Priest (7.5) An Introduction to Non-Classical Logic, ‘Many-valued Logics and Conditionals,’ summary

 

by Corry Shores

 

[Search Blog Here. Index-tags are found on the bottom of the left column.]

 

[Central Entry Directory]

[Logic and Semantics, entry directory]

[Graham Priest, entry directory]

[Priest, Introduction to Non-Classical Logic, entry directory]

 

[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

 

7

Many-valued Logics

 

 

7.5

Many-valued Logics and Conditionals

 

 

 

 

Brief summary:

(7.5.1) We will now examine the conditional operator in many-valued logics. (7.5.2) We will assess whether or not some problematic inferences using conditionals are valid in K₃, Ł₃, LP₃, and RM₃, by making a table. (In the table below, a ‘✓’ means the inference or formula is valid in the given system, and an ‘×’ means it is not valid.)

 

		K3	Ł3	LP	RM3
1	q ⊨ p ⊃ q	✔	✔	✔	×
2	¬p ⊨ p ⊃ q	✔	✔	✔	×
3	(p ∧ q) ⊃ r ⊨ (p ⊃ r) ∨ (q ⊃ r)	✔	✔	✔	✔
4	(p ⊃ q) ∧ (r ⊃ s) ⊨ (p ⊃ s) ∨ (r ⊃ q)	✔	✔	✔	✔
5	¬(p ⊃ q) ⊨ p	✔	✔	✔	✔
6	p ⊃ r ⊨ (p ∧ q) ⊃ r	✔	✔	✔	✔
7	p ⊃ q, q ⊃ r ⊨ p ⊃ r	✔	✔	×	✔
8	p ⊃ q ⊨ ¬q ⊃ ¬p	✔	✔	✔	✔
9	⊨ p ⊃ (q ∨ ¬q)	×	×	✔	×
10	⊨ (p ∧ ¬p) ⊃ q	×	×	✔	×

 

(7.5.3) Generally speaking, the many-valued logics still validate many of the problematic inferences using the conditional. (7.5.4) We have the intuitions that in finitely many-valued logics, the following two things should hold:

(i) if A (or B) is designated, so is A ∨ B

(ii) if A and B have the same value, A ≡ B must be designated (since A ≡ A is).

Only in K₃ does (ii) not hold. (7.5.5) Given these two rules, suppose we have a many-valued logic with one more formula than there are truth-values; that means a disjunction of all of its biconditionals will need to be logically valid, because at least one of them will have to have both biconditional terms with the same value and thus be designated. (7.5.6) But there are counter-examples to this claim (and in these counter-examples, the intuitive sense of the sentences does not allow for any true biconditional combinations of two different sentences, even though technically they should evaluate as true). For instance, “Consider n + 1 propositions such as ‘John has 1 hair on his head’, ‘John has 2 hairs on his head’, . . ., ‘John has n + 1 hairs on his head’. Any biconditional relating a pair of these would appear to be false. Hence, the disjunction of all such pairs would also appear to be false – certainly not logically true” (127). (So suppose we have a three-valued logic, and John has 1 hair on his head. That means “John has 2 hairs on his head if and only if John has 3 hairs on his head” is true (or at least true, or ‘designated’ whatever way), on account of both sides being false, even though the intuitive sense of the formulation would make the biconditional false (or at least senseless); for, John’s having x number of hairs on his head should not be conditional on his having x ± 1 hairs on his head. Thus, finitely many-valued logics will always be potentially vulnerable to the following problem: because the disjunction of all biconditionals should be true, then at least one must be true, meaning that in the case of propositions like “John has x number of hairs”, there must be at least one true one that reads “John has x number of hairs on his head only if John has x + 1 number of hairs on his head.” But that is senseless even though it would be evaluated as true.)

 

 

 

 

 

 

Contents

 

7.5.1

[Turning to the Conditional in Many-Valued Logics]

 

7.5.2

[A Table of Problematic Conditional Inferences in Many-Valued Logics]

 

7.5.3

[Evaluating the Many-Valued Logics]

 

7.5.4

[Some Disjunction and Biconditional Rules for Many-Valued Logics to Show Why Many Problematic Conditional Inferences are Inevitable in Them]

 

7.5.5

[The Disjunction of all Biconditionals as a Logical Truth]

 

7.5.6

[Counter-Examples to This Claim]

 

 

 

 

 

 

Summary

 

7.5.1

[Turning to the Conditional in Many-Valued Logics]

 

[We will now examine the conditional operator in many-valued logics.]

 

[(ditto)]

Further details of the properties of ∧, ∨ and ¬ in the logics we have just met will emerge in the next chapter. For the present, let us concentrate on the conditional.

