I'm studying the subject myself from the book "A Friendly Introduction to Mathematical Logic" and I could not understand a corollary about the sentences saying that:

"If D is a sentence in the language L and A is an L-structure, either A |= D[s] for all assignment functions s, or A |= D[s] for no assignment function s."

In a L-language, a sentence is a formula without any free variables ... and we only have the universal quantifier for transforming a formula into a sentence. So, I think that the first part of the corollary refers to the "true" sentence and the second part refers to the "false" sentence ... but what about the third option (not available in the corollary) in which some assginment functions satisfy the formula and some others don't?

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    $\begingroup$ The point is that since $D$ has no free variables at all, it doesn't depend on the variable assignment. $\endgroup$ – Noah Schweber Nov 8 '18 at 19:38
  • $\begingroup$ But, it continues by saying that "Notice that if D is a sentence, then A |= D if and only if A |= D[s] for any assignment function s. In this case we will say that the sentence D is true in A." So, I think that this is the first part of the corollary above ... and I think that the second part of the corollary means "D is false in A". Am I wrong? $\endgroup$ – Alp Nov 8 '18 at 20:06
  • $\begingroup$ Yup, that's right. $\endgroup$ – Noah Schweber Nov 8 '18 at 20:52
  • $\begingroup$ Thanks a lot for your help. I want to clarify my original question, so let me give a real example. Let's say that we have a propositional function (formula), Hx. We have a universe, {1, 2, 3, 4} ... an our proposition (sentence) is All(Hx). In this case, we have four variable assignment functions for x, s[x|a] where a is € of our universe. For the sentence All(Hx) to be true, the formula have to be satisfied by all the assignment functions. For the sentence All(Hx) to be false, there is no assignment functions satisfying the formula. I think that these are the mentioned in the corollary. $\endgroup$ – Alp Nov 8 '18 at 21:07
  • $\begingroup$ But what about the third case: If some assignment functions satisfy the the formula, and others don't. Of course, we will have a false sentence in this third case, but this case is not mentioned in the corollary. I could not understand here. $\endgroup$ – Alp Nov 8 '18 at 21:09

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