Priest (12.3) Introduction to Non-Classical Logic, ‘Semantics [of first classical first-order logic]’, summary

 

by Corry Shores

 

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[The following is summary of Priest’s text, which is already written with maximum efficiency. Bracketed commentary and boldface are my own, unless otherwise noted. I do not have specialized training in this field, so please trust the original text over my summarization. I apologize for my typos and other distracting mistakes, because I have not finished proofreading.]

 

 

Summary of

 

Graham Priest

 

An Introduction to Non-Classical Logic: From If to Is

 

Part II

Quantification and Identity

 

12

Classical First-order Logic

 

12.3

Semantics

 

 

 

Brief summary [much of the following is quotation]:

Regarding the semantics of classical first-order logic, we say that:

An interpretation of the language is a pair, ℑ = ⟨D, v⟩. D is a non-empty set (the domain of quantification); v is a function such that:

• if c is a constant, v(c) is a member of D

• if P is an n-place predicate, v(P) is a subset of Dⁿ

(Dⁿ is the set of all n-tuples of members of D, {⟨d₁, ..., d_n⟩: d₁, ..., d_n ∈ D}. By convention, ⟨d⟩ is just d, and so D¹ is D.)

(Priest 264)

All formulas have a truth value. To evaluate them, since they use variables which are substitutable by constants,

we extend the language to ensure that every member of the domain has a name. For all d ∈ D, we add a constant to the language, k_d, such that v(k_d) = d. The extended language is the language of ℑ, and written L(ℑ). The truth conditions for (closed) atomic sentences are:

v(Pa₁ ... a_n) = 1 iff  ⟨v(a₁),  ..., v(a_n)⟩ ∈ v(P) (otherwise it is 0)

The truth conditions for the connectives are as in the propositional case (1.3.2).

(Priest 265)

v(¬A) = 1 if v(A) = 0, and 0 otherwise.
v(A ∧ B) = 1 if v(A) = v(B) = 1, and 0 otherwise.
v(A ∨ B) = 1 if v(A) = 1 or v(B) = 1, and 0 otherwise.
v(A ⊃ B) = 1 if v(A) = 0 or v(B) = 1, and 0 otherwise.
v(A ≡ B) = 1 if v(A) = v(B), and 0 otherwise.

(Priest 5)

For the quantifiers:

v(∀xA) = 1 iff for all d ∈ D, v(A_x(k_d)) = 1 (otherwise it is 0)

v(∃xA) = 1 iff for some d ∈ D, v(A_x(k_d)) = 1 (otherwise it is 0)

(Priest 265)

We define validity in the following way:

Validity is a relationship between premises and conclusions that are closed formulas, and is defined in terms of the preservation of truth in all interpretations, thus: Σ ⊨ A iff every interpretation that makes all the members of Σ true makes A true.

(Priest 265)

We find the following equivalences in classical first-order logic:

v(¬∃xA) = v(∀x¬A)

v(¬∀xA) = v(∃x¬A)

v(¬∃x(Px ∧ A)) = v(∀x(Px ⊃ ¬A)

v(¬∀x(Px ⊃ A) = v(∃x(Px∧¬A))

(Priest 265)

And lastly we note [something regarding denotation, namely] that

If C is some set of constants such that every object in the domain has a name in C, then:

v(∀xA) = 1 iff for all c ∈ C, v(A_x(c)) = 1 (otherwise it is 0)

v(∃xA) = 1 iff for some c ∈ C, v(A_x(c)) = 1 (otherwise it is 0)

(Priest 265)

 

 

 

Contents

 

12.3.1

[Interpretations of Constants and Predicates]

 

12.3.2

[Truth Evaluation for Classical First-Order Logic]

 

12.3.3

[Validity in Classical First-Order Logic]

 

12.3.4

[Some Equivalences]

 

12.3.5

[Constants, Denotation, and Quantification]

 

 

 

 

Summary

 

12.3.1

[Interpretations of Constants and Predicates]

 

[In the following we will look at interpretations for classical first-order logic. To give illustration to some of these concepts, let me first take an example from David Agler’s Symbolic Logic: Syntax, Semantics, and Proof section 6.4.2. We have the following domain:

D = {Alfred, Bill, Corinne}

(Agler 265)

We have an interpretation function I. It assigns objects in the domain to names (considered constants also, I think).