(125)

[contents]

 

 

 

 

 

 

7.5.2

[A Table of Problematic Conditional Inferences in Many-Valued Logics]

 

[We will assess whether or not some problematic inferences using conditionals are valid in K₃, Ł₃, LP₃, and RM₃, by making a table.]

 

[The issue of problematic inferences using the conditional is something we have explored quite a bit in previous sections. See sections 1.6, 1.7, 1.8, 1.9, 1.10, 4.5, 4.6, 4.8, 4.9, and 5.2. Priest now summarizes many of these problematic inferences that use the conditional in a table where we can see whether or not they are are valid in K₃, Ł₃, LP₃, and RM₃. Recall that part of this evaluation involves the designated value, which is the truth-preserving value (like 1 in classical logic and i and 1 in LP) and which is not the same in all of these systems. In the table below, a ‘✓’ means the inference or formula is valid in the given system, and an ‘×’ means it is not valid.]

In past chapters, we have met a number of problematic inferences concerning conditionals. The following table summarises whether or not they hold in the various logics we have looked at. (A tick means yes; a cross means no.)

|

		K3	Ł3	LP	RM3
1	q ⊨ p ⊃ q	✔	✔	✔	×
2	¬p ⊨ p ⊃ q	✔	✔	✔	×
3	(p ∧ q) ⊃ r ⊨ (p ⊃ r) ∨ (q ⊃ r)	✔	✔	✔	✔
4	(p ⊃ q) ∧ (r ⊃ s) ⊨ (p ⊃ s) ∨ (r ⊃ q)	✔	✔	✔	✔
5	¬(p ⊃ q) ⊨ p	✔	✔	✔	✔
6	p ⊃ r ⊨ (p ∧ q) ⊃ r	✔	✔	✔	✔
7	p ⊃ q, q ⊃ r ⊨ p ⊃ r	✔	✔	×	✔
8	p ⊃ q ⊨ ¬q ⊃ ¬p	✔	✔	✔	✔
9	⊨ p ⊃ (q ∨ ¬q)	×	×	✔	×
10	⊨ (p ∧ ¬p) ⊃ q	×	×	✔	×

 

(1) and (2) we met in 1.7, and (3)–(5) we met in 1.9, all in connection with the material conditional. (6)–(8) we met in 5.2, in connection with conditional logics. (9) and (10) we met in 4.6, in connection with the strict conditional. The checking of the details is left as a (quite lengthy) exercise. For K₃, a generally good strategy is to start by assuming that the premises take the value 1 (the only designated value), and recall that, in K₃, if a conditional takes the value 1, then either its antecedent takes the value 0 or the consequent takes the value 1. For L₃, it is similar, except that a conditional with value 1 may also have antecedent and consequent with value i. For LP, a generally good strategy is to start by assuming that the conclusion takes the value 0 (the only undesignated value), and recall that, in LP, if a conditional takes the value 0, then the antecedent takes the value 1 and the consequent takes the value 0. For RM₃, it is similar, except that if a conditional has value 0, the antecedent and consequent may also take the values 1 and i, or i and 0, respectively. And recall that classical inputs (1 or 0) always give the classical outputs.

(125-126)

[contents]

 

 

 

 

 

 

7.5.3

[Evaluating the Many-Valued Logics]

 

[Generally speaking, the many-valued logics still validate many of the problematic inferences using the conditional.]

 

[If we look at the table, we can see that there are many check-marks, which means that these problematic inferences are commonly valid in the many-valued logics. If if we ignore conditionals with an enthymematic ceteris paribus clause (see section 5.2), all these many-valued logics still “suffer from some of the same problems as the material conditional” (126). Priest then writes: “K₃ and Ł₃ also suffer from some of the problems that the strict conditional does. In particular, even though (10) tells us that (p ∧ ¬p) ⊃ q is not valid in these logics, contradictions still entail everything, since p ∧ ¬p can never assume a designated value. By contrast, this is not | true of LP (as we saw in 7.4.4) ...” (126-127). I am not sure I get that, but maybe the ideas are the following. One of the problems of the strict conditional is the explosion of contradictions, where contradictions entail everything (see especially section 4.8). But, as we can see from row 10, this is not valid in K₃ and Ł₃. Let us consider the evaluation of ‘⊨ (p ∧ ¬p) ⊃ q’ in K₃ and also ‘p ∧ ¬p ⊨ q’ in K₃, because I am guessing that is the distinction here. Recall from section 7.3.2 that the evaluation for negation, conjunction, and the conditional are the following.

