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Knowing $$A \subseteq B \Leftrightarrow (x\in A \Longrightarrow x\in B)$$ $$C \subseteq D \Leftrightarrow (x\in C \Longrightarrow x\in D)\\$$ I tried to prove, that for any $y\in A$ and $z\in D$, given formula $$A \setminus D \subseteq B\setminus C$$ is true, but I got stuck almost at the beginning, when I tried to show $$(y,z)\in A\setminus D\equiv \{y\in A\}\wedge \{z\notin D\}$$ and I am not really sure what should I do now. Although I know, that if $x\notin D$, then $x\notin C$, but it really leads me to nowhere. I would like to see a formal proof of this. Thanks in advance.

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    $\begingroup$ It's supposed to be $y\in A\setminus D\iff \{y\in A\}\wedge \{y\notin D\}$. No $z$ anywhere. $\endgroup$ – Arthur Nov 25 '18 at 14:47
  • $\begingroup$ Thank you, I was trying to prove it the same way as I did for $A\cup C \subseteq B\cup D$, that's why I used two elements. $\endgroup$ – whiskeyo Nov 25 '18 at 15:01
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The problem is in your assumption: $$(y,z) \in A \setminus D$$

When showing membership for a non-relational set we suppose a single element in the set, not an ordered pair. This is how it should go: $$x\in (A\setminus D)\implies x\in A \land x \notin D$$

Notice, $C\subseteq D \implies (\forall x)[(x \in C)\rightarrow (x \in D)]$, consequently, by modus tollens $x\notin D \implies x\notin C$

Since $A \subseteq B \land C \subseteq D$, then: $$(x\in A \implies x\in B) \land (x\notin D \implies x\notin C)$$

Therefore, $$x \in A \setminus D \implies x\in B \setminus C$$ $$A\setminus D \subseteq B\setminus C$$

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$C \subseteq D \Leftrightarrow D^{\complement} \subseteq C^{\complement}$

So $A \setminus D = A \cap D^{\complement} \subseteq B \cap C^{\complement} = B \setminus C$ as $A \subseteq B$ as well, and if we replace two sets in the intersection by posibly larger sets, we get a possibly larger set.

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