Proving $(A cup B) backslash (A cap B) = (A cup C) backslash (A cap C) implies B = C$
$begingroup$
Indirect proof. Assume that $B neq C$. Therefore I assume without loss of generality that $exists x (x in B land x notin C)$. This leaves us with two possible cases:
$x in A$. But then $x notin (A cup B) backslash (A cap B)$ and $x in (A cup C) backslash (A cap C)$ which contradicts the given equality.
$x notin A$. But then $x in (A cup B) backslash (A cap B)$ and $x notin (A cup C) backslash (A cap C)$ which contradicts the given equality as well.
Therefore $B = C$.
Is this proof complete and correct?
discrete-mathematics elementary-set-theory
edited Jan 2 at 10:51
Asaf Karagila♦
309k33441775
asked Jan 2 at 10:44
KethoKetho
234
$endgroup$
add a comment |
$begingroup$
Indirect proof. Assume that $B neq C$. Therefore I assume without loss of generality that $exists x (x in B land x notin C)$. This leaves us with two possible cases:
$x in A$. But then $x notin (A cup B) backslash (A cap B)$ and $x in (A cup C) backslash (A cap C)$ which contradicts the given equality.
$x notin A$. But then $x in (A cup B) backslash (A cap B)$ and $x notin (A cup C) backslash (A cap C)$ which contradicts the given equality as well.
Therefore $B = C$.
Is this proof complete and correct?
discrete-mathematics elementary-set-theory
edited Jan 2 at 10:51
Asaf Karagila♦
309k33441775
asked Jan 2 at 10:44
KethoKetho
234
$endgroup$
2
$begingroup$
Correct........
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:47
$begingroup$
Looks good to me!
$endgroup$
– Erik Parkinson
Jan 2 at 10:47
$begingroup$
You only assume thatBhas any elements at all.
$endgroup$
– Johannes Kuhn
Jan 2 at 10:58
1
$begingroup$
@JohannesKuhn In the case where $Bsupsetneq C$ there surely will be some elements in $B$.
$endgroup$
– Graham Kemp
Jan 2 at 11:06
2
$begingroup$
Some additional information: LHS is called symmetric difference between $A$ and $B$ denoted by $ADelta B$. Given $ADelta B=ADelta C $ we get $ ADelta (ADelta B)=ADelta (ADelta C) $ and the two side of the s equation are precisely $B$ and $C$!
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:07
add a comment |
$begingroup$
Indirect proof. Assume that $B neq C$. Therefore I assume without loss of generality that $exists x (x in B land x notin C)$. This leaves us with two possible cases:
$x in A$. But then $x notin (A cup B) backslash (A cap B)$ and $x in (A cup C) backslash (A cap C)$ which contradicts the given equality.
$x notin A$. But then $x in (A cup B) backslash (A cap B)$ and $x notin (A cup C) backslash (A cap C)$ which contradicts the given equality as well.
Therefore $B = C$.
Is this proof complete and correct?
discrete-mathematics elementary-set-theory
edited Jan 2 at 10:51
Asaf Karagila♦
309k33441775
asked Jan 2 at 10:44
KethoKetho
234
$endgroup$
Indirect proof. Assume that $B neq C$. Therefore I assume without loss of generality that $exists x (x in B land x notin C)$. This leaves us with two possible cases:
$x in A$. But then $x notin (A cup B) backslash (A cap B)$ and $x in (A cup C) backslash (A cap C)$ which contradicts the given equality.
$x notin A$. But then $x in (A cup B) backslash (A cap B)$ and $x notin (A cup C) backslash (A cap C)$ which contradicts the given equality as well.
Therefore $B = C$.
Is this proof complete and correct?
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
edited Jan 2 at 10:51
Asaf Karagila♦
309k33441775
asked Jan 2 at 10:44
KethoKetho
234
edited Jan 2 at 10:51
Asaf Karagila♦
309k33441775
asked Jan 2 at 10:44
KethoKetho
234
edited Jan 2 at 10:51
Asaf Karagila♦
309k33441775
edited Jan 2 at 10:51
Asaf Karagila♦
309k33441775
edited Jan 2 at 10:51
Asaf Karagila♦
309k33441775
309k33441775
asked Jan 2 at 10:44
KethoKetho
234
asked Jan 2 at 10:44
KethoKetho
234
asked Jan 2 at 10:44
KethoKetho
234
234
2
$begingroup$
Correct........
