Proving set identities: empty set case.
I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$
I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.
We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.
But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.
In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?
If so, some suggestions as to how to incorporate them into proofs would be helpful :-).
elementary-set-theory proof-writing
asked 1 hour ago
E-mu
405312
add a comment |
I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$
I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.
We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.
But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.
In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?
If so, some suggestions as to how to incorporate them into proofs would be helpful :-).
elementary-set-theory proof-writing
asked 1 hour ago
E-mu
405312
2
Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
– DonAntonio
1 hour ago
Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
– Mauro ALLEGRANZA
44 mins ago
In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
– Mauro ALLEGRANZA
41 mins ago
add a comment |
I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$
I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.
We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.
But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.
In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?
If so, some suggestions as to how to incorporate them into proofs would be helpful :-).
elementary-set-theory proof-writing
asked 1 hour ago
E-mu
405312
I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$
I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.
We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.
But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.
In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?
If so, some suggestions as to how to incorporate them into proofs would be helpful :-).
elementary-set-theory proof-writing
elementary-set-theory proof-writing
asked 1 hour ago
E-mu
405312
asked 1 hour ago
E-mu
405312
asked 1 hour ago
E-mu
405312
asked 1 hour ago
E-mu
405312
asked 1 hour ago
E-mu
405312
405312
2
Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
– DonAntonio
1 hour ago
Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
– Mauro ALLEGRANZA
44 mins ago
In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
– Mauro ALLEGRANZA
41 mins ago
add a comment |
2
Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
– DonAntonio
1 hour ago
Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
– Mauro ALLEGRANZA
44 mins ago
In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
– Mauro ALLEGRANZA
41 mins ago
2
2
Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
– DonAntonio
1 hour ago
Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
– DonAntonio
1 hour ago
Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
– Mauro ALLEGRANZA
44 mins ago
Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
– Mauro ALLEGRANZA
44 mins ago
In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
– Mauro ALLEGRANZA
41 mins ago
In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
– Mauro ALLEGRANZA
41 mins ago
add a comment |
3 Answers
3
active
oldest
votes
In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that
for every $x$, if $xin X$, then $xin Y$.
Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when
either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true
If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.
The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.
answered 1 hour ago
egreg
177k1484198
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
– E-mu
13 mins ago
add a comment |
If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.
It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
answered 1 hour ago
drhab
96.7k544128
add a comment |
Since the empty set is a subset of any set, there is no need of including that in a formal proof.
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .
answered 1 hour ago
Mohammad Riazi-Kermani
40.5k42058
add a comment |
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3 Answers
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3 Answers
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active
oldest
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active
oldest
votes
active
oldest
votes
In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that
for every $x$, if $xin X$, then $xin Y$.
Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when
either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true
If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.
The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.
answered 1 hour ago
egreg
177k1484198
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
– E-mu
13 mins ago
add a comment |
In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that
for every $x$, if $xin X$, then $xin Y$.
Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when
either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true
If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.
The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.
answered 1 hour ago
egreg
177k1484198
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
– E-mu
13 mins ago
add a comment |
In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that
for every $x$, if $xin X$, then $xin Y$.
Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when
either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true
If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.
The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.
answered 1 hour ago
egreg
177k1484198
In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that
for every $x$, if $xin X$, then $xin Y$.
Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when
either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true
If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.
The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.
answered 1 hour ago
egreg
177k1484198
answered 1 hour ago
egreg
177k1484198
answered 1 hour ago
egreg
177k1484198
answered 1 hour ago
egreg
177k1484198
177k1484198
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
– E-mu
13 mins ago
add a comment |
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
– E-mu
13 mins ago
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
– E-mu
13 mins ago
Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
– E-mu
13 mins ago
add a comment |
If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.
It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
answered 1 hour ago
drhab
96.7k544128
add a comment |
If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.
It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
answered 1 hour ago
drhab
96.7k544128
add a comment |
If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.
It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
answered 1 hour ago
drhab
96.7k544128
If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.
It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.
So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.
answered 1 hour ago
drhab
96.7k544128
edited 1 hour ago
answered 1 hour ago
drhab
96.7k544128
answered 1 hour ago
drhab
96.7k544128
answered 1 hour ago
drhab
96.7k544128
96.7k544128
add a comment |
add a comment |
Since the empty set is a subset of any set, there is no need of including that in a formal proof.
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .
answered 1 hour ago
Mohammad Riazi-Kermani
40.5k42058
add a comment |
Since the empty set is a subset of any set, there is no need of including that in a formal proof.
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .
answered 1 hour ago
Mohammad Riazi-Kermani
40.5k42058
add a comment |
Since the empty set is a subset of any set, there is no need of including that in a formal proof.
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .
answered 1 hour ago
Mohammad Riazi-Kermani
40.5k42058
Since the empty set is a subset of any set, there is no need of including that in a formal proof.
However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .
answered 1 hour ago
Mohammad Riazi-Kermani
40.5k42058
answered 1 hour ago
Mohammad Riazi-Kermani
40.5k42058
answered 1 hour ago
Mohammad Riazi-Kermani
40.5k42058
answered 1 hour ago
Mohammad Riazi-Kermani
40.5k42058
40.5k42058
add a comment |
add a comment |
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Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
– DonAntonio
1 hour ago
Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
– Mauro ALLEGRANZA
44 mins ago
In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
– Mauro ALLEGRANZA
41 mins ago