Mixing
Is the codomain of a surjective lattice-homomorphism Heyting if the domain is Heyting?
Clash Royale CLAN TAG#URR8PPP
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Let $L$ and $L'$ be lattices and let $F:Lto L'$ be a surjective lattice-homomorphism in the sense that $F$ respects $wedge$ and $vee$.
I managed to prove for several properties of $L$ that they will be inherited by $L'$.
Let me mention some:
- if $L$ is bounded then $L'$ is bounded (and $F$ preserves $0$ and $1$).
- if $L$ is distributive then $L'$ is distributive.
- if $L$ is Boolean then $L'$ is Boolean (and $F$ preserves complements).
Now my question:
(1) Do we also have: "if $L$ is Heyting then $L'$ is Heyting"?
(2) And if so then will $F$ preserve $ato b$ in the sense that $F(ato b)=F(a)to F(b)$?
Actually the question can be reformulated as:
Is it true that $F(a)wedge F(x)leq F(b)implies F(x)leq F(ato b)$?
It is clear to me that the inverse of this implication is true.
It would not surprise me if the answer is "no", so let me conclude with requesting for a counterexample if that is indeed the case.
Thank you in advance.
lattice-orders
asked Aug 3 at 10:02
drhab
85.8k540118
add a comment |Â
up vote
4
down vote
favorite
Let $L$ and $L'$ be lattices and let $F:Lto L'$ be a surjective lattice-homomorphism in the sense that $F$ respects $wedge$ and $vee$.
I managed to prove for several properties of $L$ that they will be inherited by $L'$.
Let me mention some:
- if $L$ is bounded then $L'$ is bounded (and $F$ preserves $0$ and $1$).
- if $L$ is distributive then $L'$ is distributive.
- if $L$ is Boolean then $L'$ is Boolean (and $F$ preserves complements).
Now my question:
(1) Do we also have: "if $L$ is Heyting then $L'$ is Heyting"?
(2) And if so then will $F$ preserve $ato b$ in the sense that $F(ato b)=F(a)to F(b)$?
Actually the question can be reformulated as:
Is it true that $F(a)wedge F(x)leq F(b)implies F(x)leq F(ato b)$?
It is clear to me that the inverse of this implication is true.
It would not surprise me if the answer is "no", so let me conclude with requesting for a counterexample if that is indeed the case.
Thank you in advance.
lattice-orders
asked Aug 3 at 10:02
drhab
85.8k540118
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $L$ and $L'$ be lattices and let $F:Lto L'$ be a surjective lattice-homomorphism in the sense that $F$ respects $wedge$ and $vee$.
I managed to prove for several properties of $L$ that they will be inherited by $L'$.
Let me mention some:
- if $L$ is bounded then $L'$ is bounded (and $F$ preserves $0$ and $1$).
- if $L$ is distributive then $L'$ is distributive.
- if $L$ is Boolean then $L'$ is Boolean (and $F$ preserves complements).
Now my question:
(1) Do we also have: "if $L$ is Heyting then $L'$ is Heyting"?
(2) And if so then will $F$ preserve $ato b$ in the sense that $F(ato b)=F(a)to F(b)$?
Actually the question can be reformulated as:
Is it true that $F(a)wedge F(x)leq F(b)implies F(x)leq F(ato b)$?
It is clear to me that the inverse of this implication is true.
It would not surprise me if the answer is "no", so let me conclude with requesting for a counterexample if that is indeed the case.
Thank you in advance.
lattice-orders
asked Aug 3 at 10:02
drhab
85.8k540118
Let $L$ and $L'$ be lattices and let $F:Lto L'$ be a surjective lattice-homomorphism in the sense that $F$ respects $wedge$ and $vee$.
I managed to prove for several properties of $L$ that they will be inherited by $L'$.
Let me mention some:
- if $L$ is bounded then $L'$ is bounded (and $F$ preserves $0$ and $1$).
- if $L$ is distributive then $L'$ is distributive.
- if $L$ is Boolean then $L'$ is Boolean (and $F$ preserves complements).
Now my question:
(1) Do we also have: "if $L$ is Heyting then $L'$ is Heyting"?
(2) And if so then will $F$ preserve $ato b$ in the sense that $F(ato b)=F(a)to F(b)$?
