ponens as well).

Answer 3

$C\subset A \iff \forall x\in C x\in A$

$x\in A\setminus B \iff x\in A \land x \notin B \Rightarrow x\in A$

Keep caution that $C\subset A$ is not equivalent to $C\in A$

Answer 4

[Edited]

The expression :

$\forall x(x\mid x \in (A-B) \to x \in A).\tag{$i$}$

would read maybe : for all $x$, the set of all $x$ such that if $x$ belongs to $A-B$ then x belongs to $A$

The expression " for all x" needs a sentence in its scope, not a set. Note also that the set builder notation already contains an implicit " for all $x$ ".

It seems to me that what you wanted to do was to show that the set of all $x$ satisfying the inclusion condition was identical to the universal set $U$ ; which is equivalent to saying that $A-B$ is included in $A$ , or, that any $x$ belonging to $A-B$ is also a member of $A$.

What you could do, in order to follow this line of thought, is this, using

$\bullet$ $ (x\in S \land x\notin S) \equiv x\in \emptyset$ ( For all set $S$)

and

$\bullet$ $ (x\in \emptyset \land x\in S) \equiv x\in\emptyset$ ( For all set $S$)

and finally the idea that

$\bullet$ $ \forall (x) \phi(x) \equiv [\{x|\neg \phi(x)\}=\emptyset] $ :

$\{x| \neg ( x\in A-B \rightarrow x\in A)\}$

$=\{x| \neg [ (x\in A \land x\notin B) \rightarrow x\in A]\}$

$=\{x| \neg [ \neg (x\in A \land x\notin B)\land x\notin A]\}$

$=\{x| \neg [ \neg (x\in A \land x\notin A\land x\notin B]\}$

$=\{x| \neg [ \neg (x\in\emptyset \land x\notin B]\}$

$=\{x| \neg [ \neg x\in\emptyset]\}$

$=\{x| x\in\emptyset\}$

$= \emptyset$

since the set of all $x$ that do not satisfy the inclusion condition ( namely : $ x\in A - B \rightarrow x\in A$ ) is equal to $\emptyset$, it means that any $x$ satisfies this condition :

$\forall (x) [ x\in A - B \rightarrow x\in A ] $

and therefore that

$\ A-B \subseteq A$.

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My first attempt made use of a universal set U , but was not reliable. See comments below.