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 unfortunate mistakes, because I have not finished proofreading, and I also have not finished learning all the basics of these logics.]
Summary of
Graham Priest
An Introduction to Non-Classical Logic: From If to Is
Part II:
Quantification and Identity
21
Many-valued Logics
21.4
Some 3-valued Logics
Brief summary:
(21.4.1) We can evaluate quantification in 3-valued logics in the following way. We rank the values as: 0 < i < 1. And: “v(∀xA) = Glb({v(Ax(kd)) : d ∈ D}); and because this set is finite (it can have at most three members), and the values are linearly ordered, the greatest lower bound is the minimum (Min) of these values. Similarly, v(∃xA) is the maximum (Max) of the values in the set. Thus, ∀xA takes the value 1 if all instantiations with the constants kd take the value 1; it takes the value 0 if some instantiation takes the value 0; otherwise it takes the value i. Dually, ∃xA takes the value 1 if some instantiation with a constant kd takes the value 1; it takes the value 0 if all instantiations take the value 0; otherwise it takes the value i” (459). (21.4.2) As there will be no variance in the 3-valued logics we will deal with here among their D, V , and f values, we can simply regard their semantic structures as being “of the form ⟨D, v⟩, where D is the domain of quantification, and v assigns a denotation to each constant and predicate” (459). (21.4.3) As we will not in this chapter examine the tableau systems for these logics, we need to argue directly for the validity of inferences. (21.4.4) Priest next gives an example proof argument using reductio. (21.4.5) Priest next makes an example proof argument by contraposition. (21.4.6) Priest next shows how to show invalid inferences by constructing counter-models by trial and error.
[Evaluating Quantified Formulas in 3-Valued Logics]
[Abbreviating the Semantic Structure]
[The Lack of Tableaux]
[Reductio Example Proof]
[Contraposition Example Proof]
[Counter-Model Construction by Trial and Error. Example of Such.]
Summary
[Evaluating Quantified Formulas in 3-Valued Logics]
[We can evaluate quantification in 3-valued logics in the following way. We rank the values as: 0 < i < 1. And: “v(∀xA) = Glb({v(Ax(kd)) : d ∈ D}); and because this set is finite (it can have at most three members), and the values are linearly ordered, the greatest lower bound is the minimum (Min) of these values. Similarly, v(∃xA) is the maximum (Max) of the values in the set. Thus, ∀xA takes the value 1 if all instantiations with the constants kd take the value 1; it takes the value 0 if some instantiation takes the value 0; otherwise it takes the value i. Dually, ∃xA takes the value 1 if some instantiation with a constant kd takes the value 1; it takes the value 0 if all instantiations take the value 0; otherwise it takes the value i” (459).]
[In the previous section 21.3, we discussed how we might evaluate universally and particularly quantified formulas in a many-valued logic. In section 21.3.2 and section 21.3.3, Priest explained that:
“In classical logic, the universal quantifier acts essentially like a conjunction over all the members of the domain. So ∀xA is something like Ax(kd1) ∧ Ax(kd2) ∧ . . . , where d1, d2, . . . are all the members of the domain.” “Dually, the particular quantifier is something like a disjunction over all members of the domain: ∃xA is Ax(kd1) ∧ Ax(kd2) ∨ . . .”. We would expect that the universal and particular quantifiers behave the same way many-valued logics.
(Priest, parts in quotation his, p.458, section 21.3.2)
We can use greatest lower bound (Glb) of conjuncts (that is, the greatest truth-value that is less than or equal to the values assigned to the conjuncts) and least upper bound (Lub) of disjuncts (that is, the least truth-value greater than or equal to the value assigned to either disjunct) in order to evaluate universal and particular quantification, respectively: we define “f∀(X) as Glb(X), so that v(∀xA) is the greatest lower bound of {v(Ax(kd)): d ∈ D}” and we “define f∃(X) as Lub(X), and v(∃xA) is the least upper bound of {v(Ax(kd)): d ∈ D}.”
(Priest, parts in quotation his, p.458, section 21.3.3)
Priest will now look at how exactly we evaluate quantified formulas in 3-valued logics. In section 7.3 we examined the non-quantified, propositional forms of the 3-valued logics K3 and Ł3, and in section 7.4 we examined LP and RM3. Recall now from the previous section 21.3.3 that after ranging the truth-values in a sequential order,
We can use greatest lower bound (Glb) of conjuncts (that is, the greatest truth-value that is less than or equal to the values assigned to the conjuncts) and least upper bound (Lub) of disjuncts (that is, the least truth-value greater than or equal to the value assigned to either disjunct) in order to evaluate universal and particular quantification, respectively: we define “f∀(X) as Glb(X), so that v(∀xA) is the greatest lower bound of {v(Ax(kd)): d ∈ D}” and we “define f∃(X) as Lub(X), and v(∃xA) is the least upper bound of {v(Ax(kd)): d ∈ D}.”
(From the brief summary to section 21.3.3, with quotation Priest’s, p.458).
In the 3-valued logics we mentioned above (K3, Ł3, LP and RM3), recall that the third value is i. Priest says now that the ranking would place it in the middle: 0 < i < 1. Then, we can use the conjunction and disjunction evaluation procedures we saw in section 21.3.2 and section 21.3.3 to evaluate quantified formulas in three valued logics, namely:
v(∀xA) = Glb({v(Ax(kd)) : d ∈ D})
(or, the value of the universally quantified formula is the Min value of the substituted formulas)
and
v(∃xA) is the maximum (Max) of the values in the set.
