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Say that $x \notin A \lor x \in B$. Is $x\in(A\setminus B)^c$ ?

The answer is apparently yes, but I can't think of what logical intermediary steps I must take in order to reach this conclusion. Would kindly appreciate your assistance.

EDIT: I will explain where I'm coming from. The original starting point was $x \in(A^c\cup B) \implies x\in A^c\lor x\in B \implies x \notin A \lor x\in B \implies help? $

And from here I do not know which deductions I can make. I am not trying to use De Morgan's laws, I am trying to get to the answer using the basic building blocks. I would kindly appreciate it if someone continues the implications but goes through the exact steps so I can see exactly how it works.

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  • $\begingroup$ There is a mathjax typo which I cant figure out how to correct. $\endgroup$
    – insipidintegrator
    Commented Jun 14, 2022 at 20:00
  • $\begingroup$ @insipidintegrator Yepp, me neither. $\endgroup$
    – RandomUser
    Commented Jun 14, 2022 at 20:01
  • $\begingroup$ Uhh wait. What is the expression supposed to mean? Union or intersection or subtraction or what? $\endgroup$
    – insipidintegrator
    Commented Jun 14, 2022 at 20:02
  • $\begingroup$ Fixed it I believe $\endgroup$
    – Lorago
    Commented Jun 14, 2022 at 20:03
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    $\begingroup$ @insipidintegrator I am quite the beginner so I am not really familiar with your notation but I would guess what you said is equivalent. $\endgroup$
    – RandomUser
    Commented Jun 14, 2022 at 20:07

2 Answers 2

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$(A\setminus B)^c=(A \cap B^c)^c=A^c \cup B$. The first equation comes from the definition of $\setminus$. The second equation is de Morgan's law.

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  • $\begingroup$ @RandomUser Note that $x \in A^c \cup B$, just means $x \notin A$ OR $x \in B$. Not explained in the answer, but when first encountering sets and their operation, this may help you understand. $x \in A^c$ means x is in anything other than A. $\endgroup$
    – amWhy
    Commented Jun 14, 2022 at 20:26
  • $\begingroup$ Thank you for chiming in. $\endgroup$
    – RandomUser
    Commented Jun 15, 2022 at 8:58
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Say that $x \notin A \lor x \in B$. Is $x\in(A\setminus B)^c$ ?

I am not trying to use De Morgan's laws, I am trying to get to the answer using the basic building blocks.

Basically, you are trying to re-invent the wheel. That is, you are trying to use the same ideas that led to the De Morgan laws.

Okay, so before you can analyze under what circumstances $x \in (A \setminus B)^c$, you have to have a clear understanding of what $(A \setminus B)^c$ represents.

For any set $(S)$, $(S)^c$ represents all of the elements that are not in the set $S$.

So, $(A \setminus B)^c$ represents all of the elements that are not in $(A \setminus B)$. So, exactly which elements are in $(A \setminus B)$. By definition, it is all of the elements that satisfy both of the following constraints:

  • Constraint-1: The element is in $(A)$.
  • Constraint-2: The element is not in $(B)$.

Since an element is in $(A \setminus B)$ if and only if it satisfies both of the above constraints, this means that if it is not the case that the element satisfies both of the above constraints, then it is not the case that the element is in $(A \setminus B).$

This means that an element is not in $(A \setminus B)$ if and only if it fails to satisfy either Constraint-1 or Constraint-2.

This means that an element is in $(A \setminus B)^c$ if and only if the element fails to satisfy either Constraint-1 or Constraint-2.

So, $(A \setminus B)^c$ represents the union of all elements $(x)$ such that $(x)$ either:

  • fails to satisfy Constraint-1 $~\iff x \not\in A$
  • fails to satisfy Constraint-2 $~\iff x \in B$.

So,

$$x \in (A \setminus B)^c \iff $$

$$\{ ~ [x \not\in A] ~~~\text{or}~~~ [x \in B] ~\}.$$

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  • $\begingroup$ Your answer really helped me a lot and now I see it, but I'm just trying to figure out how I would've intuitively known that $x\notin A \lor x\in B \implies x\notin (A\setminus B)$ without resorting to deconstructing the conclusion. I guess it was just not possible for me unless I would've remembered by heart that $x\notin A \lor x\in B$ is the opposite of $x\in (A\setminus B)$ on a definitional level. $\endgroup$
    – RandomUser
    Commented Jun 15, 2022 at 8:56
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    $\begingroup$ @RandomUser In my opinion, when studying Math, the approach of "remembering by heart" is a recipe for disaster. For the alternative approach that I recommend, see this answer. Although the linked answer specifically mentions Calculus, my intention is that the linked answer should represent my personal opinion (only) as to what is the best approach for studying any area of Math. $\endgroup$
    – user2661923
    Commented Jun 15, 2022 at 11:11
  • $\begingroup$ Thank you very much. $\endgroup$
    – RandomUser
    Commented Jun 15, 2022 at 12:26

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