Universal set proof [discrete mathematics]Sets problem with upper boundsDiscrete Math - Set Theory - Power SetExtend an acyclic relation to an orderingIs this a valid proof for the following: Let $A$ and $B$ be sets. Show in general that $overlineA times B neq overlineAtimes overlineB$Why does universal generalization work? (the rule of inference)Proof by Cases [discrete mathematics]Direct proof of Universal Set [discrete mathematics]Draw a pentagon with any 2 angle sums being less than $216^o$Indirect Proof [discrete mathematics]Big-$O$ verification [discrete mathematics]
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Universal set proof [discrete mathematics]
Sets problem with upper boundsDiscrete Math - Set Theory - Power SetExtend an acyclic relation to an orderingIs this a valid proof for the following: Let $A$ and $B$ be sets. Show in general that $overlineA times B neq overlineAtimes overlineB$Why does universal generalization work? (the rule of inference)Proof by Cases [discrete mathematics]Direct proof of Universal Set [discrete mathematics]Draw a pentagon with any 2 angle sums being less than $216^o$Indirect Proof [discrete mathematics]Big-$O$ verification [discrete mathematics]
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I've come across challenge proof question in my discrete mathematics textbook that I'm trying to solve for practice but unfortunately it does not have a solution. Any help with a reasonable explanation or solution so that I can understand where to start and verify my work would be greatly appreciated:
Suppose that $mathcal U$ is the universal set, and that $A$, $B$ and
$C$ are three arbitrary sets of elements of $mathcal U$. Prove that if $A - B subseteq C$, then $A - C subseteq B$.
Thank you!
discrete-mathematics proof-verification proof-writing proof-explanation recreational-mathematics
asked Mar 16 at 21:48
Joe BidenJoe Biden
25
$endgroup$
add a comment |
$begingroup$
I've come across challenge proof question in my discrete mathematics textbook that I'm trying to solve for practice but unfortunately it does not have a solution. Any help with a reasonable explanation or solution so that I can understand where to start and verify my work would be greatly appreciated:
Suppose that $mathcal U$ is the universal set, and that $A$, $B$ and
$C$ are three arbitrary sets of elements of $mathcal U$. Prove that if $A - B subseteq C$, then $A - C subseteq B$.
Thank you!
discrete-mathematics proof-verification proof-writing proof-explanation recreational-mathematics
asked Mar 16 at 21:48
Joe BidenJoe Biden
25
$endgroup$
add a comment |
$begingroup$
I've come across challenge proof question in my discrete mathematics textbook that I'm trying to solve for practice but unfortunately it does not have a solution. Any help with a reasonable explanation or solution so that I can understand where to start and verify my work would be greatly appreciated:
Suppose that $mathcal U$ is the universal set, and that $A$, $B$ and
$C$ are three arbitrary sets of elements of $mathcal U$. Prove that if $A - B subseteq C$, then $A - C subseteq B$.
Thank you!
discrete-mathematics proof-verification proof-writing proof-explanation recreational-mathematics
asked Mar 16 at 21:48
Joe BidenJoe Biden
25
$endgroup$
I've come across challenge proof question in my discrete mathematics textbook that I'm trying to solve for practice but unfortunately it does not have a solution. Any help with a reasonable explanation or solution so that I can understand where to start and verify my work would be greatly appreciated:
Suppose that $mathcal U$ is the universal set, and that $A$, $B$ and
$C$ are three arbitrary sets of elements of $mathcal U$. Prove that if $A - B subseteq C$, then $A - C subseteq B$.
Thank you!
discrete-mathematics proof-verification proof-writing proof-explanation recreational-mathematics
discrete-mathematics proof-verification proof-writing proof-explanation recreational-mathematics
asked Mar 16 at 21:48
Joe BidenJoe Biden
25
asked Mar 16 at 21:48
Joe BidenJoe Biden
25
asked Mar 16 at 21:48
Joe BidenJoe Biden
25
asked Mar 16 at 21:48
Joe BidenJoe Biden
25
asked Mar 16 at 21:48
Joe BidenJoe Biden
25
25
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $A cap B^complement subset C$, we must have $C^complement subset (A cap B^complement)^complement = A^complement cup B$. Therefore,
$$A cap C^complement subset A cap (A^complement cup B) = A cap B subset B.$$
answered Mar 16 at 22:01
Gary MoonGary Moon
77616
$endgroup$
$begingroup$
I'm not quite sure I understand your notation, what does BC mean?
