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jkorb / KI1V13001-Inleiding-Logica

This is the source material for the course "Inleiding Logica" (KI1V13001) as taught at Utrecht University for the BSc "Kunstmatige Intelligentie"
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Ambiguity in 6.5.2 b #104

Closed DorusKeijzer closed 3 years ago

DorusKeijzer commented 3 years ago

Another possible problem that should be pushed for next year:

When discussing this exercise in the work groups, we were confused why 6.5.2b was incorrect:

image

We initially parsed this statement as "For all formulas phi in gamma and for all valuations v, ...", thinking that the "for each" phrase at the beginning of the sentence distributes to both the "phis in gamma" and the "valuation v" part of the phrase: "for each formula, ..., and for each valuation", so we concluded that this must mean that gamma is unsatisfiable. When we found out that our answer to this question was wrong, we took another look and found we were supposed to parse it as if "for each" didn't distribute to the valuation: "for each formula,... and the one valuation ... ", which would not necessarily make gamma unsatisfiable.

Part of the confusion is that a few lines earlier, we're confronted with 6.5.1 f, where "for all" does indeed distribute to both the formulas and the valuations.

image

(here, f is correct according to the answers)

With this one it is obvious that it does distribute, because it uses "for all" and "valuations" (plural) rather than "valuation", though it does set a precedent that terms such as "for all" can distribute. 6.5.2b on the other hand, uses "for each", which means we cannot look at whether the nouns are plural to determine whether the term distributes because "for each" always requires the nouns it refers to to be singular.

jkorb commented 3 years ago

Thanks for this @DorusKeijzer ! I think, however, you guys missed the real reason why (b) is not a correct answer for 6.5.2. What if the set is empty? First, note that the empty set is satisfiable. To see this, note that, trivially, all valuations make all members of the empty set true. So any valuation v can be the witness of the satisfiability of the empty set. At the same time, the empty set satisfies condition (b): again trivially, all formulas in the empty set are false under any given valuation. We get that (b) doesn't imply that Gamma is satisfiable.

I could add a hint in the solutions that help seeing this. What do you think?

DorusKeijzer commented 3 years ago

right! I hadn't considered that at all. I will report this to the person who asked the question.

I think a hint might indeed be in place here, or instead, this explanation could be provided in appendix F.

jkorb commented 3 years ago

Great! I'll add an explanation/hint in the solution :)