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Suppose $x \in A \backslash C$. This means that $x \in A$ and $x \notin C$. Suppose $x \notin B$. Then $x \in A \backslash B$, so since $A \backslash B ⊆ C$, $x \in C$. But this contradicts the fact that $x \notin C$. Therefore $x ∈ B$. Thus, if $x ∈ A \backslash C$ then $x ∈ B$.

Im not sure how showing $x\in C$ shows that $x\in B$.

If we work backwards, showing that $x\in C$ contradicts one of the givens that $x\in A\backslash C$ so its not the case that $x\in A\backslash C$ which was part of the original conditional statement. This is where im lost because this would make the conditional statement false.

HeroZhang001
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Not Friedrich gauss
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  • I believe you used a proof by contradiction. You can show using logic tables that showing such a contradiction makes the claim that $x\in A\setminus B$ false and $x\in B$ true – moboDawn_φ May 15 '23 at 01:01

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Suppose $x \in A-C$. Then, $x \in A$ and $x \notin C$. Since $C$ contains $A-B$, $x \notin A-B$.

Since $x \in A$ and $x \notin A-B$, hence $x \in B$.

Category_Theorist
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  • ah i get it, if $x\notin$ B and $x\in$ A then we can infer that $x\in$ A\B . But we are given that A\B ⊆ C which tells us that $x\in$ C. But this contradicts that $x\notin$ C. thus the chain of logic resulting from $x\notin$ B cannot be the case because of the contradiction. So it must be the case that x is in B. – Not Friedrich gauss May 15 '23 at 01:35