7 Aug 2017

Priest (7.3) An Introduction to Non-Classical Logic, ‘The 3-valued Logics of Kleene and Łukasiewicz,’ summary

 

by Corry Shores

 

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

 

 

 

Summary of

 

Graham Priest

 

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

 

7. Many-valued Logics

 

7.3. The 3-valued Logics of Kleene and Łukasiewicz

 

 

 

Brief summary:

The structure of many-valued logics can be formulated as:

V, D, {fc; c C}⟩

V is the set of assignable truth values. D is the set of designated values, which are those that are preserved in valid inferences (like 1 for classical bivalent logic).  C is the set of connectives. c is some particular connective. And fc is the truth function corresponding to some connective, and it operates on the truth values of the formula in question. In classical logic: the assignable truth values of V are true and false, or 1 and 0; the designated values are just 1, and the connectives are: f¬, f, f, f. [A B we are defining as (A B) ∧ (B A)] And finally, the connective functions operate on truth values in accordance with certain rules (displayed often as the truth tables for connectives that we are familiar with. [This was covered in the previous section.] K3 (strong Kleene three-valued logic) and Ł3 are two sorts of three-valued logics. Both keep D as {1}, and both extend V to {1, i, 0}. 1 is true , 0 is false, and i is neither true nor false. We have the same connectives (excluding for simplicity the biconditional), and the connective functions associated with them are defined in the following way. K3 in particular has these assignments for the connective functions:

 

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 1 1
i 1 i i
0 1 i 0

 

f 1 i 0
1 1 i 0
i 1 i i
0 1 1 1

 

One problem with K3 is that every formula can obtain the undesignated value i by assigning all of its propositional parameters the value of i. (Simply look at the tables above where the input values are i. You will see in all cases the output is i too.) This means there are no logical truths in K3. One remedy for making the law of identity a logical truth is by changing the value assignment for the conditional such that when both antecedent and consequent are i, the whole conditional is 1.

 

f 1 i 0
1 1 i 0
i 1 1 i
0 1 1 1

 

This new system, where everything else is identical to K3except for the above alternate valuation for the conditional, is called Ł3 (Łukasiewicz’ three-valued logic).

 

 

 

Summary

 

7.3.1

[We will now examine examples of 3-valued logics in the language of classical propositional calculus by modifying the structure ⟨V, D, {fc; c C}⟩]

 

 

[Recall from section 7.2 the general structure of many-valued logics:

V, D, {fc; c C}⟩

V is the set of assignable truth values. D is the set of designated values, which are those that are preserved in valid inferences (like 1 for classical bivalent logic).  C is the set of connectives. c is some particular connective. And fc is the truth function corresponding to some connective, and it operates on the truth values of the formula in question. We will modify these components to generate different sorts of many-valued logics.] We will use the structure from section 7.2 to create examples of 3-valued logics. Our language will be classical propositional calculus [see section 1.2].

 

 

7.3.2

[We begin with a 3-valued logic. Its truth values are true (1), false (0), and neither true nor false (i). Its designated value is just 1. The negation connective function inverts the formula’s value, unless it is i, in which case it remains i. For conjunction, the only instance it is 1 is if both conjuncts are 1. And if at least one conjunct is 0, then the conjunction is 0. If one or both conjuncts are i, then the conjunction is i. A disjunction is 0 only when both disjuncts are 0, and it is 1 whenever at least one disjunct is 1. And it is i if both disjuncts are i or if one is i and the other is 0. A conditional is 0 only when the antecedent is 1 and the consequent are 0. It is 1 whenever the consequent is 1 or the antecedent 0. The conditional is i whenever: {a}  the antecedent is 1 and the consequent is i, {b} when the antecedent is i and the consequent is i, and {c} when the antecedent is i and the consequent is 0.]

 

We begin with a simple 3-valued logic. We will have three values, 1 for True, 2 for False, and 0 for neither True nor False.