I (a) Alfred
I (b) Bill
I (c) Corinne

That is, Alfred in D is assigned to ‘a,’ Bill in D is assigned to ‘b,’ and Corinne in D is assigned to ‘c.’

(Agler 265)

Then we have n-place predicates:

Sx: x is short

(Agler 265)

Lxy: x loves y

(Agler 265)

To these n-place predicates, the interpretation function I assigns sets of n-tuples:

I(Sx): {⟨Alfred⟩, ⟨Bill⟩}

I(Lxy): {⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩}

(Agler 265, angle-brackets for the singletons added by me for my own convenience, but perhaps erroneously. See the text of this section explaining their omission.)

Agler then describes how we make the truth-value assignments with the v valuation function:

We will call an interpretation where a truth value is assigned to a wff a valuation (v). For this, let ‘R’ be an n-place predicate, let ‘a₁,’ ... ‘a_n’ be a finite set of names in RL, and let ‘a_i’ be a randomly selected name.

(1) v(Ra_i) = T if and only if the interpretation of ‘a_i’ is in ‘R.’

This says that we can assign a value of true to ‘Ra_i’ if and only if our interpretation of ‘a_i’ is in the interpretation of ‘R.’ To consider this concretely, we examine two examples. First, consider the following wff:

Sa

| Let’s say that ‘Sa’ is the predicate logic translation of Alfred is short. According to (1), ‘Sa’ is true if and only if ‘a’ is in ‘S,’ that is, if and only if an interpretation of the predicate ‘short’ includes an interpretation of the name Alfred. Earlier, we said that

I (Sx): objects in D that are short (i.e., {Alfred, Bill})

And so, ‘Sa’ is true in the model since Alfred belongs to the collection of objects that are short.

(Agler 266-267)

I am a little confused about the use of Ra_i, because this seems to be a 1-place predicate only. I wonder if we can modify this for larger valences. So maybe:

v(Ra₁, ... a_n) = T if and only if the n-tupled interpretations of a₁, ... a_n are in ‘R.’

Suppose we have

v(Lac)

The interpretation of a and c would be Alfred and Corinnne. But recall the interpretation for L

I(Lxy): {⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩}

Here we have Alfred and Corinne. But they are not in the same order as Lac. I am not sure how to formally work with this. I think back when I made the summary, I had:

I(Lac) = {⟨Alfred, Corinne⟩}

This is not written in the Agler text, but it is based on the above formulation for I(Lxy) and this line:

I (Sx): objects in D that are short (i.e.,{Alfred, Bill})

(Agler 266)

But maybe for Priest’s formulations we are not allowed to use the v function for a specific formula with constants, and rather the formulations need to refer to the interpretations of the constants taken by themselves but not to formulas with constants in them. This is why I wrote for the Agler reformulation:

the n-tupled interpretations of a₁, ... a_n

meaning that the interpretations made in that order and then combined into an n-tuple. I am not sure if that works, but note that in the next section, 12.3.2, the evaluation is written as:

v(Pa₁ ... a_n) = 1 iff  ⟨v(a₁),  ..., v(a_n)⟩ ∈ v(P) (otherwise it is 0)

where it says

⟨v(a₁),  ..., v(a_n)⟩

is what I mean I suppose by the n-tupled interpretations of a₁, ... a_n. Here Priest is not saying

v(Pa₁ ... a_n) = 1 iff  v(Pa₁ ... a_n) ∈ v(P) (otherwise it is 0)

