Prove that either $aleq x$ or $aleq -x$ if and only if $a$ is an atom.
$begingroup$
Let $(mathbb{B},wedge,vee,-,0,1,leq)$ be a Boolean algebra. I wish to prove the following:
Claim: $a$ is an atom if and only if for each $xinmathbb{B}$ either $aleq -x$, or $aleq x$ (exclusively)
I managed to prove the $Rightarrow$ part:
Assume that both $aleq -x$ and $aleq x$ hold, then $aleq xwedge(-x)=0$, so $aleq 0$, but then $a=0$, so $a$ is not an atom.
I am stuck on proving the second part, can someone help?
order-theory boolean-algebra lattice-orders
asked Dec 8 '18 at 19:31
Michal DvořákMichal Dvořák
955416
$endgroup$
add a comment |
$begingroup$
Let $(mathbb{B},wedge,vee,-,0,1,leq)$ be a Boolean algebra. I wish to prove the following:
Claim: $a$ is an atom if and only if for each $xinmathbb{B}$ either $aleq -x$, or $aleq x$ (exclusively)
I managed to prove the $Rightarrow$ part:
Assume that both $aleq -x$ and $aleq x$ hold, then $aleq xwedge(-x)=0$, so $aleq 0$, but then $a=0$, so $a$ is not an atom.
I am stuck on proving the second part, can someone help?
order-theory boolean-algebra lattice-orders
asked Dec 8 '18 at 19:31
Michal DvořákMichal Dvořák
955416
$endgroup$
add a comment |
$begingroup$
Let $(mathbb{B},wedge,vee,-,0,1,leq)$ be a Boolean algebra. I wish to prove the following:
Claim: $a$ is an atom if and only if for each $xinmathbb{B}$ either $aleq -x$, or $aleq x$ (exclusively)
I managed to prove the $Rightarrow$ part:
Assume that both $aleq -x$ and $aleq x$ hold, then $aleq xwedge(-x)=0$, so $aleq 0$, but then $a=0$, so $a$ is not an atom.
I am stuck on proving the second part, can someone help?
order-theory boolean-algebra lattice-orders
asked Dec 8 '18 at 19:31
Michal DvořákMichal Dvořák
955416
$endgroup$
Let $(mathbb{B},wedge,vee,-,0,1,leq)$ be a Boolean algebra. I wish to prove the following:
Claim: $a$ is an atom if and only if for each $xinmathbb{B}$ either $aleq -x$, or $aleq x$ (exclusively)
I managed to prove the $Rightarrow$ part:
Assume that both $aleq -x$ and $aleq x$ hold, then $aleq xwedge(-x)=0$, so $aleq 0$, but then $a=0$, so $a$ is not an atom.
I am stuck on proving the second part, can someone help?
order-theory boolean-algebra lattice-orders
order-theory boolean-algebra lattice-orders
asked Dec 8 '18 at 19:31
Michal DvořákMichal Dvořák
955416
asked Dec 8 '18 at 19:31
Michal DvořákMichal Dvořák
955416
asked Dec 8 '18 at 19:31
Michal DvořákMichal Dvořák
955416
asked Dec 8 '18 at 19:31
Michal DvořákMichal Dvořák
955416
asked Dec 8 '18 at 19:31
Michal DvořákMichal Dvořák
955416
955416
add a comment |
add a comment |
1 Answer
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$begingroup$
Assume that $a$ is not an atom. As you observed, $a=0$ is no good, so the indirect assumption means that there exists some $b$ with $0<b<a$.
Clearly, $aleq b$ is false.
If $aleq -b$, then $b=bwedge aleq bwedge -b=0$, a contradiction with $x=b$.
