I have to prove that. While I know this is true by thinking about it I'm having a lot of trouble actually writing the proof how can I prove $A\setminus(B\setminus C) = (A \setminus B) \cup (A \cap C)$  first of all it's true?
123354
I have to prove that. While I know this is true by thinking about it I'm having a lot of trouble actually writing the proof
3633
Show that  $A \setminus ( B \setminus C ) \equiv ( A \setminus B) \cup ( A \cap C )$  This is an exercise I was trying to do ( not homework ) and I got stuck as follows:  Working from $A \setminus ( B \setminus C )$:  $x \in A \wedge x \notin ( B \setminus C )$ $x \in A \wedge ((x \in B \wedge x \in C) \vee (x \notin B \wedge x \in C) \vee (x \notin B \wedge x \notin C) )$  At this point I'm not sure how to proceed.  Working from the other side $ ( A \setminus B) \cup ( A \cap C ) $:  $ x \in ( A \setminus B) \cup ( A \cap C ) $  $ x \in ( A \setminus B) \vee x \in ( A \cap C ) $  $ ( x \in A \wedge x \notin B ) \vee ( x \in A \wedge x \notin C ) $  $ x \in A \wedge (x \notin B \vee x \notin C ) $  $ x \in A \wedge x \notin ( B \cap C ) $  $ A \setminus ( B \cap C ) $  I could also use some formating tips like how display logical not and how to get the lines to display properly without blank lines inbetween. :-)
91768
I want to prove the following so I decide to rewrite the RHS  $$A \setminus (B\setminus C) = (A\setminus B)\cap (A \cup C)$$ $$ A\setminus(B\setminus C) = A\setminus (B \cap C^c) = A\cap (B\cap C^c)^c = A\cap(B^c\cup C) = (A\cap B^c)\cup(A\cap C) $$ there is way that $(A\cap B^c)\cup(A\cap C)\rightarrow (A\setminus B)\cup(A\cap C) = (A\setminus B)\cap (A \cup C)$?  thanks. EDIT  LHS: $$x\in A \vee x\in A\cap C \rightarrow x\in A \wedge x\in C$$ RHS: $$x\in A \wedge x\in A\cup C \rightarrow x\in A \vee x\in C$$  how it can help me?
89618
Please proof that $A\cap B=A\setminus(A\setminus B)$ this really gave me sleepless night. I tried using the set intersection properties but I still got confused.
87908
Prove that $f(z) = e^z$ maps $\mathbb{C} \setminus B $ onto $\mathbb{C} \setminus \{0\}$ where $B$ is a bounded subset of $\mathbb{C}$.  I have no idea how to even attempt this question so I would greatly appreciate some hints.
21960
If $A, B, C$ are sets prove that $(A\setminus B)\setminus C\subseteq A\setminus(B\setminus C)$. Find a description of when it is that we have equality, and give an example where the inclusion is strict.   I'm not sure how to do this one I know that to prove that two sets $X$ and $Y$ are equal, we need to show that if $x$ is any element of $X$, then $x\in Y$, and that if $x$ is any element of $Y$, then $x\in X$: the first of these means that $X\subseteq Y$ and the second that $Y\subseteq X$, so the two together establish that $X=Y$. I'm not sure how to use this for a question and how to finish it.
3765
First part :  I want to prove the following De Morgan's law :  ref.(dfeuer)  $A\setminus (B \cap C) = (A\setminus B) \cup (A\setminus C) $    Second part:  Prove that $(A\setminus B) \cup (A\setminus C) = A\setminus (B \cap C) $    Proof:  Let $y\in (A\setminus B) \cup (A\setminus C)$     $(A\setminus B) \cup (A\setminus C) = (y \in A\;  \land y \not\in  B\;) \vee (y \in A\; \land y \not\in C\;)$      $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= y \in A\ \land (  y \not\in B\; \vee \; y \not\in C ) $      $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= y \in A \land  (\lnot ( y\in B) \lor \lnot( y\in C) )$   According to De-Morgan's theorem :  $\lnot( B \land C) \Longleftrightarrow (\lnot B \lor \lnot C)$ thus     $y \in A \land (\lnot ( y\in B) \lor \lnot( y\in C) ) = y \in A \land y \not\in (B \land C)$   We can conclude that  $(A\setminus B) \cup (A\setminus C) = A\setminus (B \cap C)$
63087
Suppose $A,B,C$ are sets.  Prove that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$.  I proved that for $x\in A\setminus B$, $x\in A\cup C$ and for $x\in B\setminus A$, $x\in B\cup C$. How do I show that $A\cup C=B\cup C$, and that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$?
