1 Jun 2000

RECORD of pre-revised version of post from 10-Jan-2018 entitled: Priest (1.1) “Multiple Denotation, Ambiguity, and the Strange Case of the Missing Amoeba”, ‘1.1. Classical Semantics’, summary

 

[actual publication date of this record post: 10-Jan-2018]

 

[The following preserves the pre-revised version of the entry

Priest (1.1) “Multiple Denotation, Ambiguity, and the Strange Case of the Missing Amoeba”, ‘1.1. Classical Semantics’, summary

which was revised 10-Jan-2018]

 

 

 

by Corry Shores

 

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[Priest, “Multiple Denotation, Ambiguity, and the Strange Case of the Missing Amoeba”, entry directory]

 

[The following is meant to be summary, but is in large part my best guesswork of the text’s meaning, made on the basis of other texts by Priest and using examples from Agler. So please consult the original text rather than trust my notes below. Also, proofreading is incomplete, so you will find typos and other mistakes.]

 

 

 

 

Summary of

 

Graham Priest

 

“Multiple Denotation, Ambiguity, and the Strange Case of the Missing Amoeba”

 

1

Formal Considerations

 

1.1

Classical Semantics

 

 

Brief summary (quotation is provided as my interpretations can be wrong):

We outline a classical semantics for a first-order logic, which is formulated for a discussion on denotation. In this classical semantics, a truth-evaluating interpretation function v assigns to each sentence either 1 or 0. We calculate the values of formulas built up by means of logical operators by using the truth-functions for the operators: ⊖ (negation);  ⊗ (conjunction); and ⊕ (disjunction). An interpretation in our classical first-order language has the structure  ⟨D, d⟩. D is the domain of objects, and d is a function that assigns objects to constants, and it assigns n-tuples (of domain members) to n-place predicates:

d is a function such that for each constant, c, d(c) ∈ D, and for each n-place predicate, P, d(P) ⊆ Dn

(361)

Function s assigns objects in the domain to variables, and the vs function assigns a truth-value to formulas containing constants or variables by determining whether or not the d function (for constants) or the s function (for variables) yields objects that lie in the interpretation for the predicate:

Truth values are now assigned relative to an evaluation of the free variables, s (that is, a map from the variables into D). For atomic formulas:

vs(Pt1 ... tn) = 1 iff ⟨f(t1), ..., f(tn)⟩ ∈ d(P)

vs(Pt1 ... tn) = 0 iff ⟨f(t1), ..., f(tn)⟩ ∉ d(P)

where f(ti) is d(ti) if ti is a constant, and s(ti) if ti is a variable.

(361-362)

Quantified formulas are evaluated by determining the truth values for the variable-substitutions in the sentences. A universal quantification is 1 if the truth valuation for all the substitutions yields 1 in each case, and it is 0 otherwise. [An existential quantification is 1 if the truth valuation of one or more substitutions yields 1, and it is o otherwise]:

For the quantifiers, if we represent the natural generalisations of ⊕ and ⊗ by the same signs (so that ⊕X=1 iff 1 ∈ X, etc.), then the truth conditions of the quantifiers are as follows:

vs(∀vα) = ⊗{vs(v/a)(α); aD}

vs(∃vα) = ⊕{vs(v/a)(α); aD}

where s(v/a) is that evaluation of the variables that is the same as s, except that its value at v is a.

(362)

Validity is truth preservation in all interpretations and variable-evaluations.

 

 

 

 

Contents

 

1.1.1

[The Interpretation Function. The Truth-Functions for Logical Operators. Validity]

 

1.1.2

[First-Order Language. Assignment and Valuation Functions. Evaluation Rules. Validity.]

 

 

 

Summary

 

1

Formal Considerations

 

1.1

Classical Semantics

 

1.1.1

[The Interpretation Function. The Truth-Functions for Logical Operators. Validity]

 

[Our interpretation function v assigns to each sentence either 1 or 0. These two values are then calculated for complex formulas using the truth-functions for logical operators: ⊖ for f¬ (negation);  ⊗ for f (conjunction); and ⊕ for f (disjunction). Validity is truth preservation: an inference is valid only if when all its premises are true, so too is the conclusion.]

