Not logged in. · Lost password · Register Member since Oct 2014 54 posts 2019-01-08, 12:27   #1   Subject: completeness I thought completeness is this H |= A then H  |- A In exercise 9.3 it is : H |= A IFF H  |- A Are we actually supposed to show both directions ? Do I miss something ? Or is this an error?
 Member since Oct 2016 806 posts 2019-01-08, 13:24   #2   Good question! Technically, the direction "If H |= A then H  |- A" is what's called completeness. The other direction is called soundness. However, any calculus which is *unsound* is basically useless and inadequate for any purposes (recall what it would mean if a calculus is unsound: Then I can prove things that aren't true!), so we only ever consider calculi that are sound. Hence, *given* that a calculus is (supposed to be) sound anyway, the two statements "If H |= A then H  |- A" and "H |= A IFF H  |- A" are equivalent.
 Member since Oct 2016 806 posts 2019-01-08, 13:27   #3   In reply to post #1 Possibly relevant: You may accordingly *without loss of generality* assume the calculus in your proof to be sound Member since Nov 2018 44 posts 2019-01-09, 12:00   #4   Subject: Completeness and Model Existence A little aside: one direction of the Model Existence Theorem (If you cannot derive a contradiction from H then H has a model) is equivalent to the strict definition of completeness (H |= A then H  |- A). This is a very useful equivalence which is often used to prove completeness of a logic: the proof then boils down to picking an arbitrary consistent set of formulae and constructing a model for it (the "canonical" model).
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