answered Dec 8 '18 at 19:42
A. PongráczA. Pongrácz
5,9731929
$endgroup$
$begingroup$
Actually could you elaborate the $b=bwedge aleq b wedge -b = 0$ line? I am not really sure what you actually did there.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 19:45
$begingroup$
The definition of $leq$ is $uleq v$ iff there exists $w$ such that $u=vwedge w$. Hence, if this is the case, and $t$ is any element, then $twedge uleq twedge v$ (trivial by the axioms of Boolean algebras, the same $w$ is a witness). What I did was using this identity for the inequality $aleq -b$ by "wedging" both sides with $b$, obtaining $bwedge aleq bwedge -b$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 19:49
1
$begingroup$
You can easily prove the equivalence from the axioms. Your implies mine: $a=w$ is a witness. Try the other direction yourself.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:09
$begingroup$
I don't understand the part $b=bwedge a$, then sure, this is a contradiction because $bleq 0$, so $b=0$, so contradiction.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 20:21
1
$begingroup$
Again, cf. the definition of smaller... By assumption, $b<a$, so $b=bwedge a$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:39
|
show 1 more comment
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$begingroup$
Assume that $a$ is not an atom. As you observed, $a=0$ is no good, so the indirect assumption means that there exists some $b$ with $0<b<a$.
Clearly, $aleq b$ is false.
If $aleq -b$, then $b=bwedge aleq bwedge -b=0$, a contradiction with $x=b$.
answered Dec 8 '18 at 19:42
A. PongráczA. Pongrácz
5,9731929
$endgroup$
$begingroup$
Actually could you elaborate the $b=bwedge aleq b wedge -b = 0$ line? I am not really sure what you actually did there.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 19:45
$begingroup$
The definition of $leq$ is $uleq v$ iff there exists $w$ such that $u=vwedge w$. Hence, if this is the case, and $t$ is any element, then $twedge uleq twedge v$ (trivial by the axioms of Boolean algebras, the same $w$ is a witness). What I did was using this identity for the inequality $aleq -b$ by "wedging" both sides with $b$, obtaining $bwedge aleq bwedge -b$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 19:49
1
$begingroup$
You can easily prove the equivalence from the axioms. Your implies mine: $a=w$ is a witness. Try the other direction yourself.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:09
$begingroup$
I don't understand the part $b=bwedge a$, then sure, this is a contradiction because $bleq 0$, so $b=0$, so contradiction.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 20:21
1
$begingroup$
Again, cf. the definition of smaller... By assumption, $b<a$, so $b=bwedge a$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:39
|
show 1 more comment
$begingroup$
Assume that $a$ is not an atom. As you observed, $a=0$ is no good, so the indirect assumption means that there exists some $b$ with $0<b<a$.
Clearly, $aleq b$ is false.
If $aleq -b$, then $b=bwedge aleq bwedge -b=0$, a contradiction with $x=b$.
answered Dec 8 '18 at 19:42
A. PongráczA. Pongrácz
5,9731929
$endgroup$
$begingroup$
Actually could you elaborate the $b=bwedge aleq b wedge -b = 0$ line? I am not really sure what you actually did there.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 19:45
$begingroup$
The definition of $leq$ is $uleq v$ iff there exists $w$ such that $u=vwedge w$. Hence, if this is the case, and $t$ is any element, then $twedge uleq twedge v$ (trivial by the axioms of Boolean algebras, the same $w$ is a witness). What I did was using this identity for the inequality $aleq -b$ by "wedging" both sides with $b$, obtaining $bwedge aleq bwedge -b$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 19:49
1
$begingroup$
You can easily prove the equivalence from the axioms. Your implies mine: $a=w$ is a witness. Try the other direction yourself.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:09
$begingroup$
I don't understand the part $b=bwedge a$, then sure, this is a contradiction because $bleq 0$, so $b=0$, so contradiction.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 20:21
1
$begingroup$
Again, cf. the definition of smaller... By assumption, $b<a$, so $b=bwedge a$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:39
|
show 1 more comment
$begingroup$
Assume that $a$ is not an atom. As you observed, $a=0$ is no good, so the indirect assumption means that there exists some $b$ with $0<b<a$.
Clearly, $aleq b$ is false.
If $aleq -b$, then $b=bwedge aleq bwedge -b=0$, a contradiction with $x=b$.
answered Dec 8 '18 at 19:42
A. PongráczA. Pongrácz
5,9731929
$endgroup$
Assume that $a$ is not an atom. As you observed, $a=0$ is no good, so the indirect assumption means that there exists some $b$ with $0<b<a$.
Clearly, $aleq b$ is false.