V = {1, i, 0}

D = {1}

 

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 1 1
i 1 i i
0 1 i 0

 

f 1 i 0
1 1 i 0
i 1 i i
0 1 1 1

(122)

 
 
 

7.3.3

[These connective function evaluations are the same for classical evaluations whenever the input values are classical, that is, being either 1 or 0. They also can have classical outputs (1 or 0) in some cases where we have sufficient input determinations to make the classical evaluation. So in disjunction, if our inputs are 1 and i, we still say its output is 1, because all that is needed in a classical evaluation is for one disjunct to be true. However, in a conjunction, where one conjunct is 1 and the other is i, we do not have enough input determination to make the classical evaluation, because we would say it is 1 if both are 1 and it is 0 if at least one is 0. But here, since one is i, we cannot fulfill that classical criteria, and thus we give it the non-classical value i.]

 

Priest has us note that when the input values are either of the two classical values, 1 or 0, then the output matches the classical evaluation for those inputs. [Priest’s next point seems to be that we can understand how we get i output values as being the result of when we have insufficient input information to make a determinate classical evaluation. He has us consider the case of conjunction, AB. The classical evaluation would say that it is true only when both conjuncts are true, and it is false when at least one is false. Suppose A is false. That means the conjunction is false no matter the value of B. But suppose A is true and B is neither true nor false. Since B’s value is not one of the classical values, and since its classical value needs to be determined in order to make a classical valuation of the whole conjunction, we need to say that the conjunction is also neither valued.]

Note that if the inputs of any of these functions are classical (1 or 0), the output is exactly the same as in the classical case. We compute the other entries as follows. Take A B as an example. If A is false, then, whatever B is, this is (classically) sufficient to make A B false. In particular, if B is neither true nor false, A B is false. If A is true, on the other hand, and B is neither true nor false, there is insufficient information to compute the (classical) value of A B; hence, A B is neither true nor false. Similar reasoning justifies all the other entries.

(122)

 

 

7.3.4

[This logic is called strong Kleene 3-valued logic: K3.]

 

This logic is often called strong Kleene 3-valued logic, written K3. There is a weak logic that “is the same as K3, except that, for every truth function, if any input is i, so is the output. [That would seem to be similar to “Bochvar’s three-valued semantics” that we examined in Nolt’s Logics section 15.2.

15.2.a  15.2.b

(Nolt 408)

]

 

 

7.3.5

[We can then evaluate for validity by looking for instances where the premises have the designated value, 1, while the conclusion does not. We find for example that: pq K3 ¬q⊃¬p.]

 

[We can then do evaluations for validity, looking for sets of valuations for propositional parameters that make the premises have the designated value 1 and the conclusion not have 1. In the example given, there is no row where that is the case, so it is valid in K3.]

The following table verifies that p qK3 ¬ q ⊃ ¬p:

p  q pq ¬q ¬p
1  1 1 0 1 0
1  i i i i 0
1  0 0 1 0 0
1 1 0 1 i
i  i i i i i
0 i 1 i i
0  1 1 0 1 1
 i 1 i 1 1
0  0 1 1 1 1

In the last three columns, the first number is the value of ¬q; the last number is that of ¬p, and the central number (printed in bold) is the value of the whole formula. As can be seen, there is no interpretation where the premise is designated, that is, has the value 1, and the conclusion is not.

(123, here instead of bold, it is in red. Note also, in the part reading ⊨K3 , the ‘3’ is additionally subscripted:

image

)

 

 

7.3.6

[We can test for the validity of a formula by assigning the whole formula an undesignated value and then looking to see if there is any valuation for the propositional parameters that can produce that undesignated value. This is looking for a counter-model.]

 

[Recall from Agler’s Symbolic Logic section 3.6 his “forcing” truth table method for testing for validity. Priest will do something similar here, but he is working with a formula rather than an argument.

p ⊃ (qp)

(I am not sure yet how to understand a valid formula, since validity was defined using premises and conclusion, where here there is just a singular formula. In section 7.2.3 of this Priest text it was written: “an inference is semantically valid just if there is no interpretation that assigns all the premises a value in D, but assigns the conclusion a value not in D” ((121)). In his Logics, Nolt defines a valid formula in the following way:

DEFINITION

A valid formula is a formula true on all of its valuations.