So probably a formulation like

I(Lac) = {⟨Alfred, Corinne⟩}

would not work here. (Also, it brings out my confusion between the uses of the v function.) At any rate, supposing that we have determined the n-tuples corresponding to the predicate formulations, we would see that in our example

{⟨Alfred, Corinne⟩}

is not a member of

I(Lxy): {⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩}

Thus

v(Lac) = F

Now let us turn to Priest’s mode of formulation. He writes:

An interpretation of the language is a pair, ℑ = ⟨D, v⟩. D is a non-empty set (the domain of quantification); v is a function such that:

• if c is a constant, v(c) is a member of D

• if P is an n-place predicate, v(P) is a subset of Dⁿ

(Dⁿ is the set of all n-tuples of members of D, {⟨d₁, ..., d_n⟩: d₁, ..., d_n ∈ D}. By convention, ⟨d⟩ is just d, and so D¹ is D.)

(Priest 264. Note, the first formulation should look like:

)

I am not certain, but what Priest is calling an “interpretation” seems to be what Agler was calling a “model”:

a model consists of a domain (D) and an interpretation function (I).

(Agler 264)

So returning to Priest, the domain D is the “domain of quantification”. I am assuming this is the domain of objects to which the quantifiable variables may be assigned, but I am not sure. Let us suppose that it is, as I am not sure what else it would be. (I am not sure if the constants are in this domain. From what is said later, it seems not, but I really do not know.) The v function seems to be like Agler’s I function. Again from Priest:

 v is a function such that:

• if c is a constant, v(c) is a member of D

• if P is an n-place predicate, v(P) is a subset of Dⁿ

In our Agler example,

I (c) = Corinne

So I would think that here with the Priest notation we would write:

v(c) = Corinne

or maybe

v(c) = {Corinne}

But I am not sure. The next idea we should cover is the idea behind Dⁿ. Priest says:

Dⁿ is the set of all n-tuples of members of D, {⟨d₁, ..., d_n⟩: d₁, ..., d_n ∈ D}. By convention, ⟨d⟩ is just d, and so D¹ is D.

So let us look at our Agler example:

D = {Alfred, Bill, Corinne}

D¹ = {⟨Alfred⟩, ⟨Bill⟩, ⟨Corinne⟩}

D² = {⟨Alfred, Alfred⟩, ⟨Alfred, Bill⟩, ⟨Alfred, Corinne⟩, ⟨Bill, Alfred⟩, ⟨Bill, Bill⟩, ⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩, ⟨Corinne, Bill⟩, ⟨Corinne, Corinne⟩}

As he says, “By convention, ⟨d⟩ is just d, and so D¹ is D,” so:

D¹ = {⟨Alfred⟩, ⟨Bill⟩, ⟨Corinne⟩} =

D = {Alfred, Bill, Corinne}

(I wonder if one way to understand these formulations are as Cartesian products of sets made upon iterations of themselves. Consider slide 8 of Baran Kaynak’s presentation “Classical Relations and Fuzzy Relations”

The elements in two sets A and B are given as A ={0, 1} and B ={a,b, c}.

Various Cartesian products of these two sets can
be written as shown:

A × B ={(0,a),(0,b),(0,c),(1,a),(1,b),(1,c)}

B × A ={(a, 0), (a, 1), (b, 0), (b, 1), (c, 0), (c, 1)}

A × A = A²={(0, 0), (0, 1), (1, 0), (1, 1)}

B × B = B²={(a, a), (a, b), (a, c), (b, a), (b, b), (b,
c), (c, a), (c, b), (c, c)}

(Kaynak, slide 8)

At any rate,) consider again:

D² = {⟨Alfred, Alfred⟩, ⟨Alfred, Bill⟩, ⟨Alfred, Corinne⟩, ⟨Bill, Alfred⟩, ⟨Bill, Bill⟩, ⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩, ⟨Corinne, Bill⟩, ⟨Corinne, Corinne⟩}

And recall our two-place predicate from the Agler example:

