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 NonClassical Logic: From If to Is
Part II:
Quantification and Identity
7
Manyvalued Logics
7.5
Manyvalued Logics and Conditionals
Brief summary:
(7.5.1) We will now examine the conditional operator in manyvalued logics. (7.5.2) We will assess whether or not some problematic inferences using conditionals are valid in K_{3}, Ł_{3}, LP_{3}, and RM_{3}, 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.)

 K_{3}  Ł_{3}  LP  RM_{3} 
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 manyvalued logics still validate many of the problematic inferences using the conditional. (7.5.4) We have the intuitions that in finitely manyvalued 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 K_{3} does (ii) not hold. (7.5.5) Given these two rules, suppose we have a manyvalued logic with one more formula than there are truthvalues; 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 counterexamples to this claim (and in these counterexamples, 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 threevalued 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 manyvalued 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.)
[Turning to the Conditional in ManyValued Logics]
[A Table of Problematic Conditional Inferences in ManyValued Logics]
[Evaluating the ManyValued Logics]
[Some Disjunction and Biconditional Rules for ManyValued Logics to Show Why Many Problematic Conditional Inferences are Inevitable in Them]
[The Disjunction of all Biconditionals as a Logical Truth]
[CounterExamples to This Claim]
Summary
[Turning to the Conditional in ManyValued Logics]
[We will now examine the conditional operator in manyvalued 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)
[A Table of Problematic Conditional Inferences in ManyValued Logics]
[We will assess whether or not some problematic inferences using conditionals are valid in K_{3}, Ł_{3}, LP_{3}, and RM_{3}, 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 K_{3}, Ł_{3}, LP_{3}, and RM_{3}. Recall that part of this evaluation involves the designated value, which is the truthpreserving 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.)


 K_{3}  Ł_{3}  LP  RM_{3} 
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 K_{3}, a generally good strategy is to start by assuming that the premises take the value 1 (the only designated value), and recall that, in K_{3}, if a conditional takes the value 1, then either its antecedent takes the value 0 or the consequent takes the value 1. For L_{3}, 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 RM_{3}, 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.
(125126)
[Evaluating the ManyValued Logics]
[Generally speaking, the manyvalued logics still validate many of the problematic inferences using the conditional.]
[If we look at the table, we can see that there are many checkmarks, which means that these problematic inferences are commonly valid in the manyvalued logics. If if we ignore conditionals with an enthymematic ceteris paribus clause (see section 5.2), all these manyvalued logics still “suffer from some of the same problems as the material conditional” (126). Priest then writes: “K_{3} and Ł_{3} 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) ...” (126127). 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 K_{3} and Ł_{3}. Let us consider the evaluation of ‘⊨ (p ∧ ¬p) ⊃ q’ in K_{3} and also ‘p ∧ ¬p ⊨ q’ in K_{3}, 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)
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 K_{3}, 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 K_{3}. 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 K_{3}, and it also holds in Ł_{3}. 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 counterexample 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 RM_{3}.]
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. K_{3} and Ł_{3} 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 RM_{3}.
(126127)
[Some Disjunction and Biconditional Rules for ManyValued Logics to Show Why Many Problematic Conditional Inferences are Inevitable in Them]
[We have the intuitions that in finitely manyvalued 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 K_{3} does (ii) not hold.]
[Priest will now explain why the conditional in finitely manyvalued 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 manyvalued 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 manyvalued logics we have seen, but for K_{3}, (ii) fails. ]
But there are quite general reasons as to why the conditional of any finitely manyvalued 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 K_{3}, for which (ii) fails.
(127)
[The Disjunction of all Biconditionals as a Logical Truth]
[Given these two rules, suppose we have a manyvalued logic with one more formula than there are truthvalues; 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 manyvalued 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 nvalued 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, p_{1}, p_{2}, . . . , 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 truthvalues for that system. So at least two formulas will need to have the same truthvalue. Now, given what (ii) says, that means for these two samevalued 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 manyvalued logic with one more formula than there are truthvalues, 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 nvalued logic that satisfies (i) and (ii), and consider n+1 propositional parameters, p_{1}, p_{2}, . . . , 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.
[CounterExamples to This Claim]
[But there are counterexamples to this claim (and in these counterexamples, 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 threevalued 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 manyvalued 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 manyvalued logical system with n truthvalues and n+1 propositional parameters, that the disjunction of all its biconditionals will necessarily be true. Priest then gives a counterexample. “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 manyvalued 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 threevalued 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 manyvalued 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 manyvalued 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.
From:
Priest, Graham. 2008 [2001]. An Introduction to NonClassical Logic: From If to Is, 2nd edn. Cambridge: Cambridge University.