Member since Oct 2014
54 posts

20190108, 11:27 #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
767 posts

20190108, 12:24 #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
767 posts

20190108, 12:27 #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
25 posts

20190109, 11:00 #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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