 

We first will be “formulating the semantics of the classical propositional calculus” (361). As is customary, we will “take an interpretation, v, to be a function that assigns to each sentence exactly one of the values 1 (true) or 0 (false). [See Priest, Introduction to Non-Classical Logic  section 1.3.1.] [We fundamentally must apply v to atomic formulas, because it is on their basis that we will calculate complex formulas:] “v may be thought of as defined on propositional parameters in the first instance; it is then extended to one defined on all formulas by recursive clauses” (361). [See Introduction to Non-Classical Logic  section 1.3.2.] [We will now deal with the notion of a truth-function for a logical operator. See Introduction to Non-Classical Logic  section 7.2.2. We know that negation in classical logic will “flip” the logical value assigned to a formula. We can think then of negation as operating on the truth-value. So we have formula α. We say it is true. That means the valuation function would assign it the value 1. So, v(α) = 1. Next we apply the negation operator to the formula get ¬α. With regard to its value, we see that the negation operates on the “1” value itself, to yield 0. So technically, it seems to me now at least, the truth-function for the operator (f¬) will operate on the v function, which is itself “operating” on the propositional parameter. Using the sort of notation from Introduction to Non-Classical Logic  section 7.2, we might write this like:

α

(v(α)) = 1

f¬(v(α)) = 0

In this article, I am assuming that instead of symbols like f¬, f, f, f, Priest is using another set of notation, which I am unfamiliar with:

⊖ for f¬ (negation)

⊗ for f (conjunction)

⊕ for f (disjunction)

But there is another difference between the notations in these two texts that makes me uncertain about their equivalence. Yet, looking just at negation, however, they appear similar. Here we have:

v(¬α) = ⊖ v(α)

(362)

Let’s first suppose this is equivalent to f¬ from the other text. So consider if we have:

¬α

And we want to know, what is the truth-value of ¬α?

v(¬α) = ?

We know that it will involve operating on the truth-value of the non-negated α:

v(¬α) = ...  v(α)

Suppose that the notation for the function that operates on the truth-value of α is not ¬. (Maybe it  is, but perhaps instead this mode of notation ((using ¬)) is used simply for the sentences, while the other mode of notation ((using ⊖ or f¬)) is involved not with the sentences themselves but rather with their truth-values. I say this, because otherwise, why not just use ¬ instead of ⊖?) So we would here write:

v(¬α) = ⊖ v(α)

This would mean that the truth-value of the negation of α is the negation-function’s operation on the truth-value function for α (the flipping of α’s 1 to a 0 value , or 0 to a 1, depending on α’s v value assignment). But now let me quote a bit and then I will note the discrepancy.]

Let ⊕, ⊗, ⊖  be the usual truth functions for disjunction, conjunction and negation, respectively. (Thus, 1 ⊕  1=1 , ⊖ 1=0, etc.) Then:

v(¬α) = ⊖ v(α)

v(α ∧ β) = v(α) ⊗ v(β)

v(α ∨ β) = v(α) ⊕ v(β)

(361)

[The conjunction and disjunction are not formulated with the exact same sort of structure as they are in Introduction to Non-Classical Logic  section 7.2.2. According to the structures there, the conjunction function I think would be notated like:

f(v(p), v(q))

Thus I would have expected with this alternate notation to see:

v(α ∧ β) = ⊗ (v(α),  v(β))

[In fact, later in section 1.1.2 we will see something like that.] So I could be wrong with the idea that the ⊖, ⊗, and ⊕ signs should be understood as representing functions operating on the values obtained by the v functions, like how f¬, f, f, f work in the other text. Or maybe they are equivalent,  but the way of structuring the formula is different. So perhaps:

v(α ∧ β) = v(α) ⊗ v(β)

is another way to write:

v(α ∧ β) = ⊗ (v(α),  v(β))

or

v(α ∧ β) = f(v(α),  v(β))

For now I will make that assumption, because I am not sure how else yet to construe it.] We define validity in terms of truth-preservation. [See the notion of semantic validity in Introduction to Non-Classical Logic 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)

]

Validity is defined in terms of truth-preservation. That is, Σ ⊨ α iff for every interpretation, v, if v(σ) = 1 for all σ ∈  Σ then v(α ) = 1.

(361)

[Let me quote this section in its entirety, if I may:]

In formulating the semantics of the classical propositional calculus, it is standard to take an interpretation, v, to be a function that assigns to each sentence exactly one of the values 1 (true) or 0 (false). v may be thought of as defined on propositional parameters in the first instance; it is then extended to one defined on all formulas by recursive clauses. Let ⊕, ⊗, ⊖  be the usual truth functions for disjunction, conjunction and negation, respectively. (Thus, 1 ⊕  1=1 , ⊖ 1=0, etc.) Then:

v(¬α) = ⊖ v(α)

v(α ∧ β) = v(α) ⊗ v(β)

v(α ∨ β) = v(α) ⊕ v(β)

Validity is defined in terms of truth-preservation. That is, Σ ⊨ α iff for every interpretation, v, if v(σ) = 1 for all σ ∈  Σ then v(α ) = 1.

