Set theory operations proof
$begingroup$
Demonstrate that $A cap (B cup C) = (A cap B) cup (A cap C)$
How can I proof that?
Well, if I pick up an $ x in A$ and the same $ x in B cup C$, then $x in (A cap (B cup C))$.
If $ x in A $, and if $x in B$ and $x notin C$, then the equality is true. Or, $ x in A $, and if $x notin B$ and $x in C$, then the equality is true. For last, $ x in A $, and if $x in B$ and $x in C$, then the equality is true again.
Is this enough to proof the equality?
elementary-set-theory
edited Jan 21 at 1:15
Andrés E. Caicedo
65.5k8158249
asked Jan 21 at 0:12
ArduinArduin
307
$endgroup$
add a comment |
$begingroup$
Demonstrate that $A cap (B cup C) = (A cap B) cup (A cap C)$
How can I proof that?
Well, if I pick up an $ x in A$ and the same $ x in B cup C$, then $x in (A cap (B cup C))$.
If $ x in A $, and if $x in B$ and $x notin C$, then the equality is true. Or, $ x in A $, and if $x notin B$ and $x in C$, then the equality is true. For last, $ x in A $, and if $x in B$ and $x in C$, then the equality is true again.
Is this enough to proof the equality?
elementary-set-theory
edited Jan 21 at 1:15
Andrés E. Caicedo
65.5k8158249
asked Jan 21 at 0:12
ArduinArduin
307
$endgroup$
$begingroup$
It's not really accurate to say that the "equality is true" when dealing with cases of individual members $x$ of the sets.
$endgroup$
– nathan.j.mcdougall
Jan 21 at 0:26
add a comment |
$begingroup$
Demonstrate that $A cap (B cup C) = (A cap B) cup (A cap C)$
How can I proof that?
Well, if I pick up an $ x in A$ and the same $ x in B cup C$, then $x in (A cap (B cup C))$.
If $ x in A $, and if $x in B$ and $x notin C$, then the equality is true. Or, $ x in A $, and if $x notin B$ and $x in C$, then the equality is true. For last, $ x in A $, and if $x in B$ and $x in C$, then the equality is true again.
Is this enough to proof the equality?
elementary-set-theory
edited Jan 21 at 1:15
Andrés E. Caicedo
65.5k8158249
asked Jan 21 at 0:12
ArduinArduin
307
$endgroup$
Demonstrate that $A cap (B cup C) = (A cap B) cup (A cap C)$
How can I proof that?
Well, if I pick up an $ x in A$ and the same $ x in B cup C$, then $x in (A cap (B cup C))$.
If $ x in A $, and if $x in B$ and $x notin C$, then the equality is true. Or, $ x in A $, and if $x notin B$ and $x in C$, then the equality is true. For last, $ x in A $, and if $x in B$ and $x in C$, then the equality is true again.
Is this enough to proof the equality?
elementary-set-theory
elementary-set-theory
edited Jan 21 at 1:15
Andrés E. Caicedo
65.5k8158249
asked Jan 21 at 0:12
ArduinArduin
307
edited Jan 21 at 1:15
Andrés E. Caicedo
65.5k8158249
asked Jan 21 at 0:12
ArduinArduin
307
edited Jan 21 at 1:15
Andrés E. Caicedo
65.5k8158249
edited Jan 21 at 1:15
Andrés E. Caicedo
65.5k8158249
edited Jan 21 at 1:15
Andrés E. Caicedo
65.5k8158249
65.5k8158249
asked Jan 21 at 0:12
ArduinArduin
307
asked Jan 21 at 0:12
ArduinArduin
307
asked Jan 21 at 0:12
ArduinArduin
307
307
$begingroup$
It's not really accurate to say that the "equality is true" when dealing with cases of individual members $x$ of the sets.
$endgroup$
– nathan.j.mcdougall
Jan 21 at 0:26
add a comment |
$begingroup$
It's not really accurate to say that the "equality is true" when dealing with cases of individual members $x$ of the sets.
$endgroup$
– nathan.j.mcdougall
Jan 21 at 0:26
$begingroup$
It's not really accurate to say that the "equality is true" when dealing with cases of individual members $x$ of the sets.
