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}

f_c; 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 A∧B 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 also at least false, A∧B is also at least false. So A∧B 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:

⊭_K3  p ∨ ¬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 ∧ ¬p ⊨_K3 q

p ∧ ¬p ⊭_LP  q

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

p, p ⊃ q ⊨_K3 q

p, p ⊃ q ⊭_LP  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, {f_c; 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 f_c 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}

f_c; 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 A∧B 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 also at least false, A∧B is at least false. So A∧B 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: ⊭_K3  p ∨ ¬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 ∧ ¬p ⊭_LP  q ]

 

[Let us consider p ∧ ¬p ⊨ q 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
i  1	i	i	1
i  i	i	i	i
i  0	i	i	0
0  1	1	0	1
0  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 ∧ ¬p ⊨_K3 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
i  1	i	i	1
i  i	i	i	i
i  0	i	i	0
0  1	1	0	1
0  i	1	0	i
0  0	1	0	0

 

So:

p ∧ ¬p ⊨_K3 q

p ∧ ¬p ⊭_LP  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 ∧ ¬p ⊭_LP  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 ⊃ q ⊭_LP  q ]

 

[Now let us evaluate modus ponens, p, p ⊃ q ⊨ q. 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
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

 

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 ⊃ q ⊭_LP  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
i  1	i	1	1
i  i	i	i	i
i  0	i	0	0
0  1	0	1	1
0  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 A ⊃ B 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	¬A ∨ B	A ⊃ B
1  1	0	1	1
1  i	0	1	0
1  0	0	1	0
i  1	i	1	1
i  i	i	i	i
i  0	i	i	0
0  1	1	1	1
0  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  A ⊃ B (under its new valuation) are not equivalent. When A is i and B is 0, then under my calculations, ¬A ∨ B is i and A ⊃ B 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, RM_n, 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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