(361)

[contents]

 

 

 

1.1.2

[First-Order Language. Assignment and Valuation Functions. Evaluation Rules. Validity.]

 

[We extend this semantics (for determining truth values and validity) to a first-order language (that is, one with quantified variables ranging over objects). Such an interpretation has the structure ⟨D, d⟩. D is the domain of objects, and d is a function that assigns objects to constants, and n-tuples (of domain members) to n-place predicates. We also have function s, which assigns to variables objects in the domain. The vs function assigns a truth-value to formulas containing constants or variables by determining whether or not the d function (for constants) or the s function (for variables) yields objects that lie in the interpretation for the predicate. Quantified formulas are evaluated by determining the truth values for the variable substitutions in the sentences. A universal quantification is assigned the value 1 if the truth valuation for all the substitutions yields 1 in each case, and it is 0 otherwise. (An existential quantification is 1 if the truth valuation of one or more substitutions yields 1, and it is o otherwise.) Validity is truth preservation in all interpretations and variable-evaluations.]

 

[As I understand, so far we have only defined the means by which we assign truth-values to sentences and also how we determine the validity of inferences (or of formulas taken as derivable simply by the semantics itself. See Introduction to Non-Classical Logic section 1.3.4:

A is a logical truth (tautology) (⊨ A) iff it is a semantic consequence of the empty set of premises (φA), that is, every interpretation makes A true.

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

) We will now “extend the semantics to a first order language. This is something Priest also does in his Introduction to Non-Classical Logic section 12.3. I will try to follow that text as much as possible, while also borrowing from David Agler’s Symbolic Logic: Syntax, Semantics, and Proof section 6.4.2 “Predicate Semantics.” In in Priest’s Introduction to Non-Classical Logic section 12.3.1, he writes that

An interpretation of the language is a pair, = ⟨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, ..., dn⟩: d1, ..., dn D}. By convention, ⟨d⟩ is just d, and so D1 is D.)

(Priest, Introduction to Non-Classical Logic, p.264)

In our current text, however, our notation is a little different. He writes:

To extend the semantics to a first order language, we take an interpretation to be a pair ⟨D, d⟩ where D is a non-empty domain, and d is a function such that for each constant, c, d(c) ∈ D, and for each n-place predicate, P, d(P) ⊆ Dn.

The main structure is the same. The interpretation contains the domain of objects and the function that assigns constants to items in the domain and sets of n-tuples for n-place predicates (more on that in a second). The only difference so far between

= ⟨D, v

and

[an interpretation is] a pair ⟨D, d

is that in Introduction to Non-Classical Logic, d is an item in domain D, the assignment function is called v, and the names assigned to terms are kd:

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

(Introduction to Non-Classical Logic section 12.3.2, pp. 364-365)

Or otherwise for constants assigned to items in the domain:

[...] C is some set of constants such that every object in the domain has a name in C [...]

(Introduction to Non-Classical Logic section 12.3.5, p.365)

But here, we do not specify a symbol for the items of the domain, and d is instead the assignment function for constants and predicates. Now let us look again at the line:

d is a function such that for each constant, c, d(c) ∈ D, and for each n-place predicate, P, d(P) ⊆ Dn.

(261)

The idea for predicates is explained in Introduction to Non-Classical Logic section 12.3.1. Again:

Dn is the set of all n-tuples of members of D, {⟨d1, ..., dn⟩: d1, ..., dn D}

(Priest, Introduction to Non-Classical Logic, p.264)

The idea here seems to be something like a Cartesian product (see Introduction to Non-Classical Logic section 01.1.10 and Suppes’ Introduction to Logic section 10.1) of the domain with iterations of itself. We also noted 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)

And we used an example from Agler’s Introduction to Logic section section 6.4.2. (Here I will mix Priest’s notation from Introduction to Non-Classical Logic with that of our current text. So let us here use d for the assignment function and not v.)