Actually the question can be reformulated as:
Is it true that $F(a)wedge F(x)leq F(b)implies F(x)leq F(ato b)$?
It is clear to me that the inverse of this implication is true.
It would not surprise me if the answer is "no", so let me conclude with requesting for a counterexample if that is indeed the case.
Thank you in advance.
lattice-orders
asked Aug 3 at 10:02
drhab
85.8k540118
edited 4 hours ago
asked Aug 3 at 10:02
drhab
85.8k540118
asked Aug 3 at 10:02
drhab
85.8k540118
asked Aug 3 at 10:02
drhab
85.8k540118
85.8k540118
add a comment |Â
add a comment |Â
1 Answer
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This is not a proper answer as the main question eluded me for quite some time, when trying to answer.
However, it is clear that, in the finite case, the answer is yes.
So there's the question of whether or not $f(a to b) = f(a) to f(b)$.
And the answer is that it is not necessarily so.
For a very simple counter-example, take $L$ to be the three-element chain, with $0<e<1$ and $L'$ the two-element one, $0<1$.
The map $f:Lto L'$ given by
$$f(0)=f(e)=0 quadtextandquad f(1) = 1,$$
is certainly an onto lattice homomorphism.
However,
$$f(e to 0) = f(0) = 0 neq 1 = 0 to 0 = f(e) to f(0).$$
The equalities above are easy to check, either directly by definition, or taking into account that, as you can see here, second example, in a chain, $a to b = b$ if $b < a$, and in general, if $a leq b$, then $a to b = 1$.
In a tentative to be more insightful, a good reason to come up with the example above is as follows.
As you know for every homomorphism $f:L to L'$, the kernel of $f$ is a congruence of $L$.
So the smaller question you made is equivalent to ask if $theta$ is a Heyting algebra congruence, whenever $L$ is a Heyting algebra and $theta$ is a lattice congruence on $L$.
But Heyting algebras are $1$-regular, that is, a congruence on a Heyting algebra is determined by the congruence class of $1$.
To be more precise,
$$(a,b) in theta Leftrightarrow ((a to b) wedge (b to a), 1) in theta.$$
More to the point, a Heyting algebra is subdirectly irreducible iff it has an element $e$ such that, for $x neq 1$ in that algebra, $x leq e$, and in that case,
$$mu = e,1^2 cup Delta_L$$
is the monolith, meaning that $mu leq theta$, whenever $theta$ is a non-trivial congruence ($theta neq Delta_L$).
Hence, if $(e,x) in theta neq Delta$, then $(e,1) in theta$, which doesn't happen for $theta = ker f$, in the example above.
answered yesterday
amrsa
3,2432518
Thank you. Please check my interpretation of your answer. You did not answer $(1)$ but made the remark that the answer on it for finite $L$ is: "yes". (If I interpreted correctly then why is it true for finite $L$? The fact that you use the words "it is clear that" makes me think that it is trivial somehow. Is that so?). Then you focused on $(2)$ and provided a simple and convincing counterexample. Also you made clear why this counterexample was an obvious one. I was not familiar yet with terminology like "subdirectly irreducible" and "monolith" but have sources for that and am learning.
– drhab
4 hours ago
It is not familiar stuff for me and I got interested in Heyting algebras because I was looking latedays at topoi and there they show up by the ordening of subobjects. I upvoted your answer, but (also because the main question has not been answered yet) will not accept yet. If on this site an answer stays out then I will probably give it a second try on Math.Overflow.
– drhab
4 hours ago
1
@drhab The answer for the main question is trivial (and positive) in the finite case because every distributive lattice is a Heyting algebra (that is, you can always define $a to b = maxcin L:awedge cleq b$ in a finite lattice). More generally, it can be proven that a complete distributive lattice is a Heyting algebra iff it satisfies the join-infinite distributive law: $$awedgebigvee_iin Ib_i=bigvee_iin I(a wedge b_i).$$
– amrsa
1 hour ago
1
@drhab About the counter-example, I could, of course, just stated it and prove it was one, but I though that providing information about why it was more or less obvious is much more enlightening. Thanks for the upvote. I didn't expect you to accept the answer since it is partial. Good luck!
– amrsa
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is not a proper answer as the main question eluded me for quite some time, when trying to answer.
However, it is clear that, in the finite case, the answer is yes.
So there's the question of whether or not $f(a to b) = f(a) to f(b)$.