3197
How do I prove or disprove $X \times (Y \cup Z)= (X \times Y) \cup(X\times Z)$  for all sets $X$, $Y$, and $Z$? I'm lost on the steps here.
44817
An exercise asks me to prove the following: $$A \cap (B-A) = \varnothing$$ This is what I did:  I need to prove that $A \cap (B-A) \subseteq \varnothing$ and $\varnothing \subseteq A \cap (B-A)$  The second one seems to be obvious by definition (not sure if it is fine to say that in a test).  But the first one goes like this: $$x \in A \cap (B-A)$$ $$x \in A \land x \in (B-A)$$ $$x \in A \land (x \in B \land x \notin A)$$ $$(x \in A \land x \notin A) \land x \in B$$ $$(x \in A \land x \notin A)$$ Since there is a contradiction... err... the proof... is good.  Okay, that's my problem. I think that the procedure is fine, but clearly I am unable to word it out (well, I don't even know if it is valid at all). Is it valid? How can I... "prove that my proof is correct"? I don't really see this contradiction as an obvious indicator of $A \cap (B-A) \subseteq \varnothing$
110657
Let A, B be two sets. Prove that $A \subset B \iff A \cup B = B$  I'm thinking of using disjunctive syllogism by showing that $\neg \forall Y(Y \in A).$ However, I'm not sure how the proving steps should proceed such that it leads me to that premise. Edit: Thanks for the input. FYI, I need to prove this using predicate logic.
118934
Proving $(A\cup B)\cap(B\cup C)\cap(C\cup A)=(A\cap B)\cup (A\cap C)\cup (B\cap C)$?   I managed to prove one side by saying:  Let E=LHS and F=RHS     Let $ x \in E$. Then $x \in (A\cup B) \lor \ x \in (B\cup C) \lor \ x \in (C\cup A).$ If $x\notin A$, then $x$ must be in $B$ and $C$. If $x\notin B$, then $x$ must be in $A$ and $C$. If $x \notin C$, then $x$ is in $B$ and $A$. Therefore $x \in F$ and $ E \subset F $.   I'm having trouble proving the other direction. Can I prove it in a similar way saying that let $x \in F$, if $x \notin A$, then $x$ would be in $B$ and $C$? That doesn't make sense to me. Another way I thought of solving it would be:  Let $x \in F. $ Then   if $x \in A$, then  $x \in B \cup  C$, If $x \in B$, then $x \in A \cup C$.  if $x \in C$, then $x \in B \cup A$   And then is this sufficient to show that $ x \in F, thus \ F \subset E \ and \ E=F $?
14425
I've figured out the following statement is true but I was wondering how you actually go about proving something like this?   $A \times (B \triangle C) = (A \times B) \triangle (A \times C)$
4402
I'm trying to prove $A \cap (A \cup B) = A$. I'm stuck on the last part of my proof, not sure how to show next:  $$x \in A \cap (A \cup B)$$ $$\iff x \in A \;\;\text{and}\;\; x \in A \cup B$$ $$\iff x \in A \;\;\text{and}\;\; (\text{either}\;\; x \in A \;\;\text{or}\;\; x \in B).$$
42854
Please let me know if you find my proof satisfactory.  Let $A\subset B\subset \mathbb{R}^n$, where $\bar{A}$ is the closure of $A$.  If $\vec{x}\in \bar{A}$ then $\vec{x}\in A^\circ$ or $\vec{x}\in ∂A$ (in the boundary of $A$). If $\vec{x}\in A^\circ$ then $\vec{x}\in A \implies \vec{x}\in B\subset \bar{B}$. If $\vec{x} \in ∂A$ then $B_\varepsilon(\vec{x})\cap A\ne \emptyset \ne B_\varepsilon(\vec{x})\cap B$ (for any $\varepsilon &gt; 0$). Thus, since $A\subset B$, $\vec{x}\in B \implies \bar{A}\subset \bar{B}$.
86664
Hi I'm new to set theory. I need to prove that if $A \subseteq B$, then $A \cap B = A$. I would like to do this the formal way, without a Venn diagram. How should I proceed?
45445
Let $A,B,C$ be sets such that $C\subseteq B$. Let $f: A \to B$ be a function.  Prove that $f^{-1} (B\setminus C)=A\setminus f^{-1} (C).$   I really need help with this proof problem. I'm not sure where to begin or what strategy to consider using.