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There is a counter-example againt my conjecture.

I will denote $a\not\asymp b$ iff there is a non-least element which is below both $a$ and $b$.

Take $\mathcal{P} \in F$ from the previous counter-examplethe previous counter-example. We have $$ \forall a \in \prod_{i \in \operatorname{dom} F} \operatorname{atoms} \mathcal{P}_i : a \notin \mathcal{P} . $$ Take $k = 1$.

Let $\mathcal{L} = \mathcal{P} |_{(\operatorname{dom} F) \setminus \{ k \}}$. Then $a \notin \mathrel{\upuparrows F}$ and thus $a_k \asymp \langle \upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$.

Consequently $\mathcal{P}_k \asymp \langle \upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$ and thus $\mathcal{P}_k \asymp \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L_{}}_i} \langle \upuparrows F \rangle_k a$ because $\mathcal{P}_k$ is principal.

But $\mathcal{P}_k \not\asymp \langle \upuparrows F \rangle_k \mathcal{L}$. Thus follows $\langle \upuparrows F \rangle_k \mathcal{L} \neq \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L}_i} \langle \upuparrows F \rangle_k a$.

There is a counter-example againt my conjecture.

I will denote $a\not\asymp b$ iff there is a non-least element which is below both $a$ and $b$.

Take $\mathcal{P} \in F$ from the previous counter-example. We have $$ \forall a \in \prod_{i \in \operatorname{dom} F} \operatorname{atoms} \mathcal{P}_i : a \notin \mathcal{P} . $$ Take $k = 1$.

Let $\mathcal{L} = \mathcal{P} |_{(\operatorname{dom} F) \setminus \{ k \}}$. Then $a \notin \mathrel{\upuparrows F}$ and thus $a_k \asymp \langle \upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$.

Consequently $\mathcal{P}_k \asymp \langle \upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$ and thus $\mathcal{P}_k \asymp \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L_{}}_i} \langle \upuparrows F \rangle_k a$ because $\mathcal{P}_k$ is principal.

But $\mathcal{P}_k \not\asymp \langle \upuparrows F \rangle_k \mathcal{L}$. Thus follows $\langle \upuparrows F \rangle_k \mathcal{L} \neq \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L}_i} \langle \upuparrows F \rangle_k a$.

There is a counter-example againt my conjecture.

I will denote $a\not\asymp b$ iff there is a non-least element which is below both $a$ and $b$.

Take $\mathcal{P} \in F$ from the previous counter-example. We have $$ \forall a \in \prod_{i \in \operatorname{dom} F} \operatorname{atoms} \mathcal{P}_i : a \notin \mathcal{P} . $$ Take $k = 1$.

Let $\mathcal{L} = \mathcal{P} |_{(\operatorname{dom} F) \setminus \{ k \}}$. Then $a \notin \mathrel{\upuparrows F}$ and thus $a_k \asymp \langle \upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$.

Consequently $\mathcal{P}_k \asymp \langle \upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$ and thus $\mathcal{P}_k \asymp \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L_{}}_i} \langle \upuparrows F \rangle_k a$ because $\mathcal{P}_k$ is principal.

But $\mathcal{P}_k \not\asymp \langle \upuparrows F \rangle_k \mathcal{L}$. Thus follows $\langle \upuparrows F \rangle_k \mathcal{L} \neq \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L}_i} \langle \upuparrows F \rangle_k a$.

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porton
porton
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There is a counter-example againt my conjecture.

I will denote $a\not\asymp b$ iff there is a non-least element which is below both $a$ and $b$.

Take $\mathcal{P} \in F$ from the previous counter-example. We have $$ \forall a \in \prod_{i \in \operatorname{dom} F} \operatorname{atoms} \mathcal{P}_i : a \notin \mathcal{P} . $$ Take $k = 1$.

Let $\mathcal{L} = \mathcal{P} |_{(\operatorname{dom} F) \setminus \{ k \}}$. Then $a \notin \mathrel{\upuparrows\, \uparrow F}$$a \notin \mathrel{\upuparrows F}$ and thus $a_k \asymp \langle \upuparrows\, \uparrow F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$$a_k \asymp \langle \upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$.

Consequently $\mathcal{P}_k \asymp \langle \upuparrows\, \uparrow F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$$\mathcal{P}_k \asymp \langle \upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$ and thus $\mathcal{P}_k \asymp \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L_{}}_i} \langle \upuparrows\, \uparrow F \rangle_k a$$\mathcal{P}_k \asymp \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L_{}}_i} \langle \upuparrows F \rangle_k a$ because $\mathcal{P}_k$ is principal.