 

f¬
1	0
i	i
0	1

 

f∧	1	i	0
1	1	i	0
i	i	i	0
0	0	0	0

 

f⊃	1	i	0
1	1	i	0
i	1	i	i
0	1	1	1

(122)

 

And the designated value is 1. So we need to see if there is any instance in the evaluation where the value of the whole conditional is not 1. If there is such an instance, it is invalid, and if there is no such instance, it is valid.

 

p  q	(p	∧	¬p)	⊃	q
1  1	1	0	0	1	1
1  i	1	0	0	1	i
1  0	1	0	0	1	0
i  1	i	i	i	1	1
i  i	i	i	i	i	i
i  0	i	i	i	i	0
0  1	0	0	1	1	1
0  i	0	0	1	1	i
0  0	0	0	1	1	0

 

As we can see, ‘⊨ (p ∧ ¬p) ⊃ q’ is not valid in K₃, because when p is i and q is either i or 0, then the conditional is i. Now let us evaluate p ∧ ¬p ⊨ q’ in K₃. If there are any cases where the premises are 1 and the conclusion is not 1, then it is valid. But note that if there are no cases where the premises are 1 to begin with, then it will still be valid, although “vacuously valid” (see Priest’s Logic: A Very Short Introduction, section 2.)

 

p  q	(p	∧	¬p)	⊨	q
1  1	1	0	0		1
1  i	1	0	0		i
1  0	1	0	0		0
i  1	i	i	i		1
i  i	i	i	i		i
i  0	i	i	i		0
0  1	0	0	1		1
0  i	0	0	1		i
0  0	0	0	1		0

 

As we can see, there are no cases where the premises are 1, so it is vacuously valid and thus explosion holds in K₃, and it also holds in Ł₃. But, since in LP, i is a designated value, and since when p is i and q is 0 the premises are i but the conclusion 0, that means it is not valid in LP (see section 7.4.4). Yet Priest continues, “but this is so only because modus ponens is invalid, since (p ∧ ¬p) ⊃ q is valid, as (10) shows.” I am not sure why this formula shows that modus ponens is invalid. I will need to come back to this. But for now I will note that under the exact same evaluation there is the same row as the counter-example for modus ponens, as we saw in section 7.4.5:

 

p  q	p	p ⊃ q	⊨ q
1  1	1	1	1
1  i	1	i	i
1  0	1	0	0
i  1	i	1	1
i  i	i	i	i
i  0	i	i	0
0  1	0	1	1
0  i	0	1	i
0  0	0	1	0

 

In all, the system that has the fewest valid problematic inferences using the conditional is RM₃.]

As can be seen from the number of ticks, the conditionals do not fare very well. If one’s concern is with the ordinary conditional, and not with conditionals with an enthymematic ceteris paribus clause, then one may ignore lines (6)–(8). But all the logics suffer from some of the same problems as the material conditional. K₃ and Ł₃ also suffer from some of the problems that the strict conditional does. In particular, even though (10) tells us that (p ∧ ¬p) ⊃ q is not valid in these logics, contradictions still entail everything, since p ∧ ¬p can never assume a designated value. By contrast, this is not | true of LP (as we saw in 7.4.4), but this is so only because modus ponens is invalid, since (p ∧ ¬p) ⊃ q is valid, as (10) shows. (Modus ponens is valid for the other logics, as may easily be checked.) About the best of the bunch is RM₃.

(126-127)

[contents]

 

 

 

 

 

 

7.5.4

[Some Disjunction and Biconditional Rules for Many-Valued Logics to Show Why Many Problematic Conditional Inferences are Inevitable in Them]

 

[We have the intuitions that in finitely many-valued logics, the following two things should hold:

(i) if A (or B) is designated, so is A ∨ B

(ii) if A and B have the same value, A ≡ B must be designated (since A ≡ A is).

Only in K₃ does (ii) not hold.]

[Priest will now explain why the conditional in finitely many-valued logics will probably be problematic. In this section we make the first step, so the full reasoning will not here be given. But we need to note a few important things here. First consider how disjunction works. If we can affirm either just A on the one hand or alternatively just B on the other hand, then we should be able to affirm A ∨ B, because we know one of the disjuncts would be true, which is enough for the whole disjunction to be true. Thus:

(i) if A (or B) is designated, so is A ∨ B

(127)

The next observation is that A ≡ A should be a logical truth, even in a many-valued logic. Why? The reasoning is a little tricky, but it seems to be the following. Suppose A is valued at i. Let us make that concrete with an example. We have a moving object, and for A we have, “the moving object is at point 1 at time 1.” Suppose we think that should be both true and false. Regardless, would we not still think that the following should at least also be both true and false: “If the moving object is at point 1 at time 1, then the moving object is at point 1 at time 1”? So, we are claiming that it is reasonable to say that ‘if A then A’ has designated value whenever A itself has one. Now, since A is A, then that implies the inversion of the ‘if A then A’ also holds, and so A iff A should also hold as well. Now, in this case, A will have to have the same value as itself, so both sides of the biconditional will always have the same value, and regardless, the whole biconditional will have a designated value. The next idea seems to be that we suppose we have B, but we assign it the same value as A. And then we say that the same thing should hold for the biconditional here, namely, that:

(ii) if A and B have the same value, A ≡ B must be designated (since A ≡ A is).