$endgroup$
– nathan.j.mcdougall
Jan 21 at 0:26
$begingroup$
It's not really accurate to say that the "equality is true" when dealing with cases of individual members $x$ of the sets.
$endgroup$
– nathan.j.mcdougall
Jan 21 at 0:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
That's basically correct, but it can be slightly simplified and clarified.
For the clarification, you should note that if $x in B cup C$ then it must be true that either $x in B$ or $x in C$. You haven't explicitly stated that, but you have used that fact.
For the simplification, you don't have to consider the three cases: (1) $x in B$ and $x notin C$, (2) $x in C$ and $x notin B$, and (3) $x in B$ and $x in C$. It is enough to just consider two cases: (1) $x in B$ (and we don't care whether $x$ is in $C$ or not), and (2) $x in C$ (and we don't care whether $x$ is in $B$ or not).
answered Jan 21 at 0:24
kccukccu
10.5k11228
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
That's basically correct, but it can be slightly simplified and clarified.
For the clarification, you should note that if $x in B cup C$ then it must be true that either $x in B$ or $x in C$. You haven't explicitly stated that, but you have used that fact.
For the simplification, you don't have to consider the three cases: (1) $x in B$ and $x notin C$, (2) $x in C$ and $x notin B$, and (3) $x in B$ and $x in C$. It is enough to just consider two cases: (1) $x in B$ (and we don't care whether $x$ is in $C$ or not), and (2) $x in C$ (and we don't care whether $x$ is in $B$ or not).
answered Jan 21 at 0:24
kccukccu
10.5k11228
$endgroup$
add a comment |
$begingroup$
That's basically correct, but it can be slightly simplified and clarified.
For the clarification, you should note that if $x in B cup C$ then it must be true that either $x in B$ or $x in C$. You haven't explicitly stated that, but you have used that fact.
For the simplification, you don't have to consider the three cases: (1) $x in B$ and $x notin C$, (2) $x in C$ and $x notin B$, and (3) $x in B$ and $x in C$. It is enough to just consider two cases: (1) $x in B$ (and we don't care whether $x$ is in $C$ or not), and (2) $x in C$ (and we don't care whether $x$ is in $B$ or not).
answered Jan 21 at 0:24
kccukccu
10.5k11228
$endgroup$
add a comment |
$begingroup$
That's basically correct, but it can be slightly simplified and clarified.
For the clarification, you should note that if $x in B cup C$ then it must be true that either $x in B$ or $x in C$. You haven't explicitly stated that, but you have used that fact.
For the simplification, you don't have to consider the three cases: (1) $x in B$ and $x notin C$, (2) $x in C$ and $x notin B$, and (3) $x in B$ and $x in C$. It is enough to just consider two cases: (1) $x in B$ (and we don't care whether $x$ is in $C$ or not), and (2) $x in C$ (and we don't care whether $x$ is in $B$ or not).
answered Jan 21 at 0:24
kccukccu
10.5k11228
$endgroup$
That's basically correct, but it can be slightly simplified and clarified.
For the clarification, you should note that if $x in B cup C$ then it must be true that either $x in B$ or $x in C$. You haven't explicitly stated that, but you have used that fact.
For the simplification, you don't have to consider the three cases: (1) $x in B$ and $x notin C$, (2) $x in C$ and $x notin B$, and (3) $x in B$ and $x in C$. It is enough to just consider two cases: (1) $x in B$ (and we don't care whether $x$ is in $C$ or not), and (2) $x in C$ (and we don't care whether $x$ is in $B$ or not).
answered Jan 21 at 0:24
kccukccu
10.5k11228
answered Jan 21 at 0:24
kccukccu
10.5k11228
answered Jan 21 at 0:24
kccukccu
10.5k11228
answered Jan 21 at 0:24
kccukccu
10.5k11228
10.5k11228
add a comment |
add a comment |
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$begingroup$
It's not really accurate to say that the "equality is true" when dealing with cases of individual members $x$ of the sets.
$endgroup$
– nathan.j.mcdougall
Jan 21 at 0:26