I reminded us of what is meant by an interpretation (p. 363,
text). I pointed out the exponential notation the book uses --
e.g., given a model (=structure + interpretation) * U *,

a^*U* is the element of the UD of the model * U *
associated to the constant "a" , and M^*U* is the set of ordered
n-tuples of the UD associated to the n-place predicate M.

I said that my students may use this notation if they like it, or may use "a with a bar (i.e., a short line) drawn over it" (in this file, I will type "a-bar"), M-bar instead, if they prefer, on assignments and tests. The point is to use notation for the interpretations of a, M that look LIKE a, M but are different (because after all, normally the interpretations of these symbols of PL are not just the symbols themselves).

I repeated that a set of ordered 1-tuples of elements of a UD is basically a subset of the UD (the parentheses around the entry * of the 1-tuple (*) essentially add nothing; we might as well think of (*) as *, i.e., as just an element of the UD).

I pointed out that, given a set P-bar of ordered 2-tuples of elements of
the UD in a model, we can think of that set geometrically as an *
ordered graph *, i.e., a collection of nodes (dots), with an arrow
pointing from the node labelled
* to the node labelled $ if and only if (*,$) is an element of
P-bar.

I guess one can mimic this idea with ordered 3-tuples etc., but
the geometric interpretation starts to get cumbersome -- (*,$,-)
would be drawn

node * -> node $ -> node - . (With maybe a
boundary drawn around the whole collection of three nodes and two
arrows to remind us that we are drawing a 3-tuple here and not two
ordered 2-tuples....) I suspect that Whiteley and I will not have to
ask test/exam questions involving n-place predicates when n is greater
than 2, anyway. ... I did this

**Example:** Is the following argument valid?

premise: (\forall x)(\exists y) Dxy

conclusion: (\exists y)(\forall x) Dxy

D gets interpreted in any model as D-bar, an ordered graph. The problem then translates to this: Is it true, in an arbitrary model, that the first statement below implies the second?:

For each given element of the UD, there is some element of the UD, such that the given one points to it, in the graph D-bar.

There is an element of the UD, such that all elements of the UD point to it, in the graph D-bar.

The general strategy, then, is to think of the simplest imaginable models, starting therefore with UD's having only one element (the UD always has to have at least one element in it, because we have to be able to interpret the constants of PL as elements of the UD)

Well, if our UD had only one element * and the first of the two statements above were true, then * would have to point to itself. But then the second statement is true. So let's look at a UD with two distinct elements * and $. We could have an arrow from each pointing to the other and the first sentence above would be satisfied, but not the second. Or we could have an arrow from each element pointing to itself, ditto. So:

** Answer: Invalid. ** Example: UD = {1,2},

D-bar = { (1,2), (2,1) }.

An answer like this, with all the stuff I typed out above it just going on in your head, or scribbled somewhere out of the way, or on a back of a page, is OK for a test or exam.

So, after doing this example, I suggested that people look at the
definition of satisfaction of a sentence \alpha by a model *U*,
on page 368 (except that I did not have the page number to hand in
class...), and not to worry too much about the parts of that
definition featuring the variable assignments script a.

Then I gave this

**New defn. of consistency:** A set \Gamma of PL sentences is *
consistent* (or, "satisfiable"), if it has a model, i.e., if there
is a model *U* such that *U* |= \Gamma .

Hmm... actually, now I am not sure I did all of the above in class -- but it is all stuff you should know anyway, so here it is. The following stuff I definitely did in class on 21 March:

Ex: (9.3.13 b.) Find a model for (\forall x)Kx & (\forall x)(Kx -> ¬Lx) .

Well, we need a UD and subsets K-bar and L-bar of the UD, such that both conjuncts of the above conjunction are true in the model:

Every element of the UD should be in K-bar (i.e., K-bar is the whole UD), and for all * in the UD, if * is in K-bar, then * is not in L-bar. But by the first conjunct, this gives that every element of the UD is NOT in L-bar. I.e., L-bar is the empty set. So the simplest model would seem to be

UD = {1} = K-bar, L-bar = { }.

**Example:** (9.3.13 e.) Find a model for

[ (\forall x)(\exists y)Rxy & (\forall x)(\exists y) ¬Rxy ]
& ¬(\exists x)(\forall y)Rxy .

Well, we need a (nonempty) UD and a directed graph R-bar on it, such that all three conjuncts above interpret as true in the model. So every node must point to some node. Also, every node must fail to point to some node. And also, there is no node that points to all nodes. Actually, I did not notice this in class, but e. is stupid: If the second conjunct is satisfied, the third is automatically satisfied. (It is logically equivalent to the second.)

A UD with one element will not work. But if we take two nodes and have each point to itself (having each point only to the other would also be ok), then we are in business: UD = {1,2}, R-bar = { (1,2), (2,1) } .

** Example:** (9.4.1 f.) True or false? (If true, ....; if
false, ...., as usual.):

If \alpha and \beta are model-satisfiable sentences, then so is \alpha & \beta.

Strategy: Well, there could be a model that satisfies \alpha, and a different model that satisfies \beta, yet no model that satisfies the conjunction. So I am suspicious, and I start to believe this is false, and look for a specific counterexample. Now the man in the red cap in the fifth row in class suggested taking \alpha to be A and \beta to be ¬A, and that will work, but we have not talked about the interpretation of 0-place predicates (i.e. atoms of SL, as predicates in PL). Our text briefly does this, in a footnote on p. 363. There, a 0-place predicate interprets as a set of ordered 0-tuples. Now there is only one ordered 0-tuple of elements of any given set. Here it is: ( ). I mentioned in class that if, say, the UD has exactly three elements, then there are exactly 3^1 = 3 ordered 1-tuples of elements of the UD. And there are exactly 3^2 = 9 ordered 2-tuples of elements of the UD.

Similarly, there are 3^0 = 1 ordered 0-tuples (and we wrote them all down in the above paragraph...). So there are two possible subsets of the set of all ordered 0-tuples of elements of the UD -- {( )} and { }, the empty set of 0-tuples..... If we assign a 0-place predicate letter to the former set, we think of it as interpreting as true in the model; if we assign it to the latter, we think of it as interpreting as false in the model.

So we take \alpha to be A, and \beta to be ¬A, where A is a
0-place predicate. We note that in the model:

UD = {1}, A-bar := {( )}, A interprets as true, and in the model

UD = {1}, A-bar := { }, the negation of A interprets as true.
But there is no model in which the conjunction of A and \not A
interprets as true.

We could have avoided the discussion of 0-place predicates by taking \alpha to be Cb and \beta to be ¬Cb. Then take UD to be {1} and b-bar to be 1. To satisfy Cb, take C-bar to be {1}. To satisfy \beta, take instead the model with the same UD and same b-bar, but let now C-bar be the empty set. Etc.

I think that was about all that was done 21 March.