8 Aug 2017

Priest (7.4) An Introduction to Non-Classical Logic, ‘LP and RM3,’ 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

Part I. Propositional Logic

 

7. Many-valued Logics

 

7.4. LP and RM3

 

 

 

Brief summary:

Last time we examined K3, which was defined as:

V = {1, i, 0}

D = {1}

fc; c C = {f¬, f, f, f}

(note, A B we are defining as (A B) ∧ (B A) )

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

 

Here i means neither true nor false. LP has the same structure as K3, except in LP we have D = {1, i}. And in LP, the 1 is understood to mean true and true only, 0 to mean false and false only, and i to mean both true and false. The connective functions then follow our intuition regarding this alternate sense for the V values. Suppose for AB that A is 1 and B is i. Since B is at least true, then AB is at least true. And since B is also at least false, AB is also at least false. So AB is both true and false, or i, which is what the truth tables calculate it to be. There are two notable advantages of LP over K3. {1} In LP, unlike in K3, the law of excluded middle holds:

Kp ∨ ¬p

LP  p ∨ ¬p

And the principle of explosion, or the inference rule ex falso quodlibet, are not valid in LP, unlike in K3:

p ∧ ¬pK3 q

p ∧ ¬pLP  q

But there is one disadvantage of LP compared with K3. In LP, modus ponens is not valid:

p, p qK3 q

p, p qLP  q

This can be solved by changing the evaluation for the conditional connective in the following way, thereby creating RM3:

 

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

 

 

 

 

Summary

 

7.4.1

[The only difference between LP and K3 is that D = {1, i}.]

 

[Recall from section 7.2 the structure for 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. And recall from section 7.3 the three-valued logic K3.

V = {1, i, 0}

D = {1}

fc; c C = {f¬, f, f, f}

(note, A B we are defining as (A B) ∧ (B A) )

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

 

] Now we examine another three-valued logic called LP. “This is exactly the same as K3, except that D = {1, i}” (124).

 

 

7.4.2

[In LP, the connective functions assign the same values for the inputs. However, we understand 1 to mean true and true only, 0 to mean false and false only, and i to mean both true and false. The output calculations correspond to our intuitions regarding these meanings. For example, suppose for AB that A is 1 and B is i. Since B is at least true, then AB is at least true. And since B is also at least false, AB is at least false. So AB is both true and false, or i, which is what the truth tables calculate it to be.]

 

In LP, i is thought to mean both true and false. With that in mind, we would understand 1 as true and true only and 0 as false and false only. The truth tables remain the same, and they also make intuitive sense as well. Priest has us consider conjunction for example: A B. And suppose that A is 1 and B is i. Since B is at least true, then A B is at least true. And since B is at least false, that means A B is at least false. A B thus is both true and false, or i. As we see in the truth tables, that is how it evaluates.

 

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, section 7.3)

 

Consider also if A is 0 and B is i. So, B is at least false, which means A B is at least false, because both conjuncts are false. And, B is also at least true, but still, A B is false, because it only takes one conjunct to be false for the whole conjunction to be false.

 

 

7.4.3

[Unlike in K3, the law of excluded middle holds in LP: ⊨LP  p ∨ ¬p ]

 

[Recall from section 7.3.7 that the law of excluded middle did not hold in K3: ⊭Kp ∨ ¬p . In LP, the table will be the same:

 

p ¬p p ∨ ¬p
1 0 1
i i i
0 1 1
 
But in LP, i is a designated value, so the law of excluded middle holds in LP, because no matter the input value, the output will be 1 or i, which are both designated values.]
However, the change of designated values makes a crucial difference. For example, ⊨LP  p ∨ ¬p. (Whatever value p has, p ∨ ¬p takes either the value 1 or i. Thus it is always designated.) This fails in K3, as we saw in 7.3.7.
(125)
 

 

 

7.4.4

[The principle of explosion, or the inference rule Ex Falso Quodlibet, are not valid in LP, unlike in K3:

p ∧ ¬pLP  q ]

 

[Let us consider p ∧ ¬pq for K3 and LP.

 

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

 

In K3, the only designated value is 1. There is no line where the premise is 1 and the conclusion is not 1. (In fact, there is no instance where the premise is 1 anyway. In chapter 2 of Priest’s Logic: A Short Introduction, he says that such cases are vacuously valid.) So:

p ∧ ¬pK3 q

But, in LP, we have i and 1 as our designated values. So, if we look again, we see that when p is i and q is 0, then the premises have a designated value while the conclusion does not.

 

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

 

So:

p ∧ ¬pK3 q

p ∧ ¬pLP  q

In other words, the principle of explosion, or the inference rule Ex Falso Quodlibet, are not valid in LP.]

On the other hand, p ∧ ¬pLP  q. Counter-model: v(p) = i (making v(p ∧ ¬p) = i), v(q) = 0. But p ∧ ¬p can never take the value 1 and so be designated in K3. Thus, the inference is valid in K3.

(125)

 

 

7.4.5

[But in LP, modus ponens is not valid: p, p qLP  q ]

 

[Now let us evaluate modus ponens, p, p qq. We will do so for LP.

 

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

 

Were this K3, there would be no instances where the premises are 1 and the conclusion is not. However, in LP, we are looking for where the premises are either 1 or i, and the conclusion is 0. We see that when p is i and q is 0. Thus, modus ponens is not valid in LP.]

A notable feature of LP is that modus ponens is invalid: p, p qLP  q. (Assign p the value i, and q the value 0.)

(125)

 

 

7.4.6

[Modus ponens can be regained by changing the evaluation for the conditional connective.]

 

Priest notes that we can regain modus ponens by changing the truth function assignments for the conditional in the following way:

 

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

 

[Let us check that with a truth table:

 

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

 

As we can see, there is no line where the premises are all either 1 or i and the conclusion 0.]

One way to rectify this is to change the truth function for ⊃ to the following:

 

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

 

(As in 7.3.8, the meaning of AB in LP can still be expressed by ¬A B.) Now, if A and A B have designated values (1 or i), so does B, as a moment checking the truth table verifies.

(125)

[Let us check this.

 

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

 

As in the previous section, I must be making a mistake, because again for me ¬A B and  AB (under its new valuation) are not equivalent. When A is i and B is 0, then under my calculations, ¬A B is i and AB is 0. So I am doing something wrong. Please correct me. But we do see that if A and A B have designated values (1 or i), so does B.]

 

 

7.4.7

[The above change to the conditional connective’s evaluation produces RM3.]

 

“This change gives the logic often called RM3.” [Note, in section 7.12, Priest says that “RM3 is one family of n-valued logics, RMn, related to the logic RM (R Mingle), which we will meet in chapter 10” (139)]

 

 
 
 

 

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

 

 

 
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