I may have the reasoning wrong there, but this is my guess. Priest ends by saying that this holds in all the finitely many-valued logics we have seen, but for K₃, (ii) fails. ]

But there are quite general reasons as to why the conditional of any finitely many-valued logic is bound to be problematic. For a start, if disjunction is to behave in a natural way, the inference from A (or B) to A ∨ B must be valid. Hence, we must have:

(i) if A (or B) is designated, so is A ∨ B

Also, A ≡ A ought to be a logical truth. (Even if A is neither true nor false, for example, it would still seem to be the case that if A then A, and so, that A iff A.) Hence:

(ii) if A and B have the same value, A ≡ B must be designated (since A ≡ A is).

Note that both of these conditions hold for all the logics that we have looked at, with the exception of K₃, for which (ii) fails.

(127)

[contents]

 

 

 

 

 

 

7.5.5

[The Disjunction of all Biconditionals as a Logical Truth]

 

[Given these two rules, suppose we have a many-valued logic with one more formula than there are truth-values; that means a disjunction of all of its biconditionals will need to be logically valid, because at least one of them will have to have both biconditional terms with the same value and thus be designated.]

 

[The argumentation continues to get tricky. I may have the reasoning wrong, so see the quotation below. I am guessing we are doing the following. We are dealing with finitely many-valued logics. So that means regardless of the number of values, they will have a total number of values, that we can use the variable n to signify. So first we will consider any n-valued logic that satisfies

(i) if A (or B) is designated, so is A ∨ B

and

(ii) if A and B have the same value, A ≡ B must be designated (since A ≡ A is).

Now we will consider n+1 propositional parameters, p₁, p₂, . . . , p_n+1. Recall from section 1.2 that a propositional parameter is something like a formula which in our texts above we are writing as A, B, or C. The point is that we have one more formula than the total number of possible truth-values for that system. So at least two formulas will need to have the same truth-value. Now, given what (ii) says, that means for these two same-valued formulas, their biconditional must have a designated value. And, since we now have at least one designated biconditional, that would make it that the disjunction of all biconditionals will have to be designated, by rule (i). So in sum, suppose we have a many-valued logic with one more formula than there are truth-values, that means a disjunction of all of its biconditionals will need to be logically valid, because at least one of them will have to have both biconditional terms with the same value and thus be designated.]

Now, take any n-valued logic that satisfies (i) and (ii), and consider n+1 propositional parameters, p₁, p₂, . . . , p_n+1. Since there are only n truth values, in any interpretation, two of these must receive the same value. Hence, by (ii), for some j and k, p_j ≡ p_k must be designated. But then the disjunction of all biconditionals of this form must also be designated, by (i). Hence, this disjunction is logically valid.

[contents]

 

 

 

 

 

 

7.5.6

[Counter-Examples to This Claim]

 

[But there are counter-examples to this claim (and in these counter-examples, the intuitive sense of the sentences does not allow for any true biconditional combinations of two different sentences, even though technically they should evaluate as true). For instance, “Consider n + 1 propositions such as ‘John has 1 hair on his head’, ‘John has 2 hairs on his head’, . . ., ‘John has n + 1 hairs on his head’. Any biconditional relating a pair of these would appear to be false. Hence, the disjunction of all such pairs would also appear to be false – certainly not logically true” (127). (So suppose we have a three-valued logic, and John has 1 hair on his head. That means “John has 2 hairs on his head if and only if John has 3 hairs on his head” is true (or at least true, or ‘designated’ whatever way), on account of both sides being false, even though the intuitive sense of the formulation would make the biconditional false (or at least senseless); for, John’s having x number of hairs on his head should not be conditional on his having x ± 1 hairs on his head. Thus, finitely many-valued logics will always be potentially vulnerable to the following problem: because the disjunction of all biconditionals should be true, then at least one must be true, meaning that in the case of propositions like “John has x number of hairs”, there must be at least one true one that reads “John has x number of hairs on his head only if John has x + 1 number of hairs on his head.” But that is senseless even though it would be evaluated as true.)]

 

[So as we saw in section 7.5.5, if we accept the following two intuitive notions:

(i) if A (or B) is designated, so is A ∨ B

(ii) if A and B have the same value, A ≡ B must be designated (since A ≡ A is).