I(Lxy): {⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩}

Is

{⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩}

a subset of

{⟨Alfred, Alfred⟩, ⟨Alfred, Bill⟩, ⟨Alfred, Corinne⟩, ⟨Bill, Alfred⟩, ⟨Bill, Bill⟩, ⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩, ⟨Corinne, Bill⟩, ⟨Corinne, Corinne⟩}

? Yes, it is. So recall this part of Priest’s formulation:

if P is an n-place predicate, v(P) is a subset of Dⁿ

So with the Agler example, we might write:

if L is an 2-place predicate, v(L) is a subset of D²

(Or maybe we should write:

if Lxy is an 2-place predicate, v(Lxy) is a subset of D²

) And in fact it is, as we saw. Let me now repeat the whole quotation for a clean presentation of it:]

An interpretation of the language is a pair, ℑ = ⟨D, v⟩. D is a non-empty set (the domain of quantification); v is a function such that:

• if c is a constant, v(c) is a member of D

• if P is an n-place predicate, v(P) is a subset of Dⁿ

(Dⁿ is the set of all n-tuples of members of D, {⟨d₁, ..., d_n⟩: d₁, ..., d_n ∈ D}. By convention, ⟨d⟩ is just d, and so D¹ is D.)

(Priest 264. Note, the first formulation should look like:

)

[contents]

 

 

12.3.2

[Truth Evaluation for Classical First-Order Logic]

 

[Recall from section 12.2.4 that a formula is “bound if it occurs in a context of the form ∃x ... x ... or ∀x ... x”, and it is free otherwise. And a “formula with no free variables is said to be closed”. Priest now writes:

Given an interpretation, truth values are assigned to all closed formulas.

(264)

Now, suppose we consider our Agler example

v(Lac) = F

(see Agler p.267)

Here do not have any variables, but we have a truth-value. Perhaps since there are no variables, that means there are no free variables, and thus that it would be by definition closed and hence eligible for a truth-value assignment by means of v. But I would like to mention a confusion I have regarding v. According to 12.3.1 above, v assigns either singular members of D (to constants) or n-tuple subsets of D (to n-place predicates). But here in this section, it will assign truth-values. I do not understand that yet. Either it is a different v function that assigns to formulas values from another domain (like the domain of values V in the structure ⟨V, D, {f_c; c ∈ C}⟩ from section 7.2.2.) Or perhaps “1” and “0” are members of D, and the n-tuples of Dⁿ with 1 and 0 as members are normally not extensions of predicates. I do not know how this works yet. At any rate, we suppose that by means of v we can assign 1 or o to a formula. He next writes:

To state the truth conditions, we extend the language to ensure that every member of the domain has a name.

(264)

I am not exactly sure why this is necessary. I would guess that it has something to do with evaluating quantifiers. For example, in order to evaluate all substitutions of a formula’s variable, we might need to make sure that the formula holds for all members of the domain. And maybe we also need to note that the domain members themselves cannot be substituted for the variables (we cannot say, LAlbert,Corinne for example) and instead only names or constants can (in other words, we can only say Lac in this case). Thus we need our language to include a set of constants that will fully exhaust the set of objects in the domain. This might be what is meant then in what follows here:

For all d ∈ D, we add a | constant to the language, k_d, such that v(k_d) = d. The extended language is the language of ℑ, and written L(ℑ).

(364-365)

(Note, some of these issues are discussed more directly in section 12.3.5 below.) Let us now look at how we evaluate closed atomic sentences:

v(Pa₁ ... a_n) = 1 iff  ⟨v(a₁),  ..., v(a_n)⟩ ∈ v(P) (otherwise it is 0)

And let us consider our Agler example:

v(Lxy) = 1 iff  ⟨v(x), v(y)⟩ ∈ v(L) (otherwise it is 0)

or maybe:

v(Lxy) = 1 iff  ⟨v(x), v(y)⟩ ∈ v(Lxy) (otherwise it is 0)

