$A\subseteq B\to C\setminus B\subseteq C\setminus A\,$ -- how to prove this? Given $A \subseteq B $. Prove for every set $C, C\setminus B \subseteq C \setminus A $. Logical Argument: Given: $\forall x, x \in A \rightarrow x \in B $ Goal: $\forall C \forall x , x\in C\setminus B \rightarrow x \in C \setminus A$ For arbitrary $C$ and $x$, $ x\in C\setminus B \rightarrow x \in C \setminus A$ rewriting, $ x\in C\ \wedge x \not \in B \rightarrow x \in C \wedge \not \in A$ adding, $ x\in C\ \wedge x \not \in B $ to the given. Proof by contradiction: Now the given contains, $ x \in A \rightarrow x \in B $ $ x\in C\ \wedge x \not \in B $ $x \not \in C \vee x \in A$ In the disjunction, if I choose $x \not \in C $, then it's a contradiction. If I choose $x \in A $, then agin it's a contradiction because modus ponens, it adds $x \in B$ which is again a contradction. Is there any better way to argue this?

> Is there any better way to argue this?

I would argue it directly.

**Claim:** $A\subseteq B\to C\setminus B\subseteq C\setminus A$ for all sets $C$.

_Proof_. Suppose $A\subseteq B$. Further suppose $x\in C\setminus B$, where $x$ is arbitrary. If $C=\varnothing$, then $x\in C\setminus A$ is trivially true (i.e., $C\setminus B\subseteq C\setminus A$ is vacuously true when $C=\varnothing$). Suppose, then, that $C\
eq\varnothing$ and $x\in C\setminus B$. By definition, $x\in C\setminus B$ means $x\in C\land x\
ot\in B$. Since we are given $A\subseteq B$, we observe that $x\in A\to x\in B$ is logically equivalent to $x\
ot\in B\to x\
ot\in A$ (contrapositive). Thus $x\in C\land x\
ot\in B$ implies that $x\in C\land x\
ot\in A$, and the definition of $C\setminus A$ is $x\in C\land x\
ot\in A$. Thus, $x\in C\setminus A$. Hence, if $A\subseteq B$, then $C\setminus B\subseteq C\setminus A$. $\blacksquare$