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  1. Prove that for any sets A, B and C if A is a subset of B, then A – C is a subset of B – C.
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take any $x\in A-C$. You have $x\in A$ but $x\notin C$. Since $A$ is a subset of $B$ it follows that $x\in A$ implies $x\in B$. This means $x\in B$ but $x\notin C$. You write this as $x\in B-C$. Since $x$ was taken arbitrarily you have that for all $x\in A-C$ it follows $x\in B-C$. Therefore $A-C\subset B-C$

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Here is a proof of roughly the same length, but in a different (calculational) format: $$ \begin{align} & A - C \subset B - C \\ \equiv & \;\;\;\;\;\text{"definition of $\subset$; definition of $-$, twice"} \\ & \langle \forall x :: x \in A \land x \not\in C \Rightarrow x \in B \land x \not\in C \rangle \\ \equiv & \;\;\;\;\;\text{"logic: simplify: assume part of antecedent of $\Rightarrow$ in consequent"} \\ & \langle \forall x :: x \in A \land x \not\in C \Rightarrow x \in B \rangle \\ (*) \Leftarrow & \;\;\;\;\;\text{"logic: weaken by strengthening antecedent of $\Rightarrow$"} \\ & \langle \forall x :: x \in A \Rightarrow x \in B \rangle \\ \equiv & \;\;\;\;\;\text{"definition of $\subset$"} \\ & A \subset B \\ \end{align} $$ All steps are really uncreative, except for the key step $(*)$, which is strongly suggested by the shape of the formulae.

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