Up until now we have applied the term ‘valid’ exclusively to arguments or argument forms, not to formulas. Since it is important not to confuse the formulas with argument forms, perhaps we ought to use some other term here to avoid confusion. Yet there is a substantial justification for applying the term ‘valid’ to both. For a valid formula is in effect a valid sequent with no premises. That is, we may think of a valid formula as a formula which may legitimately function as a conclusion without any premises at all. Since there is no valuation on which this “conclusion” is not true, there is no valuation on which the (nonexistent) premises are true and the “conclusion” is not true, and hence to counter-example. A valid formula may thus be regarded as a limiting case of a valid sequent.

 

Actually, in propositional logic we do have a term that substitutes nicely for ‘valid formula’. The term is ‘tautology’:

DEFINITION

A tautology is a formula whose truth table displays a column consisting entirely of ‘T’s under its main operator.

(Nolt 58)

Perhaps in our case, we are to understand a valid formula as one that is evaluated as having a designated value on all of its valuations. That would mean that if there is at least one valuation that gives it an undesignated value, then it is invalid. What we will do in the following is try to find a valuation that makes it have an undesignated value, either 0 or i.) We first suppose that the formula has an undesignated value. Suppose it is 0.

 

p

(q

p)

 

0

     

 

That could only be if the antecedent is true and consequent false.

 

p

(q

p)

1

0

 

o

 

 

So this makes the first p true. But p is also the consequent in the conditional making up the second half of the main conditional, so it should be 1 also.

 

p

(q

p)

1

0

 

o

1

 

Whenever the consequent is true, then the whole conditional is true. That means:

 

 

p

(q

p)

1

0

 

o

1?

1

 

And if it is true, then that makes the whole formula true, which contradicts our original assumption that it is false.

 

p

(q

p)

1

o

1?

 

o

1?

1

 

So if this formula is to have an undesignated value, it cannot be 0, so we will see if it can be i. Let us first recall our truth table for the conditional, and see under which cases the conditional is evaluated as i.

 

f 1 i 0
1 1 i 0
i 1 i i
0 1 1 1

 

We see that there are three cases where it is i. Let us now use the formula in question to examine these three possibilities.

 

p qp
1 i
i i
i 0

 

Now, suppose that v(p) = 1. Will that allow for v(qp) = i ? No, if we look at the truth table, whenever the consequent is 1, the whole conditional is 1.

 

f 1 i 0
1 1 i 0
i 1 i i
0 1 1 1
 
So we need to see if there is a valuation with v(p) = i that allows for v(qp) = i or 0. Let us look for our options when v(p) = i .

 

f 1 i 0
1 1 i 0
i 1 i i
0 1 1 1

 

As we can see, our only option is for (qp) = i . We can see this another way. Focus just on qp. Since v(p) = i, that means v(qp) can only be i or 1, and not 0.

 

f 1 i 0
1 1 i 0
i 1 i i
0 1 1 1

 

We already said it cannot be 1, because then it has a designated value, and we need it to have an undesignated value. So again, our only option is for v(qp) = i . The question now is, what value of q would make v(qp) = i ? We look one last time at the table, and see one possibility is v(q) = i.

 

f 1 i 0
1 1 i 0
i 1 i i
0 1 1 1

(122)

(I wonder also if this valuation works: v(p) = i and v(q) = 1. This would be:

v(p) = i

v(q) = 1

 

v(p ⊃ (qp)) = f(v(p), v(qp)) = f(v(p), f(v(q), v(p)) =

f(i, f(1, i)) = f(i, i) = i.