Then we should conclude that for any finitely many-valued logical system with n truth-values and n+1 propositional parameters, that the disjunction of all its biconditionals will necessarily be true. Priest then gives a counter-example. “Consider n + 1 propositions such as ‘John has 1 hair on his head’, ‘John has 2 hairs on his head’, . . ., ‘John has n + 1 hairs on his head’.” Priest says that any biconditional of these formulas would seem to be false, and thus the disjunction of all of them will not have any true one in it. I am not following this well, and I also do not see very how this shows us why the conditional for any finitely many-valued logic is bound to be problematic. Let me go through this as best I can. The idea originally was that in such a set of propositions, at least two would have to have the same value. So here, I would think that there would be two with the value false, and thus at least one biconditional that is true. Suppose we are using a three-valued logic with the values 0, 1, and i. And suppose John has 1 hair on his head. That means we have the following four propositions:

(1) John has 1 hair on his head.

(2) John has 2 hairs on his head.

(3) John has 3 hairs on his head.

(4) John has 4 hairs on his head.

Now, 3 and 4 are false, so their biconditional is true. So I do not see yet why “Any biconditional relating a pair of these would appear to be false.” But this is my failing. I am just not sure where I go wrong in the above explication. My best guess at the moment is that the problems come from the intuitive sense of these biconditional formulations: One of them will be, “John has 1 hair on his head if and only if John has 2 hairs on his head.” Now, as we know, this is senseless. John does in fact have 1 hair on his head, but this cannot be biconditional with him having 2 hairs on his head. By extension, even though technically “John has 3 hairs on his head if and only if John has 4 hairs on his head” is true, it is also for the same reason senseless. In other words, John’s not having some number of hairs on his head should not be biconditional on him having some other false number of hairs on his head. In other words, because every biconditional will biconditionally equate statements saying John has a different number of hairs on his head would seem on the level of its sense to necessarily always be false, even though technically they might evaluate as true whenever both sides of the biconditional are false. So on the level of sense, John’s having x number of hairs cannot be conditioned on him having x+1 number of hairs, and vice versa. That is my best guess at the moment. Still, even supposing this to be the case, I am not entirely sure why that would show there to be something problematic with conditionals in finitely many-valued logics. Is it because the biconditional is composed of conditionals, so if there is a paradox with the biconditionals there is a problem with conditionals in general? In other words, since the biconditionals imply that we will have  conditionals of the form “John has x number of hairs if John has x+1 number of hairs” where at least one biconditional combination of them will have to be true, even though we know that cannot be so in any case, given the intuitive sense of the constituent conditional sentences, that we will always have this problem with conditionals in finitely many-valued logics. Let me quote, as I am guessing very wildly here.]

But this seems entirely wrong. Consider n + 1 propositions such as ‘John has 1 hair on his head’, ‘John has 2 hairs on his head’, . . ., ‘John has n + 1 hairs on his head’. Any biconditional relating a pair of these would appear to be false. Hence, the disjunction of all such pairs would also appear to be false – certainly not logically true.

[contents]

 

 

 

 

 

 

 

 

From:

 

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

 

 

 

 

Priest (2.3) One, ‘Material Equivalence — Paraconsistent Style,’ summary

 

by Corry Shores

 

[Search Blog Here. Index-tags are found on the bottom of the left column.]

 

[Central Entry Directory]

[Logic and Semantics, entry directory]

[Graham Priest, entry directory]

[Priest, One, entry directory]

 

[The following is summary. You will find typos and other distracting mistakes, because I have not finished proofreading. Bracketed commentary is my own. Please consult the original text, as my summaries could be wrong.]

 

 

 

 

Summary of

 

Graham Priest

 

One: Being an Investigation into the Unity of Reality and of its Parts, including the Singular Object which is Nothingness

 

Part 1:

Unity

 

Ch.2

Identity and Gluons

 

2.3

Material Equivalence — Paraconsistent Style

 

 

 

 

Brief summary:

(2.3.1) Since gluons are inconsistent and since paraconsistent logics are the ones that allow for inconsistency, we need to use a paraconsistent logic for accounting for gluons. We furthermore need to understand how material equivalence works in paraconsistent logics. (2.3.2) In classical logic, formulas are either just in the true “zone” or just in the false “zone.” Whenever two formulas are in the same zone, then their material equivalence is true, and it is false otherwise. (2.3.3) In paraconsistent logics, formulas can be in both the true and the false zones. This means that a formula might be materially equivalent to a formula in the opposite zone, if at least one of the two are in both zones. But in that case, their material equivalence will be in both the true and the false zones as well. (2.3.4) Material equivalence in paraconsistent logic is reflexive and symmetric, but not transitive. (2.3.5) An “inference is valid (⊨) just if in every situation where all the premises are true (though they may be false as well), so is the conclusion” (19). Priest then provides a list of important valid inferences, along with an important invalid one (transitivity of material equivalence).