(Here I am supposing that the first v assigns 1 or 0 to formulas, and the second v assigns sets of n-tuples to n-place predicates. I mention this to compare with the Agler text, where I assigns to objects (or n-tuples of objects) in the domain and v assigns to truth values.). Let us now make the formula under truth evaluation specific:

v(Lac) = 1 iff  ⟨v(a), v(c)⟩ ∈ v(Lxy) (otherwise it is 0)

expanded further

v(Lac) = 1 iff  ⟨v(a), v(c)⟩ ∈ {⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩} (otherwise it is 0)

and expended yet further

v(Lac) = 1 iff  ⟨Alfred, Corrine⟩ ∈ {⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩} (otherwise it is 0)

Now, since

⟨Alfred, Corrine⟩ ∉ {⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩}

we would have:

v(Lac) = 0

We now will examine how to evaluate formulas made more complex by means of the connectives. He writes:

The truth conditions for the connectives are as in the propositional case (1.3.2).

(365)

Recall from section 1.3.2 the following evaluation rules:

Given an interpretation of the language, v, this is extended to a function that assigns every formula a truth value, by the following recursive clauses, which mirror the syntactic recursive clauses:
v(¬A) = 1 if v(A) = 0, and 0 otherwise.
v(A ∧ B) = 1 if v(A) = v(B) = 1, and 0 otherwise.
v(A ∨ B) = 1 if v(A) = 1 or v(B) = 1, and 0 otherwise.
v(A ⊃ B) = 1 if v(A) = 0 or v(B) = 1, and 0 otherwise.
v(A ≡ B) = 1 if v(A) = v(B), and 0 otherwise.

(5)

So suppose we take:

v(¬Lac)

Since

v(Lac) = 0

That means

v(¬Lac) = 1

And so on for the rest. Next we learn how to evaluate quantified formulas. He writes:

For the quantifiers:

v(∀xA) = 1 iff for all d ∈ D, v(A_x(k_d)) = 1 (otherwise it is 0)

v(∃xA) = 1 iff for some d ∈ D, v(A_x(k_d)) = 1 (otherwise it is 0)

(265)

Let us go through it part by part to be sure we know their senses.

v(∀xA)

Recall from section 12.2.3:

If A is any formula, and x is any variable, then ∀xA, ∃xA are formulas. I will omit outermost brackets in formulas.

(264)

So

∀xA

means the formula A, which has variable x, and we are universally quantifying over x. I am not certain, but suppose we take the Agler example from above

Lxy

Let us then write:

∀x∀yLxy

I am not sure if we can do that, but I am trying to understand what ∀xA means. So perhaps ∀x∀yLxy would mean something like, everyone loves everyone (including themselves). (Or “for all x and for all y, x loves y.”) So, ∀xA might mean, for all x, A holds. Or something like that. Then

v(∀xA)

means the truth-value that this statement would be assigned. Similarly for v(∃xA). Recall again our formulations:

v(∀xA) = 1 iff for all d ∈ D, v(A_x(k_d)) = 1 (otherwise it is 0)

v(∃xA) = 1 iff for some d ∈ D, v(A_x(k_d)) = 1 (otherwise it is 0)

The d ∈ D we already said was all the objects d in the domain D. And recall from above that k_d is a constant, which I think is like a name for the object, like “a” for “Alfred”. Now recall from section 12.2.4 that:

A_x(c) is the formula obtained by substituting c for each free occurrence of x in A.