But $\mathcal{P}_k \not\asymp \langle \upuparrows\, \uparrow F \rangle_k \mathcal{L}$$\mathcal{P}_k \not\asymp \langle \upuparrows F \rangle_k \mathcal{L}$. Thus follows $\langle \upuparrows\, \uparrow F \rangle_k \mathcal{L} \neq \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L}_i} \langle \upuparrows\, \uparrow F \rangle_k a$$\langle \upuparrows F \rangle_k \mathcal{L} \neq \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L}_i} \langle \upuparrows F \rangle_k a$.

There is a counter-example againt my conjecture.

I will denote $a\not\asymp b$ iff there is a non-least element which is below both $a$ and $b$.

Take $\mathcal{P} \in F$ from the previous counter-example. We have $$ \forall a \in \prod_{i \in \operatorname{dom} F} \operatorname{atoms} \mathcal{P}_i : a \notin \mathcal{P} . $$ Take $k = 1$.

Let $\mathcal{L} = \mathcal{P} |_{(\operatorname{dom} F) \setminus \{ k \}}$. Then $a \notin \mathrel{\upuparrows\, \uparrow F}$ and thus $a_k \asymp \langle \upuparrows\, \uparrow F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$.

Consequently $\mathcal{P}_k \asymp \langle \upuparrows\, \uparrow F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$ and thus $\mathcal{P}_k \asymp \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L_{}}_i} \langle \upuparrows\, \uparrow F \rangle_k a$ because $\mathcal{P}_k$ is principal.

But $\mathcal{P}_k \not\asymp \langle \upuparrows\, \uparrow F \rangle_k \mathcal{L}$. Thus follows $\langle \upuparrows\, \uparrow F \rangle_k \mathcal{L} \neq \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L}_i} \langle \upuparrows\, \uparrow F \rangle_k a$.

There is a counter-example againt my conjecture.

I will denote $a\not\asymp b$ iff there is a non-least element which is below both $a$ and $b$.

Take $\mathcal{P} \in F$ from the previous counter-example. We have $$ \forall a \in \prod_{i \in \operatorname{dom} F} \operatorname{atoms} \mathcal{P}_i : a \notin \mathcal{P} . $$ Take $k = 1$.

Let $\mathcal{L} = \mathcal{P} |_{(\operatorname{dom} F) \setminus \{ k \}}$. Then $a \notin \mathrel{\upuparrows F}$ and thus $a_k \asymp \langle \upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$.

Consequently $\mathcal{P}_k \asymp \langle \upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$ and thus $\mathcal{P}_k \asymp \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L_{}}_i} \langle \upuparrows F \rangle_k a$ because $\mathcal{P}_k$ is principal.

But $\mathcal{P}_k \not\asymp \langle \upuparrows F \rangle_k \mathcal{L}$. Thus follows $\langle \upuparrows F \rangle_k \mathcal{L} \neq \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L}_i} \langle \upuparrows F \rangle_k a$.

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porton
porton
  • 765
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There is a counter-example againt my conjecture.

I will denote $a\not\asymp b$ iff there is a non-least element which is below both $a$ and $b$.

Take $\mathcal{P} \in F$ from the previous counter-example. We have $$ \forall a \in \prod_{i \in \operatorname{dom} F} \operatorname{atoms} \mathcal{P}_i : a \notin \mathcal{P} . $$ Take $k = 1$.

Let $\mathcal{L} = \mathcal{P} |_{(\operatorname{dom} F) \setminus \{ k \}}$. Then $a \notin \mathrel{\upuparrows\, \uparrow F}$ and thus $a_k \asymp \langle \upuparrows\, \uparrow F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$.

Consequently $\mathcal{P}_k \asymp \langle \upuparrows\, \uparrow F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$ and thus $\mathcal{P}_k \asymp \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L_{}}_i} \langle \upuparrows\, \uparrow F \rangle_k a$ because $\mathcal{P}_k$ is principal.

But $\mathcal{P}_k \not\asymp \langle \upuparrows\, \uparrow F \rangle_k \mathcal{L}$. Thus follows $\langle \upuparrows\, \uparrow F \rangle_k \mathcal{L} \neq \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}} \operatorname{atoms} \mathcal{L}_i} \langle \upuparrows\, \uparrow F \rangle_k a$.