Q.Prove: $A=(A\cap B)\cup (A-B)$   I want to prove it using set laws.  But I am lacking of any hint to break L.H.S into any usable result. Any hints will be appreciated.
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Prove $A = (A \setminus B) \cup (A \cap B)$  Logically, this is clearly true. I can explain why: start with $A$, remove all elements in $B$ and then add in any elements in both $A$ and $B$, which restores you back to $A$. That's an explanation, but AFAIK, it's not a proof in the formal proof sense.  I submitted a proof by a truth table which considers the four possible scenarios where an element is in/not in $A$/$B$. My professor asked me to redo the proof without a truth table.  My question is what mechanisms and strategies can I use to prove this in a acceptable formal proof sense?
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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?
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I want to prove $P(A\cap B)\geqslant P(A)+P(B)-1$.  How can I simplify the following proof?  $P(A\cap B)=P(A)-P(A\cap B^\mathsf{c})=P(A)-1+P(A^\mathsf{c}\cup B)=P(A)-1+P(A^\mathsf{c})+P(B)-P(A^\mathsf{c}\cap B)=P(A)+P(B)+P(A^\mathsf{c}\cap B^\mathsf{c})-1\geqslant P(A)+P(B)-1$
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I need help with this proof:     Let $A$ and $B$ be sets. Prove that $A\cap (B\setminus A)\subseteq \varnothing$.   My problem: Since $B\setminus A$ = $B\cap A^{c}$, we can say that if $x \in B$ then $x\in B\cap A^{c}$and $x \in A^{c}$ as well. So $x\notin A$.  Where do I go from here?
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Show that $f^{-1}(A\cup B) = f^{-1}(A)\cup f^{-1}(B)$ but not necessarily   $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$.  Let $S=A\cup B$  I know that $f^{-1}(S)=\{x:f(x)\in S\}$ assuming that that $f$ is one to one. Is this true $\{x:f(x)\in S\}=\{x:f(x) \in A\}\cup\{x:f(x)\in B\}$?  Why doesn't the intersection work?  Sources : ♦ 2nd Ed, $\;$ P219 9.60(d), $\;$ Mathematical Proofs by Gary Chartrand, ♦ P214, $\;$  Theorem 12.4.#4, $\;$ Book of Proof by Richard Hammack, ♦ P257-258, $\;$ Theorem 5.4.2.#2(b), $\;$ How to Prove It by D Velleman.
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Suppose $G=H\cup K$, where $H$ and $K$ are subgroups. Show that either $H=G$ or $K=G$.  What I did: For finite $G$, if $H\neq G$ and $K\neq G$, then $|H|,|K|\le |G|/2$. But they clearly share the identity element, so $H\cup K\neq G$.  How can I extend it to $G$ infinite?
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How to show $A \times (B \cup C) = (A \times B) \cup(A \times C)$, where $A$, $B$, and $C$ are sets. i.e. If $A=\{1,2\}$, $B=\{2,3\}$ then $A \times B = \{(1,2), (1,3), (2,2), (2,3)\}$.   I started with the proposition, suppose $x$ is an element of $LHS$, i.e. $A \times (B \cup C)$. Help is appreciated! Thanks.
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We were ask to prove that $|A \cup B| = |A| + |B| - |A \cap B|‫‪$. It was easy to prove it using a Venn diagram, but I think we might be expected to do if more formally. Is there a formal way?
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It would be great if someone could verify this proof.  Theorem:  $\mathcal{P}(A \cap B) = \mathcal{P}(A) \cap \mathcal{P}(B)$  Proof:  First I prove that $\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B)$. Take any subset $X \subseteq A \cap B$. Then $X \in \mathcal{P}(A \cap B)$. Also, $X \subseteq A \wedge X \subseteq B$, meaning that $X \in \mathcal{P}(A) \wedge X \in \mathcal{P}(B)$. This, again, means that $X \in \mathcal{P}(A) \cap \mathcal{P}(B)$, and this proves the statement.  Then I prove that $\mathcal{P}(A) \cap \mathcal{P}(B) \subseteq \mathcal{P}(A \cap B)$. Again,  take any subset $X \subseteq A \cap B$. Using just the same arguments as above, this is also a truth.  The theorem follows.
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I have been stuck on this problem :   Prove :If $A$ is Open , then $A\cap \overline B \subset \overline {A \cap B}$.I have tried to use the previously proven fact that $\overline {A \cap B}\subset \overline{A} \cap \overline{B}$ , but I cannot seem to get anywhere ...In particular, I do not see how A being open is relevant here..........
8158
How to prove $( A \times B ) \cap ( C \times D ) = ( A \cap C ) \times ( B \cap D )$ ?
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If $A$ and $B$ are sets, then $A\cup B$ is defined as follows: $$A\cup B:=\{x: x\in A \, \, \,\text{ or } \, \, \, x\in B\}.$$ In ZFC, $A\cup B$ exists because $\bigcup\{A, B\}$ exists by union axiom, and it is known that axiom of union is independent of rest of ZFC. If there is a set such that $A\subseteq C$ and $B\subseteq C$, then we can define $A\cup B$ using the $C$ and separation axiom. But can I prove the existence of $C$ in ZF-Union? Thanks for any help.
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$\displaystyle B\cup (\bigcap_{i=1}^n A_i)=\bigcap_{i=1}^n (B\cup A_i)$  I was able to prove this without using induction, however I am supposed to prove it using induction. How should I go about doing so?
91065
I am trying to prove that if $A – C = B – C$ and $A\cap C = B\cap C$ then $A = B$. I have tried using Venn Diagrams as a proof technique, but we are not able to use proof by Venn Diagrams.
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I have been learning set theory from scratch again and I have a question about the distributive property for sets. It seems the only proof I found of it actually USES the very same distributive property while proving it. This seems circular to me, could someone explain why it is allowed?  This is what I am trying to prove. $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$  I got this proof from a quora answer.  Let $x \in A \cup (B \cap C)$. If $x \in A \cup (B \cap C)$ then $x$ is either in $A$ or in ($B \cap C$).  $x \in A$ or $x \in (B \cap C)$  $x \in A$ or ($x \in B$ and $x \in C$)  HERE!!! How does one get from the statement above to the one below without actually using the distributive property itself?  ($x \in A$ or $x \in B$) and ($x \in A$ or $x \in C$)  $x \in (A \cup B)$ and $x \in (A \cup C)$  $x \in (A ∪ B) \cap (A ∪ C)$  $x \in A \cup (B \cap C) \implies x \in (A \cup B) \cap (A \cup C)$.  Thanks a lot!!
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I know that they are equivalent if we have $\cap$ instead of $\cup$.  If we start from left, we have:  $X \in P(A) \cup P(B)$  $X \subset A \lor X \subset B$  If we start from right, we have:  $X \in P(A \cup B)$  $X \subset (A \cup B)$  And now, how do we proceed proving that they are not equivalent?
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I need to prove the following letting A and B be subsets of a universal set U. $$A\cup B = U \iff A^c\subset B$$ It just isn't straight forward to me.