Or if using a truth table:

 

p  q p qp
1  1 1 1
1  i 1 1
1  0 1 1
1 i i
i  i i i
0 1 1
0  1 1 0
 i 1 i
0  0 1 1

)]

In checking for validity, it may well be easier to work backwards. Consider the formula p ⊃ (q p). Suppose that this is undesignated. Then it has either the value 0 or the value i. If it has the value 0, then p has the value 1 and q p has the value 0. But if p has the value 1, so does q p. This situation is therefore impossible. If it has the value i, there are three possibilities:

 

p qp
1 i
i i
i 0

 

The first case is not possible, since if p has the value 1, so does q p. Nor is the last case, since if p has the value i, q p has value either i or 1. But the | second case is possible, namely when both p and q have the value i. Thus, ν(p) = ν(q) = i is a counter-model to p ⊃ (q p), as a truth-table check confirms. So ⊭K3 p ⊃ (q p).

(123-124)

 

 

7.3.7

[In K3, the law of excluded middle does not hold: ⊭Kp ∨ ¬p ]

 

One important thing to note about K3 is that the law of excluded middle does not hold. This is written as:

Kp ∨ ¬p

[If we look at the truth-table, we see that when p is i, then the disjunction is i.

 

p ¬p p ∨ ¬p
1 0 1
i i i
0 1 1
 

But i is not a designated value. So not all interpretations give the formula a designated value, and so the law of excluded middle does not hold in K3.] For, the counter-model is v(p) = i. [In section 6.2.8, Priest noted that the law of excluded middle does not hold in intuitionistic logic.] But, Priest explains, K3 is distinct from intuitionistic logic, which “is not the same as any finitely many-valued logic” (124). We see this later in section 7.10.8.

 

 

7.3.8

[In K3, every formula can obtain the undesignated value i by assigning all of its propositional parameters the value of i. This means there are no logical truths in K3. One remedy for making the law of identity a logical truth is by changing the value assignment for the conditional such that when both antecedent and consequent are i, the whole conditional is 1.]

 

[So a valid formula is one that under no evaluation takes an undesignated value. Now, i is an undesignated value in this three-valued logic we are examining. And, look again at the truth tables. You will see that if we assign all of a formula’s propositional parameters i, the whole formula will be i.

 

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 1 1
i 1 i i
0 1 i 0

 

f 1 i 0
1 1 i 0
i 1 i i
0 1 1 1

 

That means, there will always be a way to make a formula have an undesignated value. Consequently, under this logic, there are no logical truths, because every formula can take a value assignment that gives it the undesignated value i. This is problematic even for such obviously valid formulations like the law of identity (or principle of identity, as it is called in Russell and Whitehead’s Principia, p.99): p p. To solve this problem, we can change the truth assignments for the conditional so that when v(p)=i, then v(pp)=1.]

In fact, K3 has no logical truths at all (7.14, problem 3)! In particular, the law of identity is not valid: ⊭K3 pp. (Simply give p the value i.) This may be changed by modifying the middle entry of the truth function for ⊃, so that fbecomes:

 

f 1 i 0
1 1 i 0
i 1 1 i
0 1 1 1

 

(The meaning of A B in K3 can still be expressed by ¬AB, since this has the same truth table, as may be checked.) Now, AA always takes the value 1.
(124)

[Note: I am confused by the part that reads: “The meaning of A B in K3 can still be expressed by ¬AB, since this has the same truth table, as may be checked.” Probably my table is wrong, but by using the new conditional valuation rules, I did not get an identical evaluation when v(A)=i and v(B)=i. Under that assignment, v(AB)=1 but vAB)=i.

 

A  B ¬A ¬AB AB
1  1 0 1 1
1  i 0 1 i
1  0 0 1 0
1 i 1 1
i  i i i 1
0 i i i
0  1 1 1 1
 i 1 1 1
0  0 1 1 1

 

I welcome corrections to my mistake/misinterpretation.]

 

 

7.3.9

[This change makes the system Łukasiewicz’ Ł3.]

 

“The logic resulting from this change is one originally given by Łukasiewicz, and is often called Ł3” (124).
 
 
 
 

 

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

 

Also cited:

Nolt, John. Logics. 1997. Belmont, CA: Wadsworth.

 

Russell, Bertrand, and Alfred North Whitehead. 1963. Principia Mathematica, vol. 1. Cambridge: Cambridge University.

 
.

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