MHB Proving: Sets Proofs - Techniques & Examples

[paulmdrdo1]

prove the followinga. prove that if $A\cap B=\emptyset$, then $(A\times C)\cap (B\times C)=\emptyset$
b. $A\cup(B\cap C)=(A\cup B)\cap (A\cup C)$

i don't have any idea how i would start proving this.
can you give me some techniques on proofs.

 

[caffeinemachine]

prove the followinga. prove that if $A\cap B=\emptyset$, then $(A\times C)\cap (B\times C)=\emptyset$
b. $A\cup(B\cap C)=(A\cup B)\cap (A\cup C)$

i don't have any idea how i would start proving this.
can you give me some techniques on proofs.

a. Assume on the contrary that $(x,y)\in (A\times C)\cap (B\times C)$. Thus $x\in A\cap B$ and $y\in C$ (why?). Can you finish?

b. Let $x\in A\cup(B\cap C)$. Then $x\in A$ or $x\in B\cap C$.
Case 1. $x\in A$.
Here $x\in A\cup B$ and $x\in A\cup C$. (why?). Thus $x\in (A\cup B)\cap (A\cup C)$.

Case 2. $x\in B\cap C$.
Again, Here $x\in A\cup B$ and $x\in A\cup C$. (why?). Thus $x\in (A\cup B)\cap (A\cup C)$.

The above shows that any element of $A\cup (B\cap C)$ is in $(A\cup B)\cap(A\cup C)$. This means $A\cup (B\cap C)\subseteq (A\cup B)\cap(A\cup C)$. Can you show the reverse containment and finish?

 

[paulmdrdo1]

caffeinemachine i know nothing about formal way of proving. i can say why that is true in words or verbally but not in a generalize way. i want a formal presentation of proofs like what you are doing. can you give some tips. because every time i encounter this kind of problems i always feel discouraged.

but this is what i tried

$x\in A\cap B$ and $y\in C$ because we assume that $(x,y)\in (A\times C)\cap (B\times C) $

i don't know howfill in the other (whys)

 

[caffeinemachine]

caffeinemachine i know nothing about formal way of proving. i can say why that is true in words or verbally but not in a generalize way. i want a formal presentation of proofs like what you are doing. can you give some tips. because every time i encounter this kind of problems i always feel discouraged.

but this is what i tried

$x\in A\cap B$ and $y\in C$ because we assume that $(x,y)\in (A\times C)\cap (B\times C) $

i don't know howfill in the other (whys)

The 'second why' is quite straightforward. If you know that $x\in A$, then $x$ also lies in any superset of $A$. Isn't it? Thus $x\in A$ gives $x\in A\cup B$ since $A\cup B$ is a superset of $A$. Similarly $x\in A$ gives $A\cup C$ too.

Now try to fill in for the third why.