## 25 Jan 2018

### Priest (8.5) Introduction to Non-Classical Logic, ‘The Routley Star’, summary

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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 Non-Classical Logic: From If to Is

Part I

Propositional Logic

8

First Degree Entailment

8.5

The Routley Star

Brief summary:

FDE can be given an equivalent, two-valued, possible world semantics in which the negation is an intensional operator, meaning that it is defined by means of related possible worlds. In this case, we use Routley’s star worlds. We have a star function, *, which maps a world to is “star” or “reverse” world (and back again to the first one, if applied yet another time. That bringing back to the first is what defines the function). So for any world w, the * function gives us its companion star world w. We evaluate conjunctions and disjunctions based on values in the given world. But what is notable in Routley Star semantics is that a negated formula in a world w is valued true in that world not on the basis of it unnegated form being false in that world w, but rather on the basis of its unnegated form being false in the star world w (so a negated formula is 1 in w if its unnegated form is 0 in w*.) Validity is defined as truth preservation for all worlds and interpretations. For constructing tableaux in Routley Star logic, we designate not just the truth-value for a formula but also the world in which that formula has that value. This is especially important for negation, where the derived formulas are found in the companion star world. Here is how the tableau rules work for Routley Star logic [quoting Priest, except for the rules tables, where I add my own names and abbreviations, following David Agler]:

Nodes are now of the form A,+x or A,−x, where x is either i or i#, i being a natural number. (In fact, i will always be 0, but we set things up in a slightly more general way for reasons to do with later chapters.) Intuitively, i# represents the star world of i. Closure occurs if we have a pair of the form A,+x and A,−x. The initial list comprises a node B,+0 for every premise, B, and A,−0, where A is the conclusion. The tableau rules are as follows, where x is either i or i#, and whichever of these it is, is the other.

 Conjunction Development, True (∧D,+x) A ∧ B,+x ↓ A,+x B,+x

 Conjