Here is my question:

Give a formal set theoretic argument to prove that, for all sets $X, Y$ and $Z$, $$X\setminus(Y\setminus Z)=(X\setminus Y)\cup(X\cap Z)\;.$$

I know this is true, using the identity found here, so I'm unsure how to answer the question as it appears to be an identity in itself?!

How do you think I should answer it?

EDIT: This is my final answer (thank you for your help). Is it correct?

//My interpretation of the half of your answer $$\begin{align*} &\text{Suppose }X\setminus(Y\setminus Z)\\ &\implies a\in X\setminus(Y\setminus Z)\\ &\implies a\in X\land a\notin (Y\setminus Z) &\quad\text{[Definition of complement]}\\ &\implies (a\in X\land a\notin Y)\lor (a\in X\land (a\in Y\land a\in Z)) &\quad\text{[distributive law]}\\ &\implies X\setminus Y\cup (X\cap Y\cap Z) &\quad\text{[associative law]}\\ &\implies X\setminus Y\cup (X\cap Y\cap Z)\subseteq X\setminus Y\cup (Y\cap Z) &\quad\text{[}A\cap B\subseteq A\text{]} \end{align*}$$

//My half of the answer

$$\begin{align*} &\text{Suppose }X\setminus Y\cup (X\cap Z)\\ &\implies a\in (X\setminus Y\cup (X\cap Z)) \\ &\implies a\in (X\setminus Y)\lor a\in(X\cap Z) &\text{[definition of union]}\\ &\implies (a\in X\land a\notin Y)\lor (a\in X\land a\in Z)&\text{[def of complement and intersection]}\\ &\implies a\in X\land (a\notin Y\lor a\in Z) &\text{[Distributive Law]}\\ &\implies a\in X\land (T\land (A\notin Y\lor a\in Z))&\text{[Identity]}\\ &\implies a\in X\land ((a\notin Y\lor a\in Y)\land (a\notin Y\lor a\in Z)) &\text{[Excluded middle]}\\ &\implies a\in X\land (a\notin Y\lor(a\in Y\land a\in Z)) &\text{[Distributive law]}\\ &\implies a\in X\land a\notin(Y\setminus Z) &\text{[Definition of complement]}\\ &\implies a\in X\setminus(Y\setminus Z)&\text{[Definition of complement]}\\ &\implies (X\setminus Y)\cup (X\cap Z) \subseteq X\setminus (Y\setminus Z) \end{align*}$$

Since A ⊆ B, and B ⊆ A, sets A and B are equal.


marked as duplicate by Lord_Farin, Bruno Joyal, user7530, Dennis Gulko, Macavity Nov 16 '13 at 18:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ As usual, argue that if $a$ is an element of the left side, then $a$ is an element of the right side, and vice-versa (two separate arguments). A picture (Venn Diagram) might act as a guide. $\endgroup$ – André Nicolas May 1 '12 at 3:29
  • $\begingroup$ Yes, it's an identity; your job in this exercise is to prove that it's an identity. $\endgroup$ – Brian M. Scott May 1 '12 at 3:29
  • $\begingroup$ There are a number of write-up issues: you don't "suppose a set". So "Suppose $X\setminus(Y\seminus Z)$" is technically nonsense. You should start by supposing you have an element in $X\setminus(Y\setminus Z)$, etc. The end of the first part also suffers from this. The final line about $A$ and $B$ should have $A$ and $B$ replaced by the sets you are actually using. Whether this is acceptable to your instructor will of course depend on your instructor, but the general idea is correct. $\endgroup$ – Arturo Magidin May 1 '12 at 20:53
  • $\begingroup$ This seems to be the same question as this one. $\endgroup$ – Martin Sleziak Jul 18 '12 at 8:22

A formal set-theoretic argument would be to show that for all $a$, $$a\in X\setminus(Y\setminus Z)\iff a\in (X\setminus Y)\cup (X\cap Z).$$

A common way of doing this is by "double inclusion": show that if $a$ is in the left hand side, then it is in the right hand side (proving that the left hand side is contained in the right hand side), and then proving that if $a$ is in the right hand side, then it is in the left hand side (proving that the right hand side is contained in the left hand side) (hence proving two inclusions which together give equality).

I'll show you one, you do the other:

Suppose $a\in X\setminus(Y\setminus Z)$. Then $a\in X$ and $a\notin (Y\setminus Z)$. Since $a\notin Y\setminus Z$, then either $a\notin Y$, or $a\in Y$ and $a\in Z$; Thus, either $a\in X$ and $a\notin Y$, in which case $a\in X\setminus Y\subseteq (X\setminus Y)\cup (X\cap Z)$, or else $a\in X$ and $a\in Z$, hence $a\in X\cap Z\subseteq (X\setminus Y)\cup (X\cap Z)$. Thus, if $a\in X\setminus (Y\setminus Z)$, then $a\in (X\setminus Y)\cup (X\cap Z)$. That is, $$X\setminus(Y\setminus Z) \subseteq (X\setminus Y)\cup (X\cap Z).$$

  • $\begingroup$ thank you! so would this be correct as a final answer to the question? a∈(X(Y\Z)) a∈X ∩ a∉(Y\Z) a∈X ∩ (a∉Y ∪ (a∈Y ∩ a∈Z)) (a∈X ∩ a∉Y) ∩ (a∈X ∩ (a∈Y ∩ a∈Z)) [Distibutive Law $\endgroup$ – Danny Rancher May 1 '12 at 19:54
  • $\begingroup$ @DannyRancher: I only did half the problem; you would still need to use a similar argument to show that if $a\in (X\setminus Y)\cup(X\cap Z)$, then $a\in X\setminus(Y\setminus Z)$ (that is, prove the other inclusion). The above only proves the containment indicated at the end, not the equality of sets. $\endgroup$ – Arturo Magidin May 1 '12 at 19:56
  • $\begingroup$ sorry, im still writing my answer! i pressed enter by accident! (and now im having trouble editing the previous comment) ill edit in the question instead $\endgroup$ – Danny Rancher May 1 '12 at 19:59
  • $\begingroup$ @DannyRancher: I don't like the use of $\cap$ for both intersection and "and", nor $\cup$ for both unions an "or"s. Second, you don't want to prove a sequence of "ands", you want to prove a sequence of $\iff$s. Third, you would need to establish the equivalence $a\notin Y\setminus Z\iff (a\notin Y\lor (a\in Y\land a\in Z))$. Fourth: I can't tell if you can use the "distributive law", because I can't tell if you are trying to use it over logical connectives or set-connectives (because you use the same symbol to mean several different things). Are you not allowed to write it discoursively? $\endgroup$ – Arturo Magidin May 1 '12 at 20:03
  • $\begingroup$ I have noted your notation and used ^ and v to represent the discursive "and" and "or" respectively. I have put my edit in the original question, since I am new to math.SE and I have found the commenting method to be confusing. This is surely the best way to answer a question! I have really gone out of my way to learn how to answer the other half. $\endgroup$ – Danny Rancher May 1 '12 at 20:28

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