We have defined satisfaction in a model with a variable assignment. We have expressed formula
being satisfied by model
with variable assignment
as

Now we can also say that a formula
is satisfied by model
(not limited to a specific variable assignment) if
is satisfied by
with every variable assignment. Thus

if and only if

If no free variables occur in
(that is, if
is a sentence), then
is true in model
.
Variable assignments allow us to deal with free variables when doing the semantic analysis of a formula. For two variable assignments,
and
, satisfaction by
differs from satisfaction by
only if the formula has free variables. But sentences do not have free variables. Thus a model satisfies a sentence with at least one variable assignment if and only if it satisfies the sentence with every variable assignment. The following two definitions are equivalent:
- A sentence
is true in
if and only if there is a variable assignment
such that

- A sentence
is true in
if and only if, for every variable assignment 

The latter is just a notational variant of:
- A sentence
is true in
if and only if

On the previous page, we looked at the following model and variable assignment.
For the model







For the variable assignment



We noted the following results:








We also noted above that for sentences (though not for formulae in general), a model satisfies the sentence with at least one variable assignment if and only if it satisfies the sentence with every variable assignment. Thus the results just listed hold for every variable assignment, not just
.
Applying our definition of truth, we get:








This corresponds to the goals (1)–(8) of the previous page. We have now achieved those goals.
On Models page, we also considered an infinite model






We can reuse the same variable assignment from above, namely



On the Models page, we listed the following goals for our definitions.

This does not require our definition of truth or the definition of satisfaction; it is simply requires evaluating the exended variable assignment. We have for any
on defined on
:


We also listed the following goal on the Models page.


First we note that:




Indeed:




Because the formulae of (9) and (10) are sentences,




Applying the definition of truth, we find the goals of (9) and (10) achieved. The sentences of (9) are true and those of (10) are false.
In addition, we listed the following goal on the Models page.


Corresponding to (11):

is true if and only if, for each i a member of the domain, the following is true of at least one j a member of the domain:
![{\displaystyle \langle {\mathfrak {M_{2}}},\ \mathrm {s} [x\!:\,i,x\!:\,j]\rangle \ \vDash \mathrm {F} (x,y)\ .\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5227c573961f96feaf1fa53da99e0cfddcd15ef5)
But
was assigned the less then relation. Thus the preceding holds if and only if, for every member of the domain, there is a larger member of the domain. Given that the domain is
this is obviously true. Thus, (13) is true. Given that the formula of (11) and (12) is a sentence, we find the goal expressed as (11) to be met.
Corresponding to (12):

is true if and only if, for each i a member of the domain, the following is true of at least one j a member of the domain:
![{\displaystyle \langle {\mathfrak {M_{2}}},\ \mathrm {s} [x\!:\,i,x\!:\,j]\rangle \ \vDash \mathrm {F} (y,x)\ .\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fd7b5df0184d3bb056f21c64b08f37a8f72ebbf)
This holds if and only if, for every member of the domain, there is a smaller member of the domain. But there is no member of the domain smaller than 0. Thus (14) is false. The formula of (12) and (14) fails to be satisfied by
with variable assignment
. The formula of (12) and (14) is a sentence, so it fails to be satisfied by
with any variable assignment. The formula (a sentence) of (12) and (14) is false, and so the goal of (12) is met.