And the answer is that it is not necessarily so.
For a very simple counter-example, take $L$ to be the three-element chain, with $0<e<1$ and $L'$ the two-element one, $0<1$.
The map $f:Lto L'$ given by
$$f(0)=f(e)=0 quadtextandquad f(1) = 1,$$
is certainly an onto lattice homomorphism.
However,
$$f(e to 0) = f(0) = 0 neq 1 = 0 to 0 = f(e) to f(0).$$
The equalities above are easy to check, either directly by definition, or taking into account that, as you can see here, second example, in a chain, $a to b = b$ if $b < a$, and in general, if $a leq b$, then $a to b = 1$.
In a tentative to be more insightful, a good reason to come up with the example above is as follows.
As you know for every homomorphism $f:L to L'$, the kernel of $f$ is a congruence of $L$.
So the smaller question you made is equivalent to ask if $theta$ is a Heyting algebra congruence, whenever $L$ is a Heyting algebra and $theta$ is a lattice congruence on $L$.
But Heyting algebras are $1$-regular, that is, a congruence on a Heyting algebra is determined by the congruence class of $1$.
To be more precise,
$$(a,b) in theta Leftrightarrow ((a to b) wedge (b to a), 1) in theta.$$
More to the point, a Heyting algebra is subdirectly irreducible iff it has an element $e$ such that, for $x neq 1$ in that algebra, $x leq e$, and in that case,
$$mu = e,1^2 cup Delta_L$$
is the monolith, meaning that $mu leq theta$, whenever $theta$ is a non-trivial congruence ($theta neq Delta_L$).
Hence, if $(e,x) in theta neq Delta$, then $(e,1) in theta$, which doesn't happen for $theta = ker f$, in the example above.
answered yesterday
amrsa
3,2432518
Thank you. Please check my interpretation of your answer. You did not answer $(1)$ but made the remark that the answer on it for finite $L$ is: "yes". (If I interpreted correctly then why is it true for finite $L$? The fact that you use the words "it is clear that" makes me think that it is trivial somehow. Is that so?). Then you focused on $(2)$ and provided a simple and convincing counterexample. Also you made clear why this counterexample was an obvious one. I was not familiar yet with terminology like "subdirectly irreducible" and "monolith" but have sources for that and am learning.
– drhab
4 hours ago
It is not familiar stuff for me and I got interested in Heyting algebras because I was looking latedays at topoi and there they show up by the ordening of subobjects. I upvoted your answer, but (also because the main question has not been answered yet) will not accept yet. If on this site an answer stays out then I will probably give it a second try on Math.Overflow.
– drhab
4 hours ago
1
@drhab The answer for the main question is trivial (and positive) in the finite case because every distributive lattice is a Heyting algebra (that is, you can always define $a to b = maxcin L:awedge cleq b$ in a finite lattice). More generally, it can be proven that a complete distributive lattice is a Heyting algebra iff it satisfies the join-infinite distributive law: $$awedgebigvee_iin Ib_i=bigvee_iin I(a wedge b_i).$$
– amrsa
1 hour ago
1
@drhab About the counter-example, I could, of course, just stated it and prove it was one, but I though that providing information about why it was more or less obvious is much more enlightening. Thanks for the upvote. I didn't expect you to accept the answer since it is partial. Good luck!
– amrsa
1 hour ago
add a comment |Â
up vote
1
down vote
This is not a proper answer as the main question eluded me for quite some time, when trying to answer.
However, it is clear that, in the finite case, the answer is yes.
So there's the question of whether or not $f(a to b) = f(a) to f(b)$.
And the answer is that it is not necessarily so.
For a very simple counter-example, take $L$ to be the three-element chain, with $0<e<1$ and $L'$ the two-element one, $0<1$.
The map $f:Lto L'$ given by
$$f(0)=f(e)=0 quadtextandquad f(1) = 1,$$
is certainly an onto lattice homomorphism.
However,
$$f(e to 0) = f(0) = 0 neq 1 = 0 to 0 = f(e) to f(0).$$
The equalities above are easy to check, either directly by definition, or taking into account that, as you can see here, second example, in a chain, $a to b = b$ if $b < a$, and in general, if $a leq b$, then $a to b = 1$.
In a tentative to be more insightful, a good reason to come up with the example above is as follows.
