Satisfiability in first order languages

Question

Suppose $L_{f}$ is a 1st-order language pertaining to fields and let $R$ be the real field structure (R;0,1,+,·), that is, a structure that has 0 and 1 as the constants, and +,· be 2 place functions. Let $s$ be the "assignment" given by $s(v_{i})$ = $i+1$, where $v_{i}$ is variable of $L_{f}$ and consider the formula $\phi:\forall v_{1}((v_{3} = 0·v_{1}) \rightarrow (v_{3} = 0))$ Is it the case that $R\vDash \phi[s]$?

First off, I am confused about the terminology used. My professor does not use (I think) the most common terminology for what he calls an "assignment". I would like to know what this is most commonly called in first order logic so I can read up on it by myself.

Secondly, I have trouble understanding what exactly $R\vDash \phi[s]$ entails. This is read as "R satisfies $\phi$ with assignment $s$." Since I don't have a very good understanding of the assignment in the first place, I can't see what this is really saying.

2) Now, on a similar vein, suppose that $L$ is a first order language that has a 2 place function $f$, a 1 place predicate symbol $Q$, and a 2 place predicate symbol $P$. The "=" argument is not present within this language and constants are absent. Let $\phi$ be the formula $\forall xP(x,y) \rightarrow Q(z)$. Let $s_{1}$ and $s_{2}$ be assignments such that they agree on x,y,z. Then why does the following hold: $L \vDash \phi[s_{1}] \iff L \vDash \phi[s_{2}] $? Additionally, does $Q(x) \vDash Q(y)$?

Similarly to the first question, since I don't have an understanding of $L \vDash \phi[s$], I can't quite answer it. For the second part, how does a predicate model a predicate?

Answer 1

First, "assignment" is the usual name for a function giving values (in the underlying set of a structure) for all the variables.

Second, to determine whether a first-order formula is true or not, you need three pieces of information: (1) meanings of all the function, constant, and relation symbols in the formula, (2) a universe of discourse over which the bound variables of the formula range, and (3) values (in that universe of discourse) for all the free variables of the formula. A structure provides (1) and (2); an assignment provides (3). Regardless of terminology (for "assignment" or anything else), any definition of $\models$, meaning a formula is true in some situation, will have to involve all the necessary information: the formula and (1), (2), and (3). In your first question, $R$ is the structure giving (1) and (2), while $s$ is the assignment giving (3). I would suggest that your best bet for understanding assignments is to study carefully the definition of "Structure $\models$ formula [assignment]."

I think that will take care of most of your difficulties, until you get to the "Additionally" question at the end of question 2. Here you encounter an unfortunate notational mess. The same symbol $\models$ that is used in the context "Structure $\models$ formula [assignment]" as above, is also used in another context: "Set of formulas $\models$ formula." Here it means that the formula on the right of $\models$ is a logical consequence of the set of formulas on the left. In addition, when the set on the left consists of just one formula $\phi$, one conventionally writes just $\phi$ instead of the correct $\{\phi\}$. So your $Q(x)\models Q(y)$ means that $Q(y)$ is a logical consequence of $Q(x)$.

Finally, there are two different notions of logical consequence in the literature, which agree on sentences but not on formulas with free variables. Your $Q(x)\models Q(y)$ is, unfortunately, one of the cases where the two definitions disagree. So all I can tell you is to look very carefully at your textbook's or your professor's definition of "logical consequence" (i.e., the definition of $\models$ in this context).

Answer 2

First: see H.Enderton, A Mathematical Introduction to Logic, page 83.

The assignment function is a function:

$s : \text{Var} \to |\mathfrak A|$,

i.e. a function from the set $\text{Var}$ of all (individual) variables to the universe (or domain) $|\mathfrak A|$ of the structure $\mathfrak A$.

In other terms, it "assigns" values, i.e. objects of the domain, to variables.

Second: how does $s$ work ?

Consider the trivial example of the formula $(x=0)$ and the structure $\mathbb N$.

Does $\mathbb N \vDash (x=0)$ or not ? It depends...

We need a function $s$ assigning objects (in this case natural numbers) to variables: if $s(x)=1$, then $\mathbb N \nvDash (x=0)[s]$, while if $s(x)=0$, then $\mathbb N \vDash (x=0)[s]$.

Last: considering your example, the answer is: yes.

$\mathbb R \vDash \phi[s]$ amounts to: $\mathbb R \vDash ∀v_1((4=0.v_1) → (4=0))[s]$ (because $s(v_3)=4$), that is TRUE, because whatever is the value $a$ assigned by $s$ to $v_1$, $(4=0.a)$ will always be FALSE.