$endgroup$
– Joe Biden
Mar 16 at 22:10
$begingroup$
$B^complement$ denotes the complement of $B$ (in the universal space). So, $A - B = A cap B^complement$.
$endgroup$
– Gary Moon
Mar 16 at 22:18
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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$begingroup$
Since $A cap B^complement subset C$, we must have $C^complement subset (A cap B^complement)^complement = A^complement cup B$. Therefore,
$$A cap C^complement subset A cap (A^complement cup B) = A cap B subset B.$$
answered Mar 16 at 22:01
Gary MoonGary Moon
77616
$endgroup$
$begingroup$
I'm not quite sure I understand your notation, what does BC mean?
$endgroup$
– Joe Biden
Mar 16 at 22:10
$begingroup$
$B^complement$ denotes the complement of $B$ (in the universal space). So, $A - B = A cap B^complement$.
$endgroup$
– Gary Moon
Mar 16 at 22:18
add a comment |
$begingroup$
Since $A cap B^complement subset C$, we must have $C^complement subset (A cap B^complement)^complement = A^complement cup B$. Therefore,
$$A cap C^complement subset A cap (A^complement cup B) = A cap B subset B.$$
answered Mar 16 at 22:01
Gary MoonGary Moon
77616
$endgroup$
$begingroup$
I'm not quite sure I understand your notation, what does BC mean?
$endgroup$
– Joe Biden
Mar 16 at 22:10
$begingroup$
$B^complement$ denotes the complement of $B$ (in the universal space). So, $A - B = A cap B^complement$.
$endgroup$
– Gary Moon
Mar 16 at 22:18
add a comment |
$begingroup$
Since $A cap B^complement subset C$, we must have $C^complement subset (A cap B^complement)^complement = A^complement cup B$. Therefore,
$$A cap C^complement subset A cap (A^complement cup B) = A cap B subset B.$$
answered Mar 16 at 22:01
Gary MoonGary Moon
77616
$endgroup$
Since $A cap B^complement subset C$, we must have $C^complement subset (A cap B^complement)^complement = A^complement cup B$. Therefore,
$$A cap C^complement subset A cap (A^complement cup B) = A cap B subset B.$$
answered Mar 16 at 22:01
Gary MoonGary Moon
77616
answered Mar 16 at 22:01
Gary MoonGary Moon
77616
answered Mar 16 at 22:01
Gary MoonGary Moon
77616
answered Mar 16 at 22:01
Gary MoonGary Moon
77616
77616
$begingroup$
I'm not quite sure I understand your notation, what does BC mean?
$endgroup$
– Joe Biden
Mar 16 at 22:10
$begingroup$
$B^complement$ denotes the complement of $B$ (in the universal space). So, $A - B = A cap B^complement$.
$endgroup$
– Gary Moon
Mar 16 at 22:18
add a comment |
$begingroup$
I'm not quite sure I understand your notation, what does BC mean?
$endgroup$
– Joe Biden
Mar 16 at 22:10
$begingroup$
$B^complement$ denotes the complement of $B$ (in the universal space). So, $A - B = A cap B^complement$.
$endgroup$
– Gary Moon
Mar 16 at 22:18
$begingroup$
I'm not quite sure I understand your notation, what does BC mean?
$endgroup$
– Joe Biden
Mar 16 at 22:10
$begingroup$
I'm not quite sure I understand your notation, what does BC mean?
$endgroup$
– Joe Biden
Mar 16 at 22:10
$begingroup$
$B^complement$ denotes the complement of $B$ (in the universal space). So, $A - B = A cap B^complement$.
$endgroup$
– Gary Moon
Mar 16 at 22:18
$begingroup$
$B^complement$ denotes the complement of $B$ (in the universal space). So, $A - B = A cap B^complement$.
$endgroup$
– Gary Moon
Mar 16 at 22:18
add a comment |
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