• ⊨ A ≡ A

• A ≡ B ⊨ B ≡ A

• A, B ⊨ A ≡ B

• ¬A, ¬B ⊨ A ≡ B

• A, ¬B ⊨ ¬(A ≡ B)

• B, ¬B ⊨ A ≡ B

• A ≡ B ⊨ ¬A ≡ ¬B

• ¬A ≡ ¬B ⊨ A ≡ B

• A ≡ B, B ≡ C ⊨ (A ≡ C) ∨ (B ∧ ¬B)

• A ≡ B, B ≡ C ⊭ A ≡ C

 

 

 

 

 

 

 

Contents

 

2.3.1

[Turning to Material Equivalence in Paraconsistent Logics]

 

2.3.2

[Material Equivalence in Classical Logic]

 

2.3.3

[Material Equivalence in Paraconsistent Logic]

 

2.3.4

[Material Equivalence as Non-Transitive Although Reflexive and Symmetric]

 

2.3.5

[Validity and Important Valid and Invalid Inferences]

 

 

 

 

 

 

 

Summary

 

2.3.1

[Turning to Material Equivalence in Paraconsistent Logics]

 

[Since gluons are inconsistent and since paraconsistent logics are the ones that allow for inconsistency, we need to use a paraconsistent logic for accounting for gluons. We furthermore need to understand how material equivalence works in paraconsistent logics.]

 

[Recall from from the brief summary of section 2.1 that:

(2.1.1) The gluon is the factor that binds parts into a unity. It has the contradictory properties of both being and not being an object. We now will see how gluons bind parts into unities, which involves breaking the Bradley regress. (2.1.2) The binding action of gluons involves non-transitive identity.

(Brief summary of section 2.1)

And from section 2.2:

(2.2.1) To explain how gluons bind parts into a unified whole, we need to break the Bradley regress, which prevents gluons from simply being object-parts. (2.2.2) We might name the parts of a unified object with letters, as for example, a, b, c, and d. The gluon, symbolized 中, is what binds all the other parts into the unified whole. If the gluon were distinct from the other parts (in the sense of not being identical to them), there would always be room for another gluon to intervene between the first gluon and the given parts, which leads to the Bradley regress. To avoid it, we say that the gluon is identical to each of the parts, thereby closing those “gaps”. (2.2.3) The gluon is non-transitively identical with each and every part. That means that although each part is identical to the gluon, they are not thereby identical to one another. And, parts can themselves be composed of parts by means of another internal gluon.

x

xxxxb

xxxx||

ax=x中x=xc

xxxx||

xxxxd

xxxx

(2.2.4) Gluonic unity involves non-transitive identity, meaning that a = 中 and 中 = c, but not thereby a = c.

(Brief summary of section 2.2)

So gluons are the binding factor that unify objects, but they themselves are contradictory objects, being that they both are and are not objects. Now recall from section P.5 that paraconsistent logics allow for contradictions in that they do not enable us to derive any arbitrary formula we want, so we will need such a logic for our account of gluons. We saw also in section P.5 that in paraconsistent logics, negation has different logical properties. In classical logic, whenever negation operates on a formula, the formula it operates on will be simply true or simply false, and the negation operator will flip that value. But in paraconsistent logics, formulas can take both true and false values. Thus the negation of such a formula is also both true and false. Priest says that now to further understand the paraconsistent logic of gluons, we need to understand the logical properties of material equivalence in paraconsistent logic.]

For a start, gluons, we know, are contradictory objects, and so the account needs to be given in a paraconsistent logic, where contradictions do not explode. In Section P.5, we saw how negation works in a paraconsistent context. What we need to know now is how material equivalence (having the same truth value) works in this context.

(18)

[contents]

 

 

 

 

 

 

2.3.2

[Material Equivalence in Classical Logic]

 

[In classical logic, formulas are either just in the true “zone” or just in the false “zone.” Whenever two formulas are in the same zone, then their material equivalence is true, and it is false otherwise.]

