# Set Theory - proof using set identities

Suppose $B, C, D, E$ are sets
Prove/Disprove: $(B \backslash C)\backslash (D\backslash E) = (B\backslash D)\backslash (C\backslash E)$

Any help would be much appreciated

-
This statement is false, try $C \subsetneq B = D = E$. – dtldarek Jan 31 '13 at 7:17

These are fairly complicated-looking sets, so it may be easier to try to prove equality and see whether anything goes wrong.

Suppose that $x\in(B\setminus C)\setminus(D\setminus E)$. Then $x\in B\setminus C$, and $x\notin D\setminus E$. From the fact that $x\in B\setminus C$ we conclude that $x\in B$ and $x\notin C$. From the fact that $x\notin D\setminus E$ we conclude that either $x\notin D$, or $x\in D\cap E$.

Suppose first that $x\notin D$; then $x\in B\setminus D$. And since $x\notin C$, $x\notin C\setminus E$, and therefore $x\in(B\setminus D)\setminus(C\setminus E)$. If this were the only case, we could conclude that

$$(B\setminus C)\setminus(D\setminus E)\subseteq(B\setminus D)\setminus(C\setminus E)\;.$$

However, it isn’t: what if $x\in D\cap E$? In that case $x\notin B\setminus D$, so $x\notin(B\setminus D)\setminus(C\setminus E)$, and we cannot conclude that

$$(B\setminus C)\setminus(D\setminus E)\subseteq(B\setminus D)\setminus(C\setminus E)\;.$$

This looks like an insurmountable obstacle: to get a counterexample, we need only find $B,C,D$, and $E$ such that $(B\setminus C)\setminus(D\setminus E)$ has an element $x\in D\cap E$. That’s certainly possible if we take $D=E\ne\varnothing$. To be concrete, let $D=E=\{0\}$. We also need to be sure that $x\in B\setminus C$; a simple way to do that is to let $B=\{0\}$ as well and set $C=\varnothing$. Then

$$(B\setminus C)\setminus(D\setminus E)=\big(\{0\}\setminus\varnothing\big)\setminus\big(\{0\}\setminus\{0\}\big)=\{0\}\setminus\varnothing=\{0\}\;,$$

but

$$(B\setminus D)\setminus(C\setminus E)=\big(\{0\}\setminus\{0\}\big)\setminus\big(\varnothing\setminus\{0\}\big)=\varnothing\setminus\varnothing=\varnothing\;.$$

The statement is indeed false.

-

Left side is

$$(B \cap \bar{C}) \cap \overline{(D\cap \bar{E})}=(B \cap \bar{C}) \cap (\bar{D} \cup E)=(B\cap \bar{C}\cap \bar{D})\cup (B\cap \bar{C}\cap E)$$

-

If your system has a universe set $U$ then replace $B=E=U$ and use that $X\setminus U=\phi$ and that $X \setminus \phi=X.$ Then you have $$(U \setminus C) \setminus (D \setminus U)=(U \setminus D)\setminus(C\setminus U)$$ from which $$U \setminus C=U \setminus D.$$ In other words $C,D$ have the same complement, and you can conclude $C=D$.

So the statement is true (in the case $B=E=U$) when and only when $C=D$.

-

By definition, $A \setminus B = A \cap B^c$. So $(B \setminus C) \setminus (D \setminus E) = B \cap C^c \cap(D \cap E^c)^c = (B \cap C^c)\cap(D^c \cup E) = (B \cap C^c \cap D^c)\cup(B \cap C^c \cap E)$

Looking at the second set, $(B \setminus D) \setminus (C \setminus E) = (B \cap D^c)\cap(C^c \cup E) = (C \cap B \cap D^c) \cup(B \cap D^C \cap E)$

It should be pretty clear from here that the sets aren't equal and how to come up with a counter-example.

-