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Hello,

doesn't the intersection of α1 and α2 contradict the silent action?

Then how can it be weakly endochronous if it has contradicting rules?

in * TF "Emb. Sys. and Rob." by (440 points)

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Yes, the intersection of α1 and α2 would contradict the silent action. However, that slide just lists examples for the intersection, union and difference of actions. It does not say that these actions are contained in the table or should be added to the table.

However, I see that there is a slight mistake in the definition of weak endochrony on slide 75 since all joinable actions are also non-contradicting. As explained on the later slides 76, 77 and 78, joinable actions are already accepted without further conditions since these form already a diamond. It is just for the non-contradicting, but not joinable actions that we need to have action a1 ⊓ a2, a1 \ a2 and a2 \ a1 to construct a weak diamond. 

For the parallel-or, only the the actions a1 and a2 are overlapping. These actions are also non-contradicting since for x1, we have a2(x1)=⧇ and for x2, we have a1(x2)=⧇ and for the output, we have a1(y) · a2(y) = 1·1 = a2(y) · a1(y). The actions a1 and a2 are moreover joinable since they read different inputs. Hence, they already form a diamond, and we don't have to add or demand more actions.

The precise definition of weak endochrony means that for all overlapping rules, these are either joinable or otherwise at least non-contradicting with enough rules to construct a1 ⊓ a2, a1 \ a2 and a2 \ a1 so that we can form a weak diamond. The definition has been fixed on the slide accordingly.

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Thank you for the detailed explanation!

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