 

[In classical logic, we have two truth-value zones, one for truth and one for falsehoods. They are mutually exclusive (meaning that no formula can be found in both) and they are mutually exhaustive (meaning that a formula must be in one or the other, but not neither). This means we will find formulas A, B, C, D, etc. in one or the other zone. Now, if any two are found in the same zone, whether that be true or false, then their material equivalence will be found in the true zone. So if A and C are in the true zone, then so too is A ≡ C, and if B and D are in the false zone, then B ≡ D is then in the true zone, even though B and D are false. And since A and D are in different zones (A is in true and D is in false), then A ≡ D is in the false zone. ]

Classically, every situation partitions sentences of the language into two zones, the truths (ℑ) and the falsehoods (ℱ), the two zones being mutually exclusive and exhaustive:

 

         ℑ                      ℱ
   ____________            ____________
  /      A     \          /     B      \
x/       C      \        /      D       \
|       A≡C      |      |      A≡D       |
x\      B≡D     /        \     B≡C      /
  \____________/          \____________/

 

Sentences, A, B, C, . . .therefore find themselves in exactly one or other of the zones. If two sentences are both in the same zone, their material equivalence is in the ℑ zone; whilst if one is in one zone, and the other is in the other zone, their material equivalence is in the ℱ zone. (See the diagram above.)

(18)

[contents]

 

 

 

 

 

 

2.3.3

[Material Equivalence in Paraconsistent Logic]

 

[In paraconsistent logics, formulas can be in both the true and the false zones. This means that a formula might be materially equivalent to a formula in the opposite zone, if at least one of the two are in both zones. But in that case, their material equivalence will be in both the true and the false zones as well.]

 

[In paraconsistent logic, one same formula can be found in both the true and the false zones. That can be depicted by overlapping the zones, and formulas in the overlap are thought to be in both zones equally. Now, the material equivalence of two formulas will be in the true zone still if both are in the same zone, and it still will be in the false zone if both are in the false zone. But it gets a bit complicated. Suppose A is in the true zone, B is in the false zone, and C is in both zones. The material equivalence of A ≡ B is straightforward. It is in the false zone, because A and B are in different zones exclusively. But since C is in both zones, its material equivalences are more complicated. Since C is at least in the true zone, and since A is entirely in the true zone, then A ≡ C will at least be in the true zone. But since C is also at least in the false zone too, then A ≡ C will at least be in the false zone as well. That means A ≡ C is in both the true and the false zones, and thus it is placed in the depicted overlap (see the diagram below.) The same for B and C.]

In paraconsistent logic, everything is the same except that the ℑ and the ℱ zones may overlap.³ Thus we have the following picture:

 

           ℑ                     ℱ 
     _________________     ________________ 
    /                 \   /                 \ 
   /                   \ /                   \ 
  /         A          / \        B           \
x/                    / C \      A≡B           \
|                    | A≡C |                    |
x\                    \C≡B/                    / 
  \                    \ /                    / 
   \                   / \                   / 
    \_________________/   \_________________/

 

As before, the material equivalence of two sentences is in the ℑ zone if both are in the same zone (ℑ or ℱ ), and in the ℱ zone if they are in different zones, but now a sentence can be in both zones.

(18)

3. In some logics, they may underlap as well, so that there are things that are in neither ℑ nor ℱ; but in the logic we will be using, this is not the case.

(18)

 

[contents]

 

 

 

 

 

 

 

2.3.4

[Material Equivalence as Non-Transitive Although Reflexive and Symmetric]

 

[Material equivalence in paraconsistent logic is reflexive and symmetric, but not transitive.]

 

[In paraconsistent logic, material equivalence is reflexive and symmetric (so if A is in a particular zone, then A ≡ A is in the true zone (because A will always be in the same zone as itself), and if A ≡ B is in a particular zone, then so too is B ≡ A in that zone. (Suppose A ≡ B is in the false zone. That means A is in one zone and B is in another. So B ≡ A is also in the false zone. Suppose A ≡ B is in the true zone. That means both A and B are in the same zone. Thus So B ≡ A is in the true zone.)), but it is not transitive. Let us look again at Priest’s diagram to see why. Were material equivalence to be transitive, that would mean that if A ≡ C and C ≡ B are in the same zone, then so too should A ≡ B be in the true zone. In our counter-model, we suppose that A is just in true.

 

            ℑ                     ℱ
     _________________     ________________
    /                 \   /                 \
   /                   \ /                   \
  /         A          / \                    \
x/                    /   \                    \
|                    |     |                    |
x\                    \   /                    /
  \                    \ /                    /
   \                   / \                   /
    \_________________/   \_________________/

 

And B is just in false.