If $aleq -b$, then $b=bwedge aleq bwedge -b=0$, a contradiction with $x=b$.
answered Dec 8 '18 at 19:42
A. PongráczA. Pongrácz
5,9731929
answered Dec 8 '18 at 19:42
A. PongráczA. Pongrácz
5,9731929
answered Dec 8 '18 at 19:42
A. PongráczA. Pongrácz
5,9731929
answered Dec 8 '18 at 19:42
A. PongráczA. Pongrácz
5,9731929
5,9731929
$begingroup$
Actually could you elaborate the $b=bwedge aleq b wedge -b = 0$ line? I am not really sure what you actually did there.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 19:45
$begingroup$
The definition of $leq$ is $uleq v$ iff there exists $w$ such that $u=vwedge w$. Hence, if this is the case, and $t$ is any element, then $twedge uleq twedge v$ (trivial by the axioms of Boolean algebras, the same $w$ is a witness). What I did was using this identity for the inequality $aleq -b$ by "wedging" both sides with $b$, obtaining $bwedge aleq bwedge -b$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 19:49
1
$begingroup$
You can easily prove the equivalence from the axioms. Your implies mine: $a=w$ is a witness. Try the other direction yourself.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:09
$begingroup$
I don't understand the part $b=bwedge a$, then sure, this is a contradiction because $bleq 0$, so $b=0$, so contradiction.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 20:21
1
$begingroup$
Again, cf. the definition of smaller... By assumption, $b<a$, so $b=bwedge a$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:39
|
show 1 more comment
$begingroup$
Actually could you elaborate the $b=bwedge aleq b wedge -b = 0$ line? I am not really sure what you actually did there.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 19:45
$begingroup$
The definition of $leq$ is $uleq v$ iff there exists $w$ such that $u=vwedge w$. Hence, if this is the case, and $t$ is any element, then $twedge uleq twedge v$ (trivial by the axioms of Boolean algebras, the same $w$ is a witness). What I did was using this identity for the inequality $aleq -b$ by "wedging" both sides with $b$, obtaining $bwedge aleq bwedge -b$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 19:49
1
$begingroup$
You can easily prove the equivalence from the axioms. Your implies mine: $a=w$ is a witness. Try the other direction yourself.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:09
$begingroup$
I don't understand the part $b=bwedge a$, then sure, this is a contradiction because $bleq 0$, so $b=0$, so contradiction.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 20:21
1
$begingroup$
Again, cf. the definition of smaller... By assumption, $b<a$, so $b=bwedge a$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:39
$begingroup$
Actually could you elaborate the $b=bwedge aleq b wedge -b = 0$ line? I am not really sure what you actually did there.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 19:45
$begingroup$
Actually could you elaborate the $b=bwedge aleq b wedge -b = 0$ line? I am not really sure what you actually did there.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 19:45
$begingroup$
The definition of $leq$ is $uleq v$ iff there exists $w$ such that $u=vwedge w$. Hence, if this is the case, and $t$ is any element, then $twedge uleq twedge v$ (trivial by the axioms of Boolean algebras, the same $w$ is a witness). What I did was using this identity for the inequality $aleq -b$ by "wedging" both sides with $b$, obtaining $bwedge aleq bwedge -b$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 19:49
$begingroup$
The definition of $leq$ is $uleq v$ iff there exists $w$ such that $u=vwedge w$. Hence, if this is the case, and $t$ is any element, then $twedge uleq twedge v$ (trivial by the axioms of Boolean algebras, the same $w$ is a witness). What I did was using this identity for the inequality $aleq -b$ by "wedging" both sides with $b$, obtaining $bwedge aleq bwedge -b$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 19:49
1
1
$begingroup$
You can easily prove the equivalence from the axioms. Your implies mine: $a=w$ is a witness. Try the other direction yourself.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:09
$begingroup$
You can easily prove the equivalence from the axioms. Your implies mine: $a=w$ is a witness. Try the other direction yourself.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:09
$begingroup$
I don't understand the part $b=bwedge a$, then sure, this is a contradiction because $bleq 0$, so $b=0$, so contradiction.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 20:21
$begingroup$
I don't understand the part $b=bwedge a$, then sure, this is a contradiction because $bleq 0$, so $b=0$, so contradiction.
$endgroup$
– Michal Dvořák
Dec 8 '18 at 20:21
1
1
$begingroup$
Again, cf. the definition of smaller... By assumption, $b<a$, so $b=bwedge a$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:39
$begingroup$
Again, cf. the definition of smaller... By assumption, $b<a$, so $b=bwedge a$.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 20:39
|
show 1 more comment
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