For any sets $A$, $B$, $C$; $(A\cap B)\cup C= A\cap (B\cup C)$     I understand that this means that (A and B) or C = A and (B or C), but how would you prove or disprove these set identities. Any help would be appreciated, Thanks
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I have a set identity: $(A \cap B) \cup C = A \cap (B \cup C)$ if and only if $C \subset A$.  I started with Venn diagrams and here is the result:   It is evident that set identity is correct. So I started to prove it algebraic:  1) According to distributive law: $(A \cap B) \cup C = (A \cup C) \cap (B \cup C)$  2) ...  I stuck a little. Because $C$ is a subset of $A$. I thought of pulling out: $(B \cup C)$ but it seems wrong step to me.  How to prove this identity having in mind that $C \subset A$?  Updated  Venn diagram for $C ⊈ A$
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I'm currently working on some discrete mathematics work and I've encountered a question I'm not sure how to answer exactly. Precisely, I'm trying to prove that two power, intersected sets statements are equal to each other and my understanding of how to do that doesn't seem to be enough. The statements in question go as follows:     Determine whether, for any sets $A$ and $B$, through proof, it is true that   $$\wp(A \cap B) = \wp(A) \cap \wp(B)$$   where $\wp$ denotes a power set.   Should I make some form of example set to better understand how these statements are equivalent? Or even a venn diagram? Any help is appreciated.
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Is this true?  If $A$ and $B$ are sets, then $A=\emptyset $ if and only if $B = A \bigtriangleup B$.  If $A=\emptyset$ then $B=\emptyset$ too?  Could someone help me please?
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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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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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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:  $$x \in A \cap (A \cup B)$$ $$\iff x \in A \;\;\text{and}\;\; x \in A \cup B$$ $$\iff x \in A \;\;\text{and}\;\; (\text{either}\;\; x \in A \;\;\text{or}\;\; x \in B).$$
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$A$ and $B$ are sets. $|A \cup B| = \mathfrak{c}$, prove that $A$ or $B$ has cardinality $\mathfrak{c}$.  This is an exercise problem from my textbook. It's easy if I assume CH to be true. But how can I prove this without it? I don't know how to eliminate the case of $A$ and $B$ both has cardinality larger than that of natural numbers and strictly smaller than that of reals.
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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.
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Trying to prove $(A ∩ B) − C=(A − C) ∩ (B − C)$ Im stuck here on this problem:  $(A∩B)−C=\{x | x ∈ (A ∩ B)∧ x ∉ C\}$  $(A∩B)−C=\{x|(x ∈ A ∧ x ∈ B) ∧ x ∉ C\}$  $(A∩B)−C=\{x|(x ∈ A ∧ x ∉ C) \lor (x ∈ B ∧ x ∉ C)\}$  I'm stuck right about here because how do I change the or (v) sign to the intersection sign (∩).  Please don't link me to the other question like this, I had a typo in it.
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I have been trying to prove the following for a while now, but I can't get to the solution:  $$(A\times C)\cup(B\times D)\subseteq(A\cup B)\times(C\cup D)$$  Can anyone help me?
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How can I prove the following statements are equivalent using laws of set theory?  $\neg((A\cap B)\cup (\neg A \cap C)) = (A\cap\neg B)\cup (\neg A\cap \neg C)$  I managed to use De Morgans laws to simplify the first statement down to:  $(\neg A\cup\neg B)\cap (A \cup \neg C)$ but I have no idea where to go from here.
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$A$ and $B$ are any sets, prove that $A \smallsetminus (A \smallsetminus B) = A \cap B.$ This formula makes sense when represented on a Venn diagram, but I am having trouble with proving it mathematically.  I have tried letting $x$ be an element of $A$ and continue from there, but it doesn't seem to work out as a valid proof anyways.   Could anyone please point me in the right direction?  Many thanks.
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I just started Set theory And I have the following question. Suppose I have $2$ sets, $A$ and $B$. $A$ is a subset of $B$ (contained inside $B$), then the union of these $2$ sets is $B$ right $?$ If no, could you guys explain to me why, if yes, could you guys provide a simple proof.  Thank you for reading my post.
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I am working with proofs in discrete math. Help to prove:  For the sets $A$ and $B$, we define the symmetric difference of $A$ and $B$ to be $A \triangle  B = (A-B)\cup(B-A).$  Prove that if $A \triangle B = C \triangle B,$ then $A = C.$
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I'm taking my first course in Probability and one of my homework problems is to prove that for any two sets:  $$P(A \cup B) = P(A) + P(B) - P(A \cap B) $$  Note that the function $P$ is the probability of an event happening.  From the third axiom, we are given that for any two disjoint sets, that is $ A \cap B = \emptyset$:  $$P(A \cup B) = P(A) + P(B)$$    My thought process follows from this third axiom, but I run into a problem near the end:  Since $A$ and $B$ are disjoint sets, $P(A)=P(A-B)$ and $P(B)=P(B-A)$, so  $$P(A \cup B) = P(A) + P(B)=P(A-B)+P(B-A)$$  From this, it appears that we are essentially taking the symmetric difference of A and B.  This, to me, seems to be equivalent to:  $$P(A-B)+P(B-A)=P(A)+P(B)-P(A \cap B)$$  Hence, by my logic:  $$P(A \cup B) = P(A)+P(B)-P(A \cap B)$$  Which, in essence, doesn't care whether two sets are disjoint or not as we are subtracting the intersection of them.  If the sets are disjoint, then the intersection is merely null.  If they intersect, then we are removing the "extra pieces".  I feel like this is inadequate proof despite my ability to convince myself that it is true..  Where, if anywhere, did I go wrong?  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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Statement: Let $A$, $B$, and $C$ be sets. Then $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$.   Proof: Let $(x,y) \in ( A \setminus B) \cap (A \setminus C)$, where $ x \in A$ and $y \in (B \cap C$ by definition of intersection. So, $(x,y) \in (A \setminus B \cap C)$ which show that $(A \setminus C) \cap (A \setminus C) \subseteq A \setminus (A \cup C)$. \ Now let $(x,y) \in A \setminus(B \cup C)$, where $ x \in A $. So, $ (x,y) \in (A \setminus B) \cap (A \setminus C)$ by definition of intersection, which shows that $A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)$. Therefore, by the Axiom Set of Equality we know that $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$ .   Am I on the right track to proving that this is a true statement or is there a better way to show this?