 

            ℑ                     ℱ
     _________________     ________________
    /                 \   /                 \
   /                   \ /                   \
  /         A          / \x       B           \
x/                    /   \                    \
|                    |     |                    |
x\                    \   /                    /
  \                    \ /                    /
   \                   / \                   /
    \_________________/   \_________________/

 

But C is in both true and false:

 

            ℑ                     ℱ
     _________________     ________________
    /                 \   /                 \
   /                   \ /                   \
  /         A          / \        B           \
x/                    / C \                    \
|                    |     |                    |
x\                    \   /                    /
  \                    \ /                    /
   \                   / \                   /
    \_________________/   \_________________/

Now recall from section 2.3.2 above that:

If two sentences are both in the same zone, their material equivalence is in the ℑ zone; whilst if one is in one zone, and the other is in the other zone, their material equivalence is in the ℱ zone. (See the diagram above.)

(p.18, section 2.3.2 )

Since A is in the true zone and C is at least in the true zone, then their material equivalence is at least in the true zone. But since C is also at least in the false zone, that means they at least are also in different zones, and so their material equivalence is at least also in the false zone. In other words, it will be in the overlap zone, as it is both true and false.

 

            ℑ                     ℱ
     _________________     ________________
    /                 \   /                 \
   /                   \ /                   \
  /         A          / \        B           \
x/                    / C \                    \
|                    | A≡C |                    |
x\                    \   /                    /
  \                    \ /                    /
   \                   / \                   /
    \_________________/   \_________________/

 

The same thing applies for C and B :

 

            ℑ                     ℱ
     _________________     ________________
    /                 \   /                 \
   /                   \ /                   \
  /         A          / \        B           \
x/                    / C \                    \
|                    | A≡C |                    |
x\                    \C≡B/                    /
  \                    \ /                    /
   \                   / \                   /
    \_________________/   \_________________/

 

Now, are A and B in the same zone? No. That means their material equivalence is false.

 

            ℑ                     ℱ
     _________________     ________________
    /                 \   /                 \
   /                   \ /                   \
  /         A          / \        B           \
x/                    / C \      A≡B           \
|                    | A≡C |                    |
x\                    \C≡B/                    /
  \                    \ /                    /
   \                   / \                   /
    \_________________/   \_________________/

 

But this is unlike the classical situation. We have A ≡ C as at least true and C ≡ B also as at least true, but A ≡ B will not even be at least true. So the material equivalence does not transfer through the shared term.]

A ≡ A will always be in the ℑ zone, since A is always in the same zone as itself. If A ≡ B is in the ℑ zone, then so is B ≡ A, since these are just ways of saying that A and B are in the same zone. So equivalence is reflexive and symmetric; but it is not transitive. A and C may be in the same zone, and C and B may be in the same zone, though A and B are not, because C is in the overlap. Hence, we may have A ≡ C and C ≡ B being in the ℑ zone, without A ≡ B being so (see the diagram above). Note also that detachment for ≡ may fail: we can have C and C ≡ B in the ℑ zone without B being in it (same diagram).

[contents]

 

 

 

 

 

 

2.3.5

[Validity and Important Valid and Invalid Inferences]

 

[An “inference is valid (⊨) just if in every situation where all the premises are true (though they may be false as well), so is the conclusion” (19). Priest then provides a list of important valid inferences, along with an important invalid one (transitivity of material equivalence).]

 

[Priest next reminds us that an inference is semantically validi, symbolized as ⊨,  when the all premises are at least true and so is the conclusion. (I am guessing that the conclusion need only be at least true and thus it can be both true and false under this criteria, given that i is thought to be the designated value in glut logics. See Priest’s Introduction to Non-Classical Logic, section 7.4. Priest lastly provides a list of important valid inferences, along with an important invalid one (transitivity of material equivalence).]

For the record, here are some paraconsistent facts concerning negation, equivalence, and validity. As we noted in Section P.5, an inference is valid (⊨) just if in every situation where all the premises are true (though they may be false as well), so is the conclusion. Bearing this in mind, and remembering that a formula is in the ℑ zone iff its negation is in the ℱ zone, it is easy to check the details.

• ⊨ A ≡ A

• A ≡ B ⊨ B ≡ A

• A, B ⊨ A ≡ B

• ¬A, ¬B ⊨ A ≡ B

• A, ¬B ⊨ ¬(A ≡ B)

• B, ¬B ⊨ A ≡ B

• A ≡ B ⊨ ¬A ≡ ¬B

• ¬A ≡ ¬B ⊨ A ≡ B

• A ≡ B, B ≡ C ⊨ (A ≡ C) ∨ (B ∧ ¬B)

• A ≡ B, B ≡ C ⊭ A ≡ C

Of course, in a paraconsistent context, the truth of ¬A is compatible with that of A. So even if it is the case that ¬(A ≡ B) is true, it can still be the case that A ≡ B is true as well.

[contents]

 

 

 

 

 

 

From:

 

Priest, Graham. 2014. One: Being an Investigation into the Unity of Reality and of its Parts, including the Singular Object which is Nothingness. Oxford: Oxford University.