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:47
$begingroup$
Looks good to me!
$endgroup$
– Erik Parkinson
Jan 2 at 10:47
$begingroup$
You only assume thatBhas any elements at all.
$endgroup$
– Johannes Kuhn
Jan 2 at 10:58
1
$begingroup$
@JohannesKuhn In the case where $Bsupsetneq C$ there surely will be some elements in $B$.
$endgroup$
– Graham Kemp
Jan 2 at 11:06
2
$begingroup$
Some additional information: LHS is called symmetric difference between $A$ and $B$ denoted by $ADelta B$. Given $ADelta B=ADelta C $ we get $ ADelta (ADelta B)=ADelta (ADelta C) $ and the two side of the s equation are precisely $B$ and $C$!
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:07
add a comment |
2
$begingroup$
Correct........
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:47
$begingroup$
Looks good to me!
$endgroup$
– Erik Parkinson
Jan 2 at 10:47
$begingroup$
You only assume thatBhas any elements at all.
$endgroup$
– Johannes Kuhn
Jan 2 at 10:58
1
$begingroup$
@JohannesKuhn In the case where $Bsupsetneq C$ there surely will be some elements in $B$.
$endgroup$
– Graham Kemp
Jan 2 at 11:06
2
$begingroup$
Some additional information: LHS is called symmetric difference between $A$ and $B$ denoted by $ADelta B$. Given $ADelta B=ADelta C $ we get $ ADelta (ADelta B)=ADelta (ADelta C) $ and the two side of the s equation are precisely $B$ and $C$!
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:07
2
2
$begingroup$
Correct........
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:47
$begingroup$
Correct........
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:47
$begingroup$
Looks good to me!
$endgroup$
– Erik Parkinson
Jan 2 at 10:47
$begingroup$
Looks good to me!
$endgroup$
– Erik Parkinson
Jan 2 at 10:47
$begingroup$
You only assume that
B has any elements at all.$endgroup$
– Johannes Kuhn
Jan 2 at 10:58
$begingroup$
You only assume that
B has any elements at all.$endgroup$
– Johannes Kuhn
Jan 2 at 10:58
1
1
$begingroup$
@JohannesKuhn In the case where $Bsupsetneq C$ there surely will be some elements in $B$.
$endgroup$
– Graham Kemp
Jan 2 at 11:06
$begingroup$
@JohannesKuhn In the case where $Bsupsetneq C$ there surely will be some elements in $B$.
$endgroup$
– Graham Kemp
Jan 2 at 11:06
2
2
$begingroup$
Some additional information: LHS is called symmetric difference between $A$ and $B$ denoted by $ADelta B$. Given $ADelta B=ADelta C $ we get $ ADelta (ADelta B)=ADelta (ADelta C) $ and the two side of the s equation are precisely $B$ and $C$!
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:07
$begingroup$
Some additional information: LHS is called symmetric difference between $A$ and $B$ denoted by $ADelta B$. Given $ADelta B=ADelta C $ we get $ ADelta (ADelta B)=ADelta (ADelta C) $ and the two side of the s equation are precisely $B$ and $C$!
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:07
add a comment |
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2
$begingroup$
Correct........
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:47
$begingroup$
Looks good to me!
$endgroup$
– Erik Parkinson
Jan 2 at 10:47
$begingroup$
You only assume that
Bhas any elements at all.$endgroup$
– Johannes Kuhn
Jan 2 at 10:58
1
$begingroup$
@JohannesKuhn In the case where $Bsupsetneq C$ there surely will be some elements in $B$.
$endgroup$
– Graham Kemp
Jan 2 at 11:06
2
$begingroup$
Some additional information: LHS is called symmetric difference between $A$ and $B$ denoted by $ADelta B$. Given $ADelta B=ADelta C $ we get $ ADelta (ADelta B)=ADelta (ADelta C) $ and the two side of the s equation are precisely $B$ and $C$!
$endgroup$
– Kavi Rama Murthy
Jan 2 at 12:07