(459)
]
Let us apply these observations to the 3-valued logics we met in 7.3 and 7.4 (K3, Ł3, LP and RM3). In these logics, the natural ordering of V is the following: 0 < i < 1. And it is not difficult to check the truth tables of 7.3 to see that conjunction and disjunction behave in the appropriate way with respect to this ordering. So v(∀xA) = Glb({v(Ax(kd)) : d ∈ D}); and because this set is finite (it can have at most three members), and the values are linearly ordered, the greatest lower bound is the minimum (Min) of these values. Similarly, v(∃xA) is the maximum (Max) of the values in the set. Thus, ∀xA takes the value 1 if all instantiations with the constants kd take the value 1; it takes the value 0 if some instantiation takes the value 0; otherwise it takes the value i. Dually, ∃xA takes the value 1 if some instantiation with a constant kd takes the value 1; it takes the value 0 if all instantiations take the value 0; otherwise it takes the value i.
(459)
[Abbreviating the Semantic Structure]
[As there will be no variance in the 3-valued logics we will deal with here among their D, V , and f values, we can simply regard their semantic structures as being “of the form ⟨D, v⟩, where D is the domain of quantification, and v assigns a denotation to each constant and predicate” (459). ]
[(ditto)]
In each of the logics at hand, D, V , and the various fs are fixed, so a semantic structure can simply be taken to be of the form ⟨D, v⟩, where D is the domain of quantification, and v assigns a denotation to each constant and predicate.
(459)
[The Lack of Tableaux]
[As we will not in this chapter examine the tableau systems for these logics, we need to argue directly for the validity of inferences.]
[(ditto)]
In this chapter we will not be concerned with tableau systems for these logics. Tableau systems for some of them will emerge in the next chapter. For the present, to establish that an inference is valid, one has to argue directly.
(459)
[Reductio Example Proof]
[Priest next gives an example proof argument using reductio.]
[(ditto). (See details below).]
So, for example, here is an argument to show that
∀x(Px ⊃ Qx) ⊨ ∃xPx ⊃ ∃xQx
holds in K3 and Ł3. (You will find it useful to have the truth tables of 7.3.2 and 7.3.8 in front of you.) Consider any interpretation, and suppose that the premise is designated, that is, has the value 1. Then, for every d ∈ D, Pkd ⊃ Qkd takes the value 1. Now, suppose, for reductio, that ∃xPx ⊃ ∃xQx is | not designated. There are four possible cases in K3:
∃xPx
∃xQx
1
i
1
0
i
0
i
i
In Ł3 only the first three are possible. In the first two, there is a d ∈ D such that Pkd takes the value 1. Since Pkd ⊃ Qkd takes the value 1, so does Qkd, and so, contrary to supposition, does ∃xQx. In the third (and second), for every d ∈ D, Qkd takes the value 0. Since Pkd ⊃ Qkd takes the value 1, Pkd takes the value 0. Hence, contrary to supposition, so does ∃xPx. In the last case (for K3 only), there must be some d ∈ D such that Pkd takes the value i. But since Pkd ⊃ Qkd takes the value 1, Qkd takes the value 1, as then does ∃xQx, contrary to supposition.
(460)
[Contraposition Example Proof]
[Priest next makes an example proof argument by contraposition.]
[(ditto). (See details below).]
Here is an argument to show that the same inference holds in LP and RM3. (Again, have the tables of 7.3.2 and 7.4.6 in front of you.) We argue by contraposition. Suppose that the conclusion is not designated. Then it takes the value 0. There are three cases for RM3:
∃xPx
∃xQx
1
0
1
i
i
0
and just the first for LP. In the first, there is a d ∈ D such that Pkd takes the value 1 and Qkd takes the value 0. In this case, Pkd ⊃ Qkd takes the value 0, as, then, does ∀x(Px ⊃ Qx). In the second case, there is a d ∈ D such that Pkd takes the value 1, and for every d ∈ D, Qkd takes the value of either i or 0. But then Pkd ⊃ Qkd takes the value 0, as does ∀x(Px ⊃ Qx). For the final case, for every d ∈ D, Qkd takes the value 0, and for every d ∈ D, Pkd takes the value 0 or i, with at least one taking that value. For this d, Pkd ⊃ Qkd takes the value 0, as does ∀x(Px ⊃ Qx).
(460)
[Counter-Model Construction by Trial and Error. Example of Such.]
[Priest next shows how to show invalid inferences by constructing counter-models by trial and error.]
[(ditto). (See details below).]
To show that an inference is invalid, we have to construct a counter-model by trial and error. Thus, we show that
∃xPx ∧ ∃xQx ⊭ ∃x(Px ∧ Qx)
| in the four logics in question as follows. We need an interpretation in which ∃xPx and ∃xQx are both designated. An easy way of obtaining this (in all the logics) is to suppose that there are d1, d2 ∈ D, such that Pkd1and Qkd2take the value 1. We also need ∃x(Px ∧ Qx) to be undesignated. An easy way to obtain that is just to ensure that whenever Pkd takes the value 1, Qkd takes the value 0, and vice versa. Thus, a simple counter-model is the following: D = {∂a, ∂b}, v(a) = ∂a, v(b) = ∂b, v(P) and v(Q) are the functions depicted as follows:
v(P)
v(Q)
∂a
1
0
∂b
0
1
It is easy to see that (in all the logics at hand) in this interpretation the premise takes the value 1, and the conclusion takes the value 0. Hence, the inference is invalid.
(461)
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From:
Priest, Graham. 2008 [2001]. An Introduction to Non-Classical Logic: From If to Is, 2nd edn. Cambridge: Cambridge University.
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