Satisfaction/validity and variable assignments

Question

I sat an exam in logic a few days ago, and one of the questions I answered was on the theme of proving validity and proving when certain variable assignments satisfy certain structures in what I know as L2 (I think it's called different things, but I mean the language of predicate logic).

Now since the exam, I've been a little concerned that I may have got something wrong, but I'm unsure because in my mock exam at the start of this term, my tutor didn't mark it wrong, and this particular detail is applied inconsistently throughout the teaching material.

Basically, the textbook I use says that:

|∃vφ| with variable assignment α over L2-structure A has the truth-value T if >and only if |φ| = T with some variable assignment β over A, differing from α at most in v.

This is obvious - to express it crudely, if there is just one case where everything 'adds up', so to speak, then the existential quantifier is satisfied in itself.

My worry in a nutshell is that I never separated 'variable assignment β' from 'variable assignment α' when proving that existential quantifiers are satisfied. I would always do proofs along the following lines for a question like: Specify an L2-structure in which ∃xPx is true, and explain why the sentence is true in that structure:

Domain of discourse: {1}
|P|A = {1}

Let variable assignment α be such that |x|α = 1
1 ϵ |P|A, thus |x|α ϵ |P|A, thus |Px|A over α = T
Therefore, |∃xPx|α over A = T

Obviously, here I didn't differentiate between using variable assignment β differing at most in x and just using α.

My textbook also has this line:

The truth-value of a sentence does not depend on the variable assignment. For both α and β over A, |φ|α = |φ|β.

So basically, I'm very confused, and I'm very much worrying - if having a different variable assignment does absolutely matter, and the single case using α wrecks the proof of validity, I will have lost a whole lot of marks. I'd really appreciate some clarification. And sorry for the difference in notation - I think my university uses its own notation, weirdly...

Answer 1

Of course α is a variable assignment over A differing from α at most in x.

Thus, it works...

"At most" means that β differs from α only in x or not differs at all.

Regarding sentences, you have to keep in mind that varibale assignments are a way to give refernce to free variables in formulae. Sentences have no free occurrences of variables: thus, in a nutshell, there is nothing to "act on" with a variable assignment.

Consider e.g. the following sentence in the language of arithmetic (with domain N):

∃x (x=0).

It is satisfied by a variable assignment α whatever, because it is enough to consider a β that differs from α in the fact that assign 0 to x and we have:

|(x=0)|β=T in N.

This fact holds irerspective of α, and thus all variable assignments in N satisfy the above sentence.

The same apply, in the negative, for:

¬∃x (x=0).