D = {Alfred, Bill, Corinne}

C = {a, b, c}

d(a) = Alfred

d(b) = Bill

d(c) = Corinne

D1 = {⟨Alfred⟩, ⟨Bill⟩, ⟨Corinne⟩}

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

[Sx: x is short]

d(Sx) = {⟨Alfred⟩, ⟨Bill⟩}

[Lxy: x loves y]

d(Lxy) = {⟨Bill, Corinne⟩, ⟨Corinne, Alfred⟩}

(Note, I am not sure actually if for Sx and Lxy we use function d or s. See the material below.) In our current article, Priest again writes:

for each n-place predicate, P, d(P) ⊆ Dn

In our Agler example, a 1-place predicate is Sx, and D1 is as follows:

d(Sx) = {⟨Alfred⟩, ⟨Bill⟩}

D1 = {⟨Alfred⟩, ⟨Bill⟩, ⟨Corinne⟩}

We ask, is d(Sx) a subset of D1, formulated differently, is:

d(Sx) ⊆ D1

? Yes, we can see that if we expand them:

d(Sx) ⊆ D1

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

We see something similar for Lxy:

d(Lxy) ⊆ D2

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

The next part of our current text is not so obvious to me. It reads:

Truth values are now assigned relative to an evaluation of the free variables, s (that is, a map from the variables into D). For atomic formulas: |

vs(Pt1 ... tn) = 1 iff ⟨f(t1), ..., f(tn)⟩ ∈ d(P)

vs(Pt1 ... tn) = 0 iff ⟨f(t1), ..., f(tn)⟩ ∉ d(P)

where f(ti) is d(ti) if ti is a constant, and s(ti) if ti is a variable.

(361-362)

Let me first quote from Introduction to Non-Classical Logic where the material is the most similar, as far as I can tell:

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 dD, we add a | constant to the language, kd, such that v(kd) = d. The extended language is the language of ℑ, and written L(ℑ). The truth conditions for (closed) atomic sentences are:

v(Pa1 ... an) = 1 iff  ⟨v(a1), ..., v(an)⟩ ∈ 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(Ax(kd)) = 1 (otherwise it is 0)

v(∃xA) = 0 iff for some d D, v(Ax(kd)) = 1 (otherwise it is 0)

(Introduction to Non-Classical Logic section 12.3.2, pp.264-265)

As we can see, the evaluation structures are very similar. But there are other parts that I have trouble with. I will go part-by-part:

Truth values are now assigned relative to an evaluation of the free variables, s (that is, a map from the variables into D).

(361)

The first concept I do not know is the notion of assigning values “relative to an evaluation of the free variables”. In the first place, in Introduction to Non-Classical Logic, he writes “truth values are assigned to all closed formulas” (see above quotation), and in section 12.2.4: “A formula with no free variables is said to be closed” (264). Probably there is no discrepancy here, but I am not sure what to make of the wording in our current text. The following is my guess. We right now are dealing with the evaluation of non-quantified formulas. That means the variables would be free in them. Lxy from our Agler example would be a formula with free variables. So perhaps the idea is the following. When we made the evaluation above:

d(Sx) ⊆ D1

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

v(Sx) = 1

We were assigning truth values “relative to an evaluation of the free variables”. I am guessing that the evaluation of the free variable x is

{Alfred, Bill, Corinne}

because writes that this evaluation is “a map from the variables into D”. And to make the evaluation “relative” to them seems to mean to do it on their basis, but I am not sure. And I am still not sure what the s is however. Let us look at it again:

Truth values are now assigned relative to an evaluation of the free variables, s (that is, a map from the variables into D). For atomic formulas: |

vs(Pt1 ... tn) = 1 iff ⟨f(t1), ..., f(tn)⟩ ∈ d(P)

vs(Pt1 ... tn) = 0 iff ⟨f(t1), ..., f(tn)⟩ ∉ d(P)

where f(ti) is d(ti) if ti is a constant, and s(ti) if ti is a variable.

(361-362)

The following is my guess right now. d is a function that operates on constants and assigns to constants items in the domain. s is a function that operates on variables and assigns to them (I suppose) also items in the domain. The only difference would seem to be that s probably assigns each variable all the members of the domain while d only assigns each constant one member of the domain. But my interpretation of s is problematic, even though I think I am sticking to the text. Again: “Truth values are now assigned relative to an evaluation of the free variables, s (that is, a map from the variables into D).” I have trouble here, but as far as I can tell, s is the function that maps the variables onto the items in the domain. And also as far as I can tell, s is the “evaluation of the free variables”. And truth values are then assigned using function vs, which is the valuation of sentences whose variables have been operated on by function s. But I am guessing. So let us try to fill this out using our example:

vs(Pt1 ... tn) = 1 iff ⟨f(t1), ..., f(tn)⟩ ∈ d(P)

vs(Sa) = 1 iff ⟨d(a)⟩ ∈ d(S)

vs(Sa) = 1 iff ⟨Alfred⟩ ∈ {⟨Alfred⟩, ⟨Bill⟩}

vs(Sa) = 1

But the f(ti) is said to be s(ti) if ti is a variable. How does that work? Here is where I can see that I do not really understand function s. Let us again consider our example.