As you know for every homomorphism $f:L to L'$, the kernel of $f$ is a congruence of $L$.
So the smaller question you made is equivalent to ask if $theta$ is a Heyting algebra congruence, whenever $L$ is a Heyting algebra and $theta$ is a lattice congruence on $L$.
But Heyting algebras are $1$-regular, that is, a congruence on a Heyting algebra is determined by the congruence class of $1$.
To be more precise,
$$(a,b) in theta Leftrightarrow ((a to b) wedge (b to a), 1) in theta.$$
More to the point, a Heyting algebra is subdirectly irreducible iff it has an element $e$ such that, for $x neq 1$ in that algebra, $x leq e$, and in that case,
$$mu = e,1^2 cup Delta_L$$
is the monolith, meaning that $mu leq theta$, whenever $theta$ is a non-trivial congruence ($theta neq Delta_L$).
Hence, if $(e,x) in theta neq Delta$, then $(e,1) in theta$, which doesn't happen for $theta = ker f$, in the example above.
answered yesterday
amrsa
3,2432518
Thank you. Please check my interpretation of your answer. You did not answer $(1)$ but made the remark that the answer on it for finite $L$ is: "yes". (If I interpreted correctly then why is it true for finite $L$? The fact that you use the words "it is clear that" makes me think that it is trivial somehow. Is that so?). Then you focused on $(2)$ and provided a simple and convincing counterexample. Also you made clear why this counterexample was an obvious one. I was not familiar yet with terminology like "subdirectly irreducible" and "monolith" but have sources for that and am learning.
– drhab
4 hours ago
It is not familiar stuff for me and I got interested in Heyting algebras because I was looking latedays at topoi and there they show up by the ordening of subobjects. I upvoted your answer, but (also because the main question has not been answered yet) will not accept yet. If on this site an answer stays out then I will probably give it a second try on Math.Overflow.
– drhab
4 hours ago
1
@drhab The answer for the main question is trivial (and positive) in the finite case because every distributive lattice is a Heyting algebra (that is, you can always define $a to b = maxcin L:awedge cleq b$ in a finite lattice). More generally, it can be proven that a complete distributive lattice is a Heyting algebra iff it satisfies the join-infinite distributive law: $$awedgebigvee_iin Ib_i=bigvee_iin I(a wedge b_i).$$
– amrsa
1 hour ago
1
@drhab About the counter-example, I could, of course, just stated it and prove it was one, but I though that providing information about why it was more or less obvious is much more enlightening. Thanks for the upvote. I didn't expect you to accept the answer since it is partial. Good luck!
– amrsa
1 hour ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is not a proper answer as the main question eluded me for quite some time, when trying to answer.
However, it is clear that, in the finite case, the answer is yes.
So there's the question of whether or not $f(a to b) = f(a) to f(b)$.
And the answer is that it is not necessarily so.
For a very simple counter-example, take $L$ to be the three-element chain, with $0<e<1$ and $L'$ the two-element one, $0<1$.
The map $f:Lto L'$ given by
$$f(0)=f(e)=0 quadtextandquad f(1) = 1,$$
is certainly an onto lattice homomorphism.
However,
$$f(e to 0) = f(0) = 0 neq 1 = 0 to 0 = f(e) to f(0).$$
The equalities above are easy to check, either directly by definition, or taking into account that, as you can see here, second example, in a chain, $a to b = b$ if $b < a$, and in general, if $a leq b$, then $a to b = 1$.
In a tentative to be more insightful, a good reason to come up with the example above is as follows.
As you know for every homomorphism $f:L to L'$, the kernel of $f$ is a congruence of $L$.
So the smaller question you made is equivalent to ask if $theta$ is a Heyting algebra congruence, whenever $L$ is a Heyting algebra and $theta$ is a lattice congruence on $L$.
But Heyting algebras are $1$-regular, that is, a congruence on a Heyting algebra is determined by the congruence class of $1$.
To be more precise,
$$(a,b) in theta Leftrightarrow ((a to b) wedge (b to a), 1) in theta.$$
More to the point, a Heyting algebra is subdirectly irreducible iff it has an element $e$ such that, for $x neq 1$ in that algebra, $x leq e$, and in that case,
$$mu = e,1^2 cup Delta_L$$
is the monolith, meaning that $mu leq theta$, whenever $theta$ is a non-trivial congruence ($theta neq Delta_L$).