(264)

In our current passage, we have something similar:

A_x(k_d)

So this would seem to mean that we have substituted k_d for all free occurrences of  x in A. Let us think of our Agler examples, using a simpler one:

I (Sx) = {⟨Alfred⟩, ⟨Bill⟩}

So we might write:

S_x(a)

This would be Sa, or Alfred is short (I guess technically it means, a is a member of the interpretation of S, and we can secondarily note that a’s interpretation is Alfred. Thus Alfred by extension is short.). And so on for b and c. But Priest’s formulation is:

for all d ∈ D, v(A_x(k_d))

In our Agler example, we have three d’s: Alfred, Bill, and Corinne. And so we have three k_d’s: a, b, and c. We need then to do the truth-valuation for each of them:

v(A_x(k_d))

So:

v(S_x(a)) = 1

v(S_x(b)) = 1

v(S_x(c)) = 0

Now, again we have:

v(∀xA) = 1 iff for all d ∈ D, v(A_x(k_d)) = 1 (otherwise it is 0)

v(∃xA) = 1 iff for some d ∈ D, v(A_x(k_d)) = 1 (otherwise it is 0)

Let us formulate it for our example:

v(∀xSx) = 1 iff for all d ∈ D, v(S_x(k_d)) = 1 (otherwise it is 0)

v(∃xSx) = 1 iff for some d ∈ D, v(S_x(k_d)) = 1 (otherwise it is 0)

Since not all our v(S_x(k_d)) equal 1, that means:

v(∀xSx) = 0

and since at least one v(S_x(k_d)) equals 1, then

v(∃xSx) = 1

. Let me quote it all now in full.]

Given an interpretation, truth values are assigned to all closed formulas. To state the truth conditions, we extend the language to ensure that every member of the domain has a name. For all d ∈ D, we add a | constant to the language, k_d, such that v(k_d) = d. The extended language is the language of ℑ, and written L(ℑ). The truth conditions for (closed) atomic sentences are:

v(Pa₁ ... a_n) = 1 iff  ⟨v(a₁),  ..., v(a_n)⟩ ∈ v(P) (otherwise it is 0)

The truth conditions for the connectives are as in the propositional case (1.3.2). For the quantifiers:

v(∀xA) = 1 iff for all d ∈ D, v(A_x(k_d)) = 1 (otherwise it is 0)

v(∃xA) = 1 iff for some d ∈ D, v(A_x(k_d)) = 1 (otherwise it is 0)

(264-265)

[contents]

 

 

12.3.3

[Validity in Classical First-Order Logic]

 

[Next we define validity. Recall from section 1.3.3:

Let Σ be any set of formulas (the premises); then A (the conclusion) is a semantic consequence of Σ (Σ ⊨ A) iff there is no interpretation that makes all the members of Σ true and A false, that is, every interpretation that makes all the members of Σ true makes A true. ‘Σ ⊭ A’ means that it is not the case that Σ ⊨ A.

(Introduction to Non-Classical Logic, p.5)

Here we say something similar, except we stipulate that this holds for closed formulas. I am not sure why that it is, but probably it can be shown why it is problematic for the validity to have a non-closed formula involved. I just myself do not know the reason. Let me quote:]

Validity is a relationship between premises and conclusions that are closed formulas, and is defined in terms of the preservation of truth in all interpretations, thus: Σ ⊨ A iff every interpretation that makes all the members of Σ true makes A true.

(265)

[contents]

 

 

12.3.4

[Some Equivalences]

 

Priest will note the following equivalences:

v(¬∃xA) = v(∀x¬A)

v(¬∀xA) = v(∃x¬A)

v(¬∃x(Px ∧ A)) = v(∀x(Px ⊃ ¬A)

v(¬∀x(Px ⊃ A) = v(∃x(Px∧¬A))

(Priest 265)

