# Help with writing proofs

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.

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 $A-C\subset B-C$
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.