vs(Pt1 ... tn) = 1 iff ⟨f(t1), ..., f(tn)⟩ ∈ d(P)

vs(Sx) = 1 iff ⟨s(x)⟩ ∈ d(S)

But what is s(x)? Is it any of the items in the domain (or any of the constants)? Is it all of them? It will only be true or false depending on whether its value is contained in the interpretation of the predicate. But the predicate’s interpretation is made by function d, which assigns to predicates n-tuples from the domain. If s assigns to variables members of the domain, then formulas with variables would always be true, I would think:

vs(Pt1 ... tn) = 1 iff ⟨f(t1), ..., f(tn)⟩ ∈ d(P)

vs(Sx) = 1 iff ⟨s(x)⟩ ∈ d(S)

vs(Sx) = 1 iff {Alfred, Bill, Corinne} ∈ {⟨Alfred⟩, ⟨Bill⟩, ⟨Corinne⟩}

vs(Sx) = 1

(I also do not know how to write the output for function s. Perhaps it should separated the names, as if doing each one independently.) vs(Sx) = 1 My best guess right now is that the s function as applied in these evaluation rules is something more like a procedure. So if we substitute x with a, it will be true. But if we substitute it with c it will be false. In other words, function s never yields a set of items. It can only yield one, depending on the substitution. (So later when we have the formulation vs(v/a)(α), the idea might be that no variable is mentioned here, rather, we are saying when some one particular substitution a in alpha is made, the truth-value would then be determined accordingly. And only by secondarily collecting all the substitutions would we get all the s values. Again, these are guesses.) At any rate, let us move on.

Truth values are assigned to other sentences by the same recursive clauses for the connectives as in the propositional case (relativized to s).

I imagine the idea here is similar to what is said in Introduction to Non-Classical Logic section 12.3.2. I am not sure, but I guess simply that for example

vsSa) = 0

because

vs(Sa) = 1

But again, I do not really understand how the truth values are determined for formulas with free variables like ¬Sx, unless, again, we simply mean that depending on the substitution, the truth-value can vary. Here is the next part:

For the quantifiers, if we represent the natural generalisations of ⊕ and ⊗ by the same signs (so that ⊕X=1 iff 1 ∈ X, etc.), then the truth conditions of the quantifiers are as follows:

vs(∀vα) = ⊗{vs(v/a)(α); aD}

vs(∃vα) = ⊕{vs(v/a)(α); aD}

where s(v/a) is that evaluation of the variables that is the same as s, except that its value at v is a.

I do not know what “natural generalisation” means, but later I will consider a meaning. For now, let us first look at what results from the natural generalizations of ⊕ and ⊗:

X=1 iff 1 ∈ X

I am not sure what the X is. So far we have three capital letters: D for domain, P for predicate, and Σ for a set of formulas (see Priest’s Introduction to Non-Classical Logic section 12.2 for similar notation conventions). X is tricky because it could be either the Roman letter ex or the Greek letter chi. It would not make sense for it to be a predicate in this case, I think. But it would seem possible that it is chi standing for a set of formulas. Also, note that in Priest’s Introduction to Non-Classical Logic section 0.1.2, a set is called X:

A set, X, is a collection of objects. If the set comprises the objects a1, ... , an, this may be written as {a1, ... , an}. If it is the set of objects satisfying some condition, A(x), then it may be written as {x :A(x)}. a X means that a is a member of the set X, that is, a is one of the objects in X. aX means that a is not a member of X.

(Priest Introduction to Non-Classical Logic, p.xxii)

Now, to get that sort of a terminological interpretation to work for what is said, it would seem that more precisely X is a set of truth values. So here we might read

X=1 iff 1 ∈ X

[and 0 otherwise]

to mean that the disjunction function operator assigns the value “1” only if “1” is a member of the set of truth-values that are assigned to particular formulas. So recall from section 1.1.1 above:

v(α ∧ β) = v(α) ⊗ v(β)

v(α ∨ β) = v(α) ⊕ v(β)

(361)

And consider the pairing:

v(α ∨ β) = v(α) ⊕ v(β)

X=1 iff 1 ∈ X

My best guess at the moment is that now we are using a structure that is similar to the one used in Introduction to Non-Classical Logic and discussed in the prior section above. So I am considering the possibility that the following two formulations are equivalent:

f(v(p), v(q))

