MHB I would like to know for example how I prove or disprove this sentence ?

[MathPro17]

Hi,
I am new here ,
I have stuck to Proved or disproved This sentence :
If for any A,B,C sets exist A\C=B\C And A ∩ C =B ∩ C , then A⊆ B.

Thanks very much for help.

 

[Evgeny.Makarov]

If for any A,B,C sets exist A\C=B\C And A ∩ C =B ∩ C , then A⊆ B.

Do you mean "it is the case" instead of "exists"? That is, is the claim as follows: If $A\setminus C=B\setminus C$ and $A\cap C=B\cap C$ for some sets $A$, $B$ and $C$, then $A\subseteq B$?

Note that $X\setminus Y=X\cap\overline{Y}$, so $X=X\cap(Y\cup\overline{Y})=(X\cap Y)\cup (X\setminus Y)$. Therefore,
\[
A=(A\cap C)\cup (A\setminus C)=\text{ (by assumption) }(B\cap C)\cup (B\setminus C)=B.
\]

 

[MathPro17]

Do you mean "it is the case" instead of "exists"? That is, is the claim as follows: If $A\setminus C=B\setminus C$ and $A\cap C=B\cap C$ for some sets $A$, $B$ and $C$, then $A\subseteq B$?

Note that $X\setminus Y=X\cap\overline{Y}$, so $X=X\cap(Y\cup\overline{Y})=(X\cap Y)\cup (X\setminus Y)$. Therefore,
\[
A=(A\cap C)\cup (A\setminus C)=\text{ (by assumption) }(B\cap C)\cup (B\setminus C)=B.
\]

Yes I mean for that .
I would like to your help with another thing - to prove this :
P(A∩B)=P(A)∩P(B)
if you could prove it with:"x ∈ to P()..." ?

Thanks.

 

[Evgeny.Makarov]

I would like to your help with another thing - to prove this :
P(A∩B)=P(A)∩P(B)
if you could prove it with:"x ∈ to P()..." ?

Why don't you continue $X\in P(A\cap B)\iff\dots$ yourself? You'll need $X\subseteq A\cap B\iff X\subseteq A\land X\subseteq B$ where $\land$ means "and".

 

[MathPro17]

Why don't you continue $X\in P(A\cap B)\iff\dots$ yourself? You'll need $X\subseteq A\cap B\iff X\subseteq A\land X\subseteq B$ where $\land$ means "and".

I Mean How could I prove this : "P(A∩B)=P(A)∩P(B)" with Some X that I take ?
Thanks.

 

[Evgeny.Makarov]

I Mean How could I prove this : "P(A∩B)=P(A)∩P(B)" with Some X that I take ?

You need to show $X\in P(A\cap B)\iff X\in P(A)\cap P(B)$ for all sets $X$. Why don't you continue the series of equivalences $X\in P(A\cap B)\iff\dots$ using the definition of powerset?