Proving adsorption law
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0
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Prove the absorption law
$Acap(Acup B)=A$
My Try:
Attempt 1:
$xin Acap(Acup B)$
$xin A$ and $xin Acup Brightarrow xin A$
Thus $Acap(Acup B)subset A$
$xin Arightarrow xin A$ and $xin Acup Brightarrow xin Acap(Acup B)$
Thus $Asubset Acap(Acup B)$
So, $Acap(Acup B)=A$
Attempt 2:
$xin Acap(Acup B)iff xin A$ and $xin Acup B$
$iff xin A$ and $[xin A$ or $xin B]$
$iff xin A$
$Acap(Acup B)=A$
My Question: Which of my above attempts is most precise?
discrete-mathematics proof-verification elementary-set-theory logic
asked Dec 2 at 19:42
user982787
897
add a comment |
up vote
0
down vote
favorite
Prove the absorption law
$Acap(Acup B)=A$
My Try:
Attempt 1:
$xin Acap(Acup B)$
$xin A$ and $xin Acup Brightarrow xin A$
Thus $Acap(Acup B)subset A$
$xin Arightarrow xin A$ and $xin Acup Brightarrow xin Acap(Acup B)$
Thus $Asubset Acap(Acup B)$
So, $Acap(Acup B)=A$
Attempt 2:
$xin Acap(Acup B)iff xin A$ and $xin Acup B$
$iff xin A$ and $[xin A$ or $xin B]$
$iff xin A$
$Acap(Acup B)=A$
My Question: Which of my above attempts is most precise?
discrete-mathematics proof-verification elementary-set-theory logic
asked Dec 2 at 19:42
user982787
897
The former is more readable. Both are missing a whole bunch of words, though.
– user3482749
Dec 2 at 19:46
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove the absorption law
$Acap(Acup B)=A$
My Try:
Attempt 1:
$xin Acap(Acup B)$
$xin A$ and $xin Acup Brightarrow xin A$
Thus $Acap(Acup B)subset A$
$xin Arightarrow xin A$ and $xin Acup Brightarrow xin Acap(Acup B)$
Thus $Asubset Acap(Acup B)$
So, $Acap(Acup B)=A$
Attempt 2:
$xin Acap(Acup B)iff xin A$ and $xin Acup B$
$iff xin A$ and $[xin A$ or $xin B]$
$iff xin A$
$Acap(Acup B)=A$
My Question: Which of my above attempts is most precise?
discrete-mathematics proof-verification elementary-set-theory logic
asked Dec 2 at 19:42
user982787
897
Prove the absorption law
$Acap(Acup B)=A$
My Try:
Attempt 1:
$xin Acap(Acup B)$
$xin A$ and $xin Acup Brightarrow xin A$
Thus $Acap(Acup B)subset A$
$xin Arightarrow xin A$ and $xin Acup Brightarrow xin Acap(Acup B)$
Thus $Asubset Acap(Acup B)$
So, $Acap(Acup B)=A$
Attempt 2:
$xin Acap(Acup B)iff xin A$ and $xin Acup B$
$iff xin A$ and $[xin A$ or $xin B]$
$iff xin A$
$Acap(Acup B)=A$
My Question: Which of my above attempts is most precise?
discrete-mathematics proof-verification elementary-set-theory logic
discrete-mathematics proof-verification elementary-set-theory logic
asked Dec 2 at 19:42
user982787
897
asked Dec 2 at 19:42
user982787
897
asked Dec 2 at 19:42
user982787
897
asked Dec 2 at 19:42
user982787
897
asked Dec 2 at 19:42
user982787
897
897
The former is more readable. Both are missing a whole bunch of words, though.
– user3482749
Dec 2 at 19:46
add a comment |
The former is more readable. Both are missing a whole bunch of words, though.
– user3482749
Dec 2 at 19:46
The former is more readable. Both are missing a whole bunch of words, though.
– user3482749
Dec 2 at 19:46
The former is more readable. Both are missing a whole bunch of words, though.
– user3482749
Dec 2 at 19:46
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Instead of a point by point analysis
here is a set alegbra proof.
As U $cap$ V $subseteq$ U $subseteq$ U $cup$ V,
A $cap$ (A $cup$ B) $subseteq$ A = A $cap$ A $subseteq$ A $cap$ (A $cup$ B).
Thus A = A $cap$ (A $cup$ B).
The dual theorem A = A $cup$ (A $cap$ B)
is mostly an application of DeMorgans rules or one
could adapt the above proof to prove the dual theorem.
answered Dec 2 at 23:51
William Elliot
6,9072518
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Instead of a point by point analysis
here is a set alegbra proof.
As U $cap$ V $subseteq$ U $subseteq$ U $cup$ V,
A $cap$ (A $cup$ B) $subseteq$ A = A $cap$ A $subseteq$ A $cap$ (A $cup$ B).
Thus A = A $cap$ (A $cup$ B).
The dual theorem A = A $cup$ (A $cap$ B)
is mostly an application of DeMorgans rules or one
could adapt the above proof to prove the dual theorem.
answered Dec 2 at 23:51
William Elliot
6,9072518
add a comment |
up vote
0
down vote
Instead of a point by point analysis
here is a set alegbra proof.
As U $cap$ V $subseteq$ U $subseteq$ U $cup$ V,
A $cap$ (A $cup$ B) $subseteq$ A = A $cap$ A $subseteq$ A $cap$ (A $cup$ B).
Thus A = A $cap$ (A $cup$ B).
The dual theorem A = A $cup$ (A $cap$ B)
is mostly an application of DeMorgans rules or one
could adapt the above proof to prove the dual theorem.
answered Dec 2 at 23:51
William Elliot
6,9072518
add a comment |
up vote
0
down vote
up vote
0
down vote
Instead of a point by point analysis
here is a set alegbra proof.
As U $cap$ V $subseteq$ U $subseteq$ U $cup$ V,
A $cap$ (A $cup$ B) $subseteq$ A = A $cap$ A $subseteq$ A $cap$ (A $cup$ B).
Thus A = A $cap$ (A $cup$ B).
The dual theorem A = A $cup$ (A $cap$ B)
is mostly an application of DeMorgans rules or one
could adapt the above proof to prove the dual theorem.
answered Dec 2 at 23:51
William Elliot
6,9072518
Instead of a point by point analysis
here is a set alegbra proof.
As U $cap$ V $subseteq$ U $subseteq$ U $cup$ V,
A $cap$ (A $cup$ B) $subseteq$ A = A $cap$ A $subseteq$ A $cap$ (A $cup$ B).
Thus A = A $cap$ (A $cup$ B).
The dual theorem A = A $cup$ (A $cap$ B)
is mostly an application of DeMorgans rules or one
could adapt the above proof to prove the dual theorem.
answered Dec 2 at 23:51
William Elliot
6,9072518
answered Dec 2 at 23:51
William Elliot
6,9072518
answered Dec 2 at 23:51
William Elliot
6,9072518
answered Dec 2 at 23:51
William Elliot
6,9072518
6,9072518
add a comment |
add a comment |
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The former is more readable. Both are missing a whole bunch of words, though.
– user3482749
Dec 2 at 19:46