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Date: 2020-06-29 06:48 am (UTC)Well, ok, maybe this is a stupid question. Of course, it's just an arrow that makes the square commute.
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Date: 2020-06-29 06:49 am (UTC)no subject
Date: 2020-06-29 10:16 am (UTC)no subject
Date: 2020-06-29 10:26 am (UTC)no subject
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Date: 2020-06-29 02:04 pm (UTC)no subject
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Date: 2020-06-29 02:41 pm (UTC)no subject
Date: 2020-06-29 02:46 pm (UTC)no subject
Date: 2020-06-30 02:47 pm (UTC)no subject
Date: 2020-06-30 03:02 pm (UTC)no subject
Date: 2020-06-30 04:58 pm (UTC)no subject
Date: 2020-06-30 05:33 pm (UTC)But then if Ω can have more than two points in some logic, then the definition of ∧ does not seem complete, because it doesn't explain how to choose mappings other than (true,true)→true.
If it's only two points in Ω, there is no choice; but if it's more points, then the specified definition doesn't seem to make ∧ a unique arrow.
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Date: 2020-06-30 05:40 pm (UTC)Regarding conjunction, it classifies (true, true) inclusion to Ω. In Grothendieck topologies, classifying maps are built by Yoneda lemma.
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Date: 2020-06-30 05:49 pm (UTC)no subject
Date: 2020-06-30 05:59 pm (UTC)We can have a variety of logics.
First, 4-valued booleans. a = (0,1), b=(1,0), T = (1,1), ⊥=(0,0), and operations are defined bitwise.
Second, numbers ⊥=0, a=1/3, b=2/3, T = 1. This linear order is a bounded distributive lattice, so everything is well-defined, conjunction is just minimum. Trying to find another four-point logic...
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Date: 2020-06-30 06:04 pm (UTC)But my question is how does the definition reflect this fact? It seems to only say that T ∧ T = T - nothing about a ∧ b = ⊥, and not, say, a ∧ b = a.
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Date: 2020-06-30 08:01 pm (UTC)no subject
Date: 2020-06-30 08:17 pm (UTC)Maybe I don't get something basic. What it looks like to me, since the square has arrows (true,true) and true, it defines only the behaviour of the classifying arrow for one pair, not all possible pairs.
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Date: 2020-07-01 12:08 am (UTC)