X

The, we would understand ⊕X=1 iff 1 ∈ X as the truth-condition for making the evaluation of the formulas involved in the logical operation. So

X=1 iff 1 ∈ X

means that if either disjunct value is 1, then the disjunction function will assign the set of values 1. But there is a problem with this idea, because as far as I know (but for reasons I am unaware of) we can only have two terms in a disjunction. Yet, it would seem that the way we use this “natural generalization” of the truth functions would involve often having more than two terms or even having just one term in the set. For, he makes the formulations:

vs(∀vα) = ⊗{vs(v/a)(α); aD}

vs(∃vα) = ⊕{vs(v/a)(α); aD}

And I would think the idea is something like we make as many substitutions that are needed (being often more or less than two). I have not found a text that explains natural generalization yet, but I came across these passages in Richard Grandy’s Advanced Logic for Applications:

In the last chapter we presented several systems which are essentially equivalent to first order quantification theory with identity. In this chapter we will discuss a natural generalization of those theories which is slightly stronger than standard quantification theory. ... The reader should be careful to note that the operations in this system are generalizations of those in the previous system – they are defined on a wider domain.

(Grandy, Advanced Logic for Applications, p.151, boldface mine)

This is what I propose, then. We regard the natural generalization of ⊕ here to mean that it operates on sets of any size. And I would guess that ⊗ (normally conjunction) is defined as:

X=0 iff 0 ∈ X

[and 1 otherwise]

With that possibility in mind, let us look again at the formulations.

For the quantifiers, if we represent the natural generalisations of ⊕ and ⊗ by the same signs (so that ⊕X=1 iff 1 ∈ X, etc.), then the truth conditions of the quantifiers are as follows:

vs(∀vα) = ⊗{vs(v/a)(α); aD}

vs(∃vα) = ⊕{vs(v/a)(α); aD}

where s(v/a) is that evaluation of the variables that is the same as s, except that its value at v is a.

I am not at all sure what they mean. Let us look first at how he formulates the rules in Introduction to Non-Classical Logic section 12.3.2:

v(∀xA) = 1 iff for all d D, v(Ax(kd)) = 1 (otherwise it is 0)

v(∃xA) = 1 iff for some d D, v(Ax(kd)) = 1 (otherwise it is 0)

(264-265)

If

v(Ax(kd))

is roughly equivalent to

vs(v/a)(α)

then maybe the structures are roughly the same.

v(Ax(kd))

means the truth-value assignment for predicate A when it is substituted by one of the constants that in sum completely exhaust the domain of objects.

So maybe

vs(v/a)(α)

means the truth valuation for sentence alpha when its variable is substituted with a constant from the domain (with the rest of the formulation indicating that we do this for every object in the domain.) But there are still some confusing things. Are we to understand the v in

vs(∀vα)

as being like the x in

v(∀xA)

and the alpha to be like the capital A? (It seems not. For the conventions used in this text, I would think that the alpha is symbol for a sentence, and the capital A is a symbol for a predicate. And I would also guess that the alpha contains a variable, but it is not specified as with Ax.). Another confusing thing for me is that in

⊗{vs(v/a)(α); aD}

we typographically seem to have two very similar looking symbols, an alpha ‘α’ and an italicized Roman lowercase ‘a’.

image

Let me consider some options for interpreting this. The easiest one defines the symbols in the following way:

Interpretation 1:

v with no subscript is a variable, equivalent to x in the other formulation (and thus it would not follow the explicit definition already given). and so

vα is equivalent to ∀xA

alpha is a formula containing variable v.

italicized lower case a is a constant representing values of objects in the domain (and so under this view:)

aD is shorthand for constant a which is assigned by function d an object in D.

v/a means all the occurrences of variable v in formula alpha are replaced with a constant a.

vs is function assigning truth

So under this most likely erroneous interpretation, we would read

vs(∀vα) = ⊗{vs(v/a)(α); aD}

to mean, the value assigned to universal quantification of a formula equals the value obtained by applying the generalized conjunction function upon the set of values obtained by truth evaluating the formula under every substitution in the domain. So if we get,

{o, 1, 1}

Then it is 0. And if we get:

{1, 1, 1}

we get 1. There are problems with this interpretation, most of which I probably cannot detect, but one is that v is used ambiguously to mean either a truth evaluation function or a variable in the sentence. But, this interpretation would mostly follow the one given in Introduction to Non-Classical Logic, which I found easier to understand and which seems to have a very similar structure. But now, let us just apply the singular meaning for v that is given in the text, namely “v [... is] a function that assigns to each sentence exactly one of the values 1 (true) or 0 (false)” (361). So again, we are trying to understand:

vs(∀vα) = ⊗{vs(v/a)(α); aD}

vs(∃vα) = ⊕{vs(v/a)(α); aD}

where s(v/a) is that evaluation of the variables that is the same as s, except that its value at v is a.