Hence, if $(e,x) in theta neq Delta$, then $(e,1) in theta$, which doesn't happen for $theta = ker f$, in the example above.
answered yesterday
amrsa
3,2432518
This is not a proper answer as the main question eluded me for quite some time, when trying to answer.
However, it is clear that, in the finite case, the answer is yes.
So there's the question of whether or not $f(a to b) = f(a) to f(b)$.
And the answer is that it is not necessarily so.
For a very simple counter-example, take $L$ to be the three-element chain, with $0<e<1$ and $L'$ the two-element one, $0<1$.
The map $f:Lto L'$ given by
$$f(0)=f(e)=0 quadtextandquad f(1) = 1,$$
is certainly an onto lattice homomorphism.
However,
$$f(e to 0) = f(0) = 0 neq 1 = 0 to 0 = f(e) to f(0).$$
The equalities above are easy to check, either directly by definition, or taking into account that, as you can see here, second example, in a chain, $a to b = b$ if $b < a$, and in general, if $a leq b$, then $a to b = 1$.
In a tentative to be more insightful, a good reason to come up with the example above is as follows.
As you know for every homomorphism $f:L to L'$, the kernel of $f$ is a congruence of $L$.
So the smaller question you made is equivalent to ask if $theta$ is a Heyting algebra congruence, whenever $L$ is a Heyting algebra and $theta$ is a lattice congruence on $L$.
But Heyting algebras are $1$-regular, that is, a congruence on a Heyting algebra is determined by the congruence class of $1$.
To be more precise,
$$(a,b) in theta Leftrightarrow ((a to b) wedge (b to a), 1) in theta.$$
More to the point, a Heyting algebra is subdirectly irreducible iff it has an element $e$ such that, for $x neq 1$ in that algebra, $x leq e$, and in that case,
$$mu = e,1^2 cup Delta_L$$
is the monolith, meaning that $mu leq theta$, whenever $theta$ is a non-trivial congruence ($theta neq Delta_L$).
Hence, if $(e,x) in theta neq Delta$, then $(e,1) in theta$, which doesn't happen for $theta = ker f$, in the example above.
answered yesterday
amrsa
3,2432518
answered yesterday
amrsa
3,2432518
answered yesterday
amrsa
3,2432518
answered yesterday
amrsa
3,2432518
3,2432518
Thank you. Please check my interpretation of your answer. You did not answer $(1)$ but made the remark that the answer on it for finite $L$ is: "yes". (If I interpreted correctly then why is it true for finite $L$? The fact that you use the words "it is clear that" makes me think that it is trivial somehow. Is that so?). Then you focused on $(2)$ and provided a simple and convincing counterexample. Also you made clear why this counterexample was an obvious one. I was not familiar yet with terminology like "subdirectly irreducible" and "monolith" but have sources for that and am learning.
– drhab
4 hours ago
It is not familiar stuff for me and I got interested in Heyting algebras because I was looking latedays at topoi and there they show up by the ordening of subobjects. I upvoted your answer, but (also because the main question has not been answered yet) will not accept yet. If on this site an answer stays out then I will probably give it a second try on Math.Overflow.
– drhab
4 hours ago
1
@drhab The answer for the main question is trivial (and positive) in the finite case because every distributive lattice is a Heyting algebra (that is, you can always define $a to b = maxcin L:awedge cleq b$ in a finite lattice). More generally, it can be proven that a complete distributive lattice is a Heyting algebra iff it satisfies the join-infinite distributive law: $$awedgebigvee_iin Ib_i=bigvee_iin I(a wedge b_i).$$
– amrsa
1 hour ago
1
@drhab About the counter-example, I could, of course, just stated it and prove it was one, but I though that providing information about why it was more or less obvious is much more enlightening. Thanks for the upvote. I didn't expect you to accept the answer since it is partial. Good luck!
– amrsa
1 hour ago
add a comment |Â
Thank you. Please check my interpretation of your answer. You did not answer $(1)$ but made the remark that the answer on it for finite $L$ is: "yes". (If I interpreted correctly then why is it true for finite $L$? The fact that you use the words "it is clear that" makes me think that it is trivial somehow. Is that so?). Then you focused on $(2)$ and provided a simple and convincing counterexample. Also you made clear why this counterexample was an obvious one. I was not familiar yet with terminology like "subdirectly irreducible" and "monolith" but have sources for that and am learning.
