MHB Proving Set Operations [Set Theory]

[TheGreat]

Could someone help me in this by simplifying/Proving the equation using Theorems/ Rules on Operation of Sets (i.e Commutative property, idempotent, assoc, dist. , definition of union , def. intersection , DeMorgan ,etc.).

Letting A, B and C be three sets ..

Prove/Disprove :
View attachment 5616
Any solution would be much of help. Hoping to get answers .. Do help me :) Thank you!

I've seen similar questions in here like this:
Problem posted: Prove/Disprove : A−(B∩C) = (A−B)∩(A−C)

Hello, KOO!

We should work on one side of the equation.


\begin{array}{cccccc}<br /> 1. &amp; A -(B \cap C) &amp;&amp; 1. &amp;\text{Given} \\<br /> 2. &amp; A \cap(B\cup C)^c &amp;&amp; 2. &amp;\text{def. Subtr&#039;n} \\<br /> 3. &amp; A \cap B^c \cap C^c &amp;&amp; 3. &amp; \text{DeMorgan} \\<br /> 4. &amp; A \cap A \cap B^c \cap C^c &amp;&amp; 4. &amp; \text{Duplication} \\<br /> 5. &amp; A\cap B^c \cap A \cap C^c &amp;&amp; 5. &amp; \text{Commutative} \\<br /> 6. &amp; (A \cap B^c) \cap (A \cap C^c) &amp;&amp; 6. &amp; \text{Associative} \\<br /> 7. &amp; (A-B) \cap (A-C) &amp;&amp; 7. &amp; \text{def. Subtr&#039;n}\end{array}

Thank you once again ... Hoping I could get some answer also ..

- jer

 

[I like Serena]

Could someone help me in this by simplifying/Proving the equation using Theorems/ Rules on Operation of Sets (i.e Commutative property, idempotent, assoc, dist. , definition of union , def. intersection , DeMorgan ,etc.).

Letting A, B and C be three sets ..

Prove/Disprove :
https://www.physicsforums.com/attachments/5616
Any solution would be much of help. Hoping to get answers .. Do help me :) Thank you!

Greetings TheGreat! Welcome to MHB! (Wave)

Let's start by replacing the subtractions by regular set operations.
That is, replace $A-B$ by $A \cap B^c$, and replace $C^c - A^c$ by $C^c \cap (A^c)^c = C^c \cap A$.
It makes it a bit easier to see where the left hand expression should go.
We'll get:
$$ (A^c \cup C^c) \cap (A\cap B^c) \overset?= C^c \cap A$$

Next, let's apply distributivity to the left hand side.
That is, apply the rule that $(P\cup Q)\cap R$ is the same as $(P \cap R) \cup (Q \cap R)$.
Can you? (Wondering)

 

[TheGreat]

I am still a student and I have not yet memorized all theorems on set operations .. I will try to read more on theorems .. Anyway, thank you very much ..
If ever you could prove/disprove it with the full steps, it be very much of a big help :) thanks

 

[I like Serena]

I am still a student and I have not yet memorized all theorems on set operations .. I will try to read more on theorems .. Anyway, thank you very much ..
If ever you could prove/disprove it with the full steps, it be very much of a big help :) thanks

Sorry, but we don't usually give full solutions... we believe that is counter productive.
That's why I tried to give as many hints as possible, but we believe a little effort from the poster is required to achieve anything.
If you're not up to showing any effort, or giving any indication where you're stuck, I don't see how we can "really" help you. (Shake)

 

[TheGreat]

ah okay sorry, I am trying to solve it now . Anyway on the LHS, I can't seem to use distributive because operations used inside the parenthesis are union and intersection. How's that?

$$ (A^c \cup C^c) \cap (A\cap B^c) \overset?= C^c \cap A$$

Next, let's apply distributivity to the left hand side.
That is, apply the rule that $(P\cup Q)\cap R$ is the same as $(P \cap R) \cup (Q \cap R)$.

 

[I like Serena]

ah okay sorry, I am trying to solve it now . Anyway on the LHS, I can't seem to use distributive because operations used inside the parenthesis are union and intersection. How's that?

Greetings TheGreat! Welcome to MHB! (Wave)

Let's start by replacing the subtractions by regular set operations.
That is, replace $A-B$ by $A \cap B^c$, and replace $C^c - A^c$ by $C^c \cap (A^c)^c = C^c \cap A$.
It makes it a bit easier to see where the left hand expression should go.
We'll get:
$$ (A^c \cup C^c) \cap (A\cap B^c) \overset?= C^c \cap A$$

Next, let's apply distributivity to the left hand side.
That is, apply the rule that $(P\cup Q)\cap R$ is the same as $(P \cap R) \cup (Q \cap R)$.
Can you? (Wondering)

It means that we pick $P=A^c$, $Q=C^c$, and $R=(A\cap B^c)$.
If we do that, what would $(P \cap R) \cup (Q \cap R)$ look like? (Wondering)

 

[TheGreat]

ah okay , tried distributive and I came up with:
[$A^c$ ∩ (A ∩ $B^c$ )] ∪ [$C^c$ ∩ ( A ∩ $B^c$ )]

what should I do next. Should I use Associative Property to eliminate $A^c$ ∩ A and come up with θ ?

 

[I like Serena]

ah okay , tried distributive and I came up with:
[$A^c$ ∩ (A ∩ $B^c$ )] ∪ [$C^c$ ∩ ( A ∩ $B^c$ )]

what should I do next. Should I use Associative Property to eliminate $A^c$ ∩ A and come up with θ ?

Yep! Keep going! (Nod)

 

[solakis1]

Could someone help me in this by simplifying/Proving the equation using Theorems/ Rules on Operation of Sets (i.e Commutative property, idempotent, assoc, dist. , definition of union , def. intersection , DeMorgan ,etc.).

Letting A, B and C be three sets ..

Prove/Disprove :

Any solution would be much of help. Hoping to get answers .. Do help me :) Thank you!

I've seen similar questions in here like this:
Problem posted: Prove/Disprove : A−(B∩C) = (A−B)∩(A−C)Thank you once again ... Hoping I could get some answer also ..

- jer

1st of all if the problem is prove/disprove andvnot only prove ,you should 1st try to disprove the problem, by a counter example because that will save you a lot of unnecessary work.

2ndly the problem that you recalled should read :A−(BUC) = (A−B)∩(A−C) and not

A−(B∩C) = (A−B)∩(A−C)

 

[solakis1]

Yep! Keep going! (Nod)

And how are you going to prove that:

$$A\cap B^c\cap C^c$$ it is equal to $$A\cap C^c$$ ?

 

[solakis1]

Greetings TheGreat! Welcome to MHB! (Wave)

Let's start by replacing the subtractions by regular set operations.
That is, replace $A-B$ by $A \cap B^c$, and replace $C^c - A^c$ by $C^c \cap (A^c)^c = C^c \cap A$.
It makes it a bit easier to see where the left hand expression should go.
We'll get:
$$ (A^c \cup C^c) \cap (A\cap B^c) \overset?= C^c \cap A$$

Next, let's apply distributivity to the left hand side.
That is, apply the rule that $(P\cup Q)\cap R$ is the same as $(P \cap R) \cup (Q \cap R)$.
Can you? (Wondering)

COUNTER EXAMPLE

If in the above identity of sets we put A=B ,where A and B are non empty and let $$C^c$$ be non empty,​then the above equality is not satisfied

 

[I like Serena]

And how are you going to prove that:

$$A\cap B^c\cap C^c$$ it is equal to $$A\cap C^c$$ ?

Heh. When we have that, we have exactly what we need to set up a counter example to disprove it.