[Priest then gives the reasoning for the first two. The main ideas seem to be the following. Suppose that it is not the case that there is at least one thing for which the predicate holds. In our Agler example, suppose that there are no people who are short. That means that it is false that there is someone who is short. That furthermore means that it is false that everyone is short (this part of the reasoning I may not be following well. Given my ignorance, I have a confusion that I will mention now but correct later after I come to a proper understanding. I am speaking about the part reading “for all d in the domain of the interpretation, v(A_x(k_d)) = 0, i.e., v(¬A_x(k_d)) = 1. So, v(∀x¬A) = 1”. As far as I can tell, the first two formulations mean that the predicate does not hold for all the objects in the domain, but I would think that there could be two very different sorts of situations where that would be so. One is where it holds for none, as in our situation, and the other is where it holds for some, maybe even most. So I am not knowledgeable enough yet to interpret this. My best guess is that “for all d in the domain of the interpretation, v(A_x(k_d)) = 0” means not that it is false that the predicate holds for all members together ((such that it may still hold for at least one)) but rather that it means, given any substitution, the predicate will not hold for any of them taken individually. So in other words, “for all d in the domain of the interpretation” does not work the same way as a universal quantification sort of structure where you would say “for all x”. Maybe d is not a variable in that sense, and we are speaking of something more like a procedure where we make each d substitution and find that for every substitution, the predicate does not hold. But let me explicate my misunderstanding a little more so that I can better fix it. I do realize from Agler’s Introduction section 7.1 that ¬(∃x)P is equivalent to (∀x)¬P. And intuitively that makes sense. If there is not at least one object for which the predicate holds, then for all objects the negation of the predicate holds (If not one of the people are short, then all of them are not short). My confusion rather comes from trying to understand the senses of the formulations in the proof and then grasping how those senses relate logically. In Agler’s Introduction section 6.5, he lists

¬(∀x)Mx

as meaning “Not everything is moveable”, which from its sense I would think can mean at least one thing is not moveable. Another sense altogether would be that “nothing at all is moveable” (not even one), which Agler says is symbolized as:

(∀x)¬Mx

Also, Priest writes “v(A_x(k_d)) = 0”. In Agler’s Introduction section 6.4, he writes:

v(∀x)P = F iff for at least one ‘a’ not in ‘P’ and at least one a-variant interpretation ‘P(a/x) = F.’

(Agler 268)

So what I am not sure about is if we are to understand Priest’s “for all d in the domain of the interpretation, v(A_x(k_d)) = 0” to mean that the predicate does not hold for at least one object. Under that sense, it is possible that there is just one that it does not hold for, but it does hold for all the others, and thus it cannot be said that it holds for none. So to recap: in Priest’s explanation, where I get lost in following the senses is going from

for all d in the domain of the interpretation, v(A_x(k_d)) = 0, i.e., v(¬A_x(k_d)) = 1

(does it mean not all are short or none are short?)

to

v(∀x¬A) = 1

(No one is short)

It seems most likely that “for all d in the domain of the interpretation, v(A_x(k_d)) = 0, i.e., v(¬A_x(k_d)) = 1” means that given any item in the domain, the predicate will not hold for it (or perhaps, given any name for an item in the domain, it is not found in the interpretation of the predicate). And thus this has the sense of “the predicate holds for none (not even one).” The other three equivalences Priest leaves for exercises; but since I did not grasp the steps in the reasoning, I will try them later after I form a proper understanding]

Note that in any interpretation, we have the following:

v(¬∃xA) = v(∀x¬A)

v(¬∀xA) = v(∃x¬A)

v(¬∃x(Px ∧ A)) = v(∀x(Px ⊃ ¬A)

v(¬∀x(Px ⊃ A) = v(∃x(Px∧¬A))

For the first of these, suppose that v(¬∃xA) = 1. Then v(∃xA) = 0. So for all d in the domain of the interpretation, v(A_x(k_d)) = 0, i.e., v(¬A_x(k_d)) = 1. So, v(∀x¬A) = 1. Conversely, suppose that v(∀x¬A) = 1. Then for all d in the domain of the interpretation v(¬A_x(k_d)) = 1, that is, v(A_x(k_d)) = 0. Hence, v(∃xA) = 0, and v(¬∃xA) = 1. The other three cases are left as exercises.

