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Suppose $A \subseteq B$ and $x \in A $ and $x \notin B $ \ $C$. Prove that $x \in C$.

Basically what i need to do is to prove this by contradiction, so what i made was:

first of all, by applying the definitions of symetric diference, De Morgan and the definition of implication i found an equivalent expression for $x \notin B $ \ $C$, as $x \in B \rightarrow x \in C$.

Proof:

Suppose that $x \notin C$, since $x \in A $ and $A \subseteq B$ it follows that $x \in B$, because $x \in B \rightarrow x \in C$ it contradicts that $x \notin C$, so it follows that $x \in C$

is this correct?

  • 3
    The part of your proof that you wrote out is correct. Together with the unseen part where you reached the (correct) expression with implication, it is not as short as it could be. A small point: $B\setminus C$ is not called the symmetric difference. The symmetric difference of $X$ and $Y$, often written as $X\:\Delta\:Y$, consists of the points that are in $A\cup B$ but *not* in $A\cap B$. It can be expressed in various ways, for example $(X\setminus Y)\cup (Y\setminus X)$. Note that it is symmetric in $X$ and $Y$, while $X\setminus Y$ isn't.2012-02-01
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    Sorry, typo above, where it says $A\cup B$, read $X\cup Y$, and same with intersection.2012-02-01
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    Yes, sorry about that, you are right2012-02-01

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