– drhab
4 hours ago
It is not familiar stuff for me and I got interested in Heyting algebras because I was looking latedays at topoi and there they show up by the ordening of subobjects. I upvoted your answer, but (also because the main question has not been answered yet) will not accept yet. If on this site an answer stays out then I will probably give it a second try on Math.Overflow.
– drhab
4 hours ago
1
@drhab The answer for the main question is trivial (and positive) in the finite case because every distributive lattice is a Heyting algebra (that is, you can always define $a to b = maxcin L:awedge cleq b$ in a finite lattice). More generally, it can be proven that a complete distributive lattice is a Heyting algebra iff it satisfies the join-infinite distributive law: $$awedgebigvee_iin Ib_i=bigvee_iin I(a wedge b_i).$$
– amrsa
1 hour ago
1
@drhab About the counter-example, I could, of course, just stated it and prove it was one, but I though that providing information about why it was more or less obvious is much more enlightening. Thanks for the upvote. I didn't expect you to accept the answer since it is partial. Good luck!
– amrsa
1 hour ago
Thank you. Please check my interpretation of your answer. You did not answer $(1)$ but made the remark that the answer on it for finite $L$ is: "yes". (If I interpreted correctly then why is it true for finite $L$? The fact that you use the words "it is clear that" makes me think that it is trivial somehow. Is that so?). Then you focused on $(2)$ and provided a simple and convincing counterexample. Also you made clear why this counterexample was an obvious one. I was not familiar yet with terminology like "subdirectly irreducible" and "monolith" but have sources for that and am learning.
– drhab
4 hours ago
Thank you. Please check my interpretation of your answer. You did not answer $(1)$ but made the remark that the answer on it for finite $L$ is: "yes". (If I interpreted correctly then why is it true for finite $L$? The fact that you use the words "it is clear that" makes me think that it is trivial somehow. Is that so?). Then you focused on $(2)$ and provided a simple and convincing counterexample. Also you made clear why this counterexample was an obvious one. I was not familiar yet with terminology like "subdirectly irreducible" and "monolith" but have sources for that and am learning.
– drhab
4 hours ago
It is not familiar stuff for me and I got interested in Heyting algebras because I was looking latedays at topoi and there they show up by the ordening of subobjects. I upvoted your answer, but (also because the main question has not been answered yet) will not accept yet. If on this site an answer stays out then I will probably give it a second try on Math.Overflow.
– drhab
4 hours ago
It is not familiar stuff for me and I got interested in Heyting algebras because I was looking latedays at topoi and there they show up by the ordening of subobjects. I upvoted your answer, but (also because the main question has not been answered yet) will not accept yet. If on this site an answer stays out then I will probably give it a second try on Math.Overflow.
– drhab
4 hours ago
1
1
@drhab The answer for the main question is trivial (and positive) in the finite case because every distributive lattice is a Heyting algebra (that is, you can always define $a to b = maxcin L:awedge cleq b$ in a finite lattice). More generally, it can be proven that a complete distributive lattice is a Heyting algebra iff it satisfies the join-infinite distributive law: $$awedgebigvee_iin Ib_i=bigvee_iin I(a wedge b_i).$$
– amrsa
1 hour ago
@drhab The answer for the main question is trivial (and positive) in the finite case because every distributive lattice is a Heyting algebra (that is, you can always define $a to b = maxcin L:awedge cleq b$ in a finite lattice). More generally, it can be proven that a complete distributive lattice is a Heyting algebra iff it satisfies the join-infinite distributive law: $$awedgebigvee_iin Ib_i=bigvee_iin I(a wedge b_i).$$
– amrsa
1 hour ago
1
1
@drhab About the counter-example, I could, of course, just stated it and prove it was one, but I though that providing information about why it was more or less obvious is much more enlightening. Thanks for the upvote. I didn't expect you to accept the answer since it is partial. Good luck!
– amrsa
1 hour ago
@drhab About the counter-example, I could, of course, just stated it and prove it was one, but I though that providing information about why it was more or less obvious is much more enlightening. Thanks for the upvote. I didn't expect you to accept the answer since it is partial. Good luck!
– amrsa
1 hour ago
add a comment |Â
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