(362)

Also, s is a function that assigns objects in D to the variables. So perhaps vs means the function that assigns truth values to formulas whose variables have been given assignments by means of s. With this in mind, then,

vα

would seem to mean: for all truth valuations of alpha. And

vs(∀vα)

would mean, the truth valuation for all truth valuations of alpha, whose variables have been given assignments by means of s. I find that interpretation to be odd at first, but if that is the meaning, then perhaps the idea is the following. Let us take our example, and let us add a new predicate, Ox, meaning x is a person.

D = {Alfred, Bill, Corinne}

C = {a, b, c}

d(a) = Alfred

d(b) = Bill

d(c) = Corinne

[Sx: x is short]

s(Sx) = {⟨Alfred⟩, ⟨Bill⟩}

[Ox: x is a person]

s(Ox) = {⟨Alfred⟩, ⟨Bill⟩, ⟨Corinne⟩}

And suppose we want to know:

vs(∀vO) = ???

I think now we need first to do the vs evaluations for all members of the domain.

Truth values are now assigned relative to an evaluation of the free variables, s (that is, a map from the variables into D). For atomic formulas: |

vs(Pt1 ... tn) = 1 iff ⟨f(t1), ..., f(tn)⟩ ∈ d(P)

vs(Pt1 ... tn) = 0 iff ⟨f(t1), ..., f(tn)⟩ ∉ d(P)

where f(ti) is d(ti) if ti is a constant, and s(ti) if ti is a variable.

(361-362)

Reformulated for our example:

a [Alfred]:

vs(Oa) = 1 iff ⟨f(a)⟩ ∈ d(O)

vs(Oa) = 1 iff ⟨Alfred⟩ ∈ {⟨Alfred⟩, ⟨Bill⟩, ⟨Corinne⟩}

vs(Oa) = 1

 

b [Bill]:

vs(Ob) = 1 iff ⟨f(b)⟩ ∈ d(O)

vs(Ob) = 1 iff ⟨Bill⟩ ∈ {⟨Alfred⟩, ⟨Bill⟩, ⟨Corinne⟩}

vs(Ob) = 1

 

c [Corinne]:

vs(Oc) = 1 iff ⟨f(c)⟩ ∈ d(O)

vs(Oc) = 1 iff ⟨Corinne⟩ ∈ {⟨Alfred⟩, ⟨Bill⟩, ⟨Corinne⟩}

vs(Oc) = 1

So the idea behind ∀vα would here be that normally we would say something like ∀x(Ax). But ∀vα is understood as having the same sense, because ∀vα means, for all truth value assignments for the variable substitutions in alpha. If there is one that is false, that is equivalent to there being one substitution of x in ∀x(Ax) not holding for the predicate. I am guessing. Now recall again the full formula we are trying to fill out:

vs(∀vα) = ⊗{vs(v/a)(α); aD}

Let us now use our Agler example to compile the second half of the formula.

{vs(v/a)(α); aD}

Recall that “s(v/a) is that evaluation of the variables that is the same as s, except that its value at v is a.” Now, in this particular interpretation, we are keeping strictly with the definition of v as the truth evaluating function. So what does it mean that the value at v is a? Since v under this interpretation is not a variable, we are not substituting a constant a for a variable v. And if they are substitutable, then they would probably be of the same class. Thus a would be a truth evaluating function. (Also, v is not the d function that assigns objects in the domain to constants.) This interpretation seems to be breaking down at this point. But let us keep reworking it. Perhaps  v/a is not a substitution of function v for some function a. It reads, “s(v/a) is that evaluation of the variables that is the same as s, except that its value at v is a.” So normally vs yields a 1 or 0. vs(v/a) will also yield a 1 or a 0. What is the “its” of “its value”? It seems to be “the evaluation of the variables”. That evaluation is an s evaluation. The value being  a “at v” for s might be the truth value assigned when s operates on the variable to get particular domain object a (or domain object assigned to constant a, maybe). So then

vs(v/a)(α); aD

would mean, the truth value of sentence alpha, which has some unspecified variable(s), when its variable has been substituted for an object in the domain (with the structure involving aD perhaps implying this is for every a in the domain). Were that the case, then for our Agler example, and using the above calculations, we get:

