Lewis Harvey

2021-07-28

Let A, B, and C be sets. Show that $\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)$

Caren

If A, B, C is any set, then to prove the equation we need to use the various Seth identity laws

nick1337

To show that $\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)$, we can start by expanding the left side:
$\left(A-B\right)-C=\left\{x\in A-B\mid x\notin C\right\}$
Now, let's simplify $A-B$:
$A-B=\left\{x\in A\mid x\notin B\right\}$
Substituting this into our expression:
$\left(A-B\right)-C=\left\{x\in A\mid x\notin B\right\}-C$
Next, let's simplify $\left(A-C\right)$:
$A-C=\left\{x\in A\mid x\notin C\right\}$
We can also simplify $\left(B-C\right)$:
$B-C=\left\{x\in B\mid x\notin C\right\}$
Substituting these into the right side of the equation:
$\left(A-C\right)-\left(B-C\right)=\left\{x\in A\mid x\notin C\right\}-\left\{x\in B\mid x\notin C\right\}$
Now, let's combine the two expressions:
$\left(A-C\right)-\left(B-C\right)=\left\{x\in A\mid x\notin C\right\}\cap \left\{x\in B\mid x\in C\right\}$
To simplify this further, note that if $x\in B$ and $x\in C$, then $x\notin A-C$. Thus, we can rewrite the expression as:
$\left(A-C\right)-\left(B-C\right)=\left\{x\in A\mid x\notin C\right\}\cap \left\{x\in B\mid x\in C\right\}=\left\{x\in A\mid x\notin C\right\}$
Therefore, we have shown that $\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)$.

Don Sumner

Step 1:
Let's consider an arbitrary element $x$ in $\left(A-B\right)-C$.
$x\in \left(A-B\right)-C$
This means that $x$ belongs to the set $A-B$ but does not belong to the set $C$.
By the definition of set difference, we have:
$x\in A$ and $x\notin B$
Step 2:
Now, we want to show that $x$ also belongs to $\left(A-C\right)-\left(B-C\right)$.
$x\in \left(A-C\right)-\left(B-C\right)$
This means that $x$ belongs to the set $A-C$ but does not belong to the set $B-C$.
By the definition of set difference, we have:
$x\in A$ and $x\notin C$
Also, $x$ does not belong to $B-C$, which means $x\notin B$ or $x\in C$.
Combining these conditions, we have:
$x\in A$ and $x\notin B$ and ($x\notin B$ or $x\in C$)
Using logical equivalence, we can simplify the last condition:
$x\in A$ and $x\notin B$ and $x\in C$
Thus, we have shown that an arbitrary element $x$ in $\left(A-B\right)-C$ also belongs to $\left(A-C\right)-\left(B-C\right)$.
Step 3:
Now, let's consider an arbitrary element $y$ in $\left(A-C\right)-\left(B-C\right)$.
$y\in \left(A-C\right)-\left(B-C\right)$
This means that $y$ belongs to the set $A-C$ but does not belong to the set $B-C$.
By the definition of set difference, we have:
$y\in A$ and $y\notin C$
Also, $y$ does not belong to $B-C$, which means $y\notin B$ or $y\in C$.
Combining these conditions, we have:
$y\in A$ and $y\notin C$ and ($y\notin B$ or $y\in C$)
Using logical equivalence, we can simplify the last condition:
$y\in A$ and $y\notin B$ and $y\notin C$
This implies that $y$ belongs to $\left(A-B\right)-C$.
Therefore, we have shown that any element $y$ in $\left(A-C\right)-\left(B-C\right)$ also belongs to $\left(A-B\right)-C$.
Since we have shown that each set is a subset of the other, we can conclude that $\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)$.

RizerMix

Starting with the left-hand side of the equation, we have:
$\left(A-B\right)-C$
Using set difference, we can rewrite this as:
$\left(A\cap \stackrel{―}{B}\right)-C$
Next, applying set difference again, we get:
$\left(A\cap \stackrel{―}{B}\right)\cap \stackrel{―}{C}$
Now, let's move to the right-hand side of the equation:
$\left(A-C\right)-\left(B-C\right)$
Again, using set difference, we can express this as:
$\left(A\cap \stackrel{―}{C}\right)\cap \stackrel{―}{\left(B\cap \stackrel{―}{C}\right)}$
Expanding the complement in the second term, we have:
$\left(A\cap \stackrel{―}{C}\right)\cap \left(\stackrel{―}{B}\cup C\right)$
Now, we can distribute the intersection over the union:
$\left(A\cap \stackrel{―}{C}\right)\cap \stackrel{―}{B}\cup \left(A\cap \stackrel{―}{C}\right)\cap C$
Applying the distributive property, we obtain:
$\left(A\cap \stackrel{―}{C}\cap \stackrel{―}{B}\right)\cup \left(A\cap \stackrel{―}{C}\cap C\right)$
Finally, simplifying the intersection of a set with its complement, we have:
$\left(A\cap \stackrel{―}{C}\cap \stackrel{―}{B}\right)\cup \varnothing$
Simplifying further, we get:
$A\cap \stackrel{―}{C}\cap \stackrel{―}{B}$
Hence, we have shown that $\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)$.

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