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Question regarding mastersolution of mockexam
I’m a bit confused by one remark in the master solution of this year’s mock exam (i.e. last year’s) regarding problem 4.2. One was asked to disprove correctness of a calculus and the part that confused me was “Let A⇒B and B⇒C be valid”.
These formulas are not valid. I think the answer is a bit more subtle:
We defined correctness of a calculus via compatibility with the entailment induced by the satisfaction relation of the corresponding logical system, which for propositional logic is given on slide 295.
So in my eyes the argument should go as follows: we will find a model for which A⇒B and B⇒C are true but C is false, thus violating the entailment A⇒B and B⇒C|=C but our calculus would derive this relation, so it must be incorrect.
Or am I missing something obvious?
The solution does that: [|A|] = F, [|B|] = F, C = [|F|], so A⇒B and B⇒C are true.
C is false since the value false was assigned to it. And we can derive C from the two formulae A⇒B and B⇒C according to the solution.
So I dont see the difference. The solution just says what you are saying if I am not mistaken. Where do you think is the difference?
Hm you’re right, this is indeed what is done in the solution … but the explanation why this solves the problem is wrong: it’s not about validity of A => B and A => C but about one model satisfying it and the problem isn’t that calculi are supposed to preserve validity (hypothesis = empty set) but general entailment (nonempty hypothesis).
I think you are partly correct. A ⇒ B and B ⇒ C are not valid per the definition of the lecture notes, as M ⊨[sup]φ[/sup] A does not hold ∀φ (e.g. A: T, B: F, A ⇒ B: F).
However our calculi are supposed to preserve validity due to their assumed correctness:
Since they obviously are not our caluculus must be incorrect.