X=0 iff 0 ∈ X

[and 1 otherwise]

vs(∀vα) = ⊗{vs(v/a)(α); aD}

:

vs(∀vO) = ⊗{vs(v/a)(O); aD}

vs(∀vO) = ⊗{1, 1, 1}

vs(∀vO) = 1

Supposing we have all that right (and most likely not), the evaluations for the existential quantifier would work similarly, except we apply the rule (which is also probably wrong):

X=1 iff 1 ∈ X

[and 0 otherwise]

(The last concept is validity. Recall again from Introduction to Non-Classical Logic section 1.3.3 that:

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)

And now recall from section 12.3.3 of that text, which was more specifically about classical first-order logic, that:

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.

(Introduction to Non-Classical Logic, 265)

In our current text validity is defined:

Validity is defined in terms of truth preservation with respect to all interpretations and evaluations of the free variables.

(362)

I do not know exactly what is going on with the “all evaluations of the free variables”, but I guess the idea is that if truth is not preserved for one of the substitutions, then that inference should not be considered valid. I will now quote this section in full:]

To extend the semantics to a first order language, we take an interpretation to be a pair ⟨D, d⟩ where D is a non-empty domain, and d is a function such that for each constant, c, d(c) ∈ D, and for each n-place predicate, P, d(P) ⊆ Dn. (To keep things simple, we assume that there are no function symbols.) Truth values are now assigned relative to an evaluation of the free variables, s (that is, a map from the variables into D). For atomic formulas: |

vs(Pt1 ... tn) = 1 iff ⟨f(t1), ..., f(tn)⟩ ∈ d(P)

vs(Pt1 ... tn) = 0 iff ⟨f(t1), ..., f(tn)⟩ ∉ d(P)

where f(ti) is d(ti) if ti is a constant, and s(ti) if ti is a variable. Truth values are assigned to other sentences by the same recursive clauses for the connectives as in the propositional case (relativized to s). For the quantifiers, if we represent the natural generalisations of ⊕ and ⊗ by the same signs (so that ⊕X=1 iff 1 ∈ X, etc.), then the truth conditions of the quantifiers are as follows:

vs(∀vα) = ⊗{vs(v/a)(α); aD}

vs(∃vα) = ⊕{vs(v/a)(α); aD}

where s(v/a) is that evaluation of the variables that is the same as s, except that its value at v is a. Validity is defined in terms of truth preservation with respect to all interpretations and evaluations of the free variables.

(361-362. Note, I am not sure why f(t1) is not italicized:

image

)

[contents]

 

 

 

 

From:

 

Priest, Graham. 1995. “Multiple Denotation, Ambiguity, and the Strange Case of the Missing Amoeba.” Logique et Analyse 38: 361-373.

 

 

Also cited:

 

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

 

Grandy, Richard. 1979 [first published 1977]. Advanced Logic for Applications. Dordrecht: Reidel.

 

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

 

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

 

 

 

 

.

16 May 2000

1.1.11.2 Derivation and word class shifts, in L'Huillier, Advanced French Grammar


presentation of L'Huillier's work, by Corry Shores

[Site Topic Directory]

[Subsection headings are my own]

Monique L'Huillier

Advanced French Grammar

1. Framework

1.1 Parts of Speech

1.1.11 Synonyms, homonyms, derivation and word class shifts

1.1.11.2 Derivation and word class shifts

We may use proper or improper derivation to derive a part of speech from another one.

(i) Proper derivation

Ex,

noun: courage adjective: courageux

verb: conserver noun: conservateur ; adjective: conservateur

verb: aimer adjective: aimable

adjective: beau verb: embellir

adjective: jaune verb: jaunir

The suffixes of such words aid us in distinguishing the different parts of speech.

(ii) Improper derivation

Ex,

verb: effrayer adjective: effrayant (from present participle)

verb: passer noun: un passant (from present participle)

verb: percevoir adjective: perçu (from past participle)

verb: mourir noun: un mort (from past participle)

verb: devoir noun: un devoir (from infinitive)

adjective: rouge noun: le rouge

A word like this can be a different part of speech, depending on its syntactical context.

Ex,

Elle porte un pantalon bleu [adj].

She’s wearing bleu trousers.

Je n’aime pas le bleu.

I don’t like (the colour) blue.



L'Huillier, Monique. Advanced French Grammar. Cambridge: Cambridge University Press, 1999.
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