(265)

[contents]

 

 

12.3.5

[Constants, Denotation, and Quantification]

 

[In this part, some of the ideas will be developed later in the text, so I will not try to completely explicate it. I will instead note some notions that I think will be very useful for understanding Priest’s text, “Multiple Denotation, Ambiguity, and the Strange Case of the Missing Amoeba.” We are to think that every object in the domain is assigned a “name” (using Agler’s term) in another set of constants, C. So using Agler’s example:

D = {Alfred, Bill, Corinne}

C = {a, b, c}

v(a) = Alfred

v(b) = Bill

v(c) = Corinne

What I note is that this seems to be a matter of denotation. In other words, denotation is the extensional reference of “names” in a set of constants to the objects in the domain. At any rate, we suppose that we have a set C of such constants, and for every object in the domain there is a constant in C  assigned to it. Another important thing to note in all of this is that, as we also mentioned above, the universal and existential quantifiers do not operate on members of the domain but rather on the constants/names that are assigned to the members of the domain. This notion seems to be reinforced in the footnote. If I gather properly (but probably I am misunderstanding something), it seems to be saying that we can formulate our set C of constants such that there are some objects in the domain for which there is no constant/name for them. As such, we could have one thing in the domain for which a predicate P should hold, but since there is no name for that thing, ∃xPx is not true. So suppose instead of the above model that we subtract the c constant to get:

D = {Alfred, Bill, Corinne}

C = {a, b}

v(a) = Alfred

v(b) = Bill

And suppose further that Corinne is the only female and that the predicate for being female is F. So

v(F) = {Corinne}

But suppose we also write:

∃xFx

Our rule in this section reads:

v(∃xA) = 1 iff for some c ∈ C, v(A_x(c)) = 1 (otherwise it is 0)

Now, our only constants are a and b.

v(F_x(a)) = 0

and

v(F_x(b)) = 0

That is all our constants. So

v(∃xFx) = 0

even though the predicate F holds for at least one member of our domain, Corinne. The problem it seems is that there is no way for the existentially quantified predicate formulation to express that state of affiars, because apparently the quantifications range over variables but not domain objects (I am not sure if they range over constants. It seems not, but I do not know). The last thing I would note is something I might be misunderstanding. My sense is that the constants/names k_d must completely name the objects in the domain, but the constants in C can be stipulated either to completely make those designations (as in the text body here) or not to (as in the footnote). That is my guess at the moment.]

Note also the following. If C is some set of constants such that every object in the domain has a name in C, then:

v(∀xA) = 1 iff for all c ∈ C, v(A_x(c)) = 1 (otherwise it is 0)

v(∃xA) = 1 iff for some c ∈ C, v(A_x(c)) = 1 (otherwise it is 0)

The proof is a simple corollary of the Denotation Lemma, and is given in 12.8.4. As we will see, counter-models read off from the open branch of a tableau are of this kind. Since appropriate versions of the Denotation Lemma can be proved for all the logics we will be concerned with in this | part of the book, the same is true for all of them. I will not keep mentioning the fact.²

(265-266)

2 If C is the set of the constants in the original language (that is, unaugmented by the special constants, k_d, of 12.3.2) and the truth conditions of quantifiers are given as in 12.3.5, we obtain a notion of quantification different from the more standard one employed here, and usually called substitutional quantification. It is a feature of substitutional quantification, that something in the domain can be in v(P), and yet ∃xPx is not true, just because the object in question has no name in the interpretation. In principle, all the logics we will meet in this book could be formulated with substitution quantification but we will not pursue this.

(266)

[contents]

 

 

 

 

 

From:

 

Priest, Graham. 2008 [2001]. An Introduction to Non-Classical Logic: From If to Is, 2nd edn. Cambridge: Cambridge University.

 

 

Also cited:

 

Agler, David. 2013. Symbolic Logic: Syntax, Semantics, and Proof. New York: Rowman & Littlefield.

 

Kaynak, Baran. 2011. “Classical Relations and Fuzzy Relations.” Slide presentation. Available at:

https://www.slideshare.net/barankaynak/classical-relations-and-fuzzy-relations

Slide 8:

https://www.slideshare.net/barankaynak/classical-relations-and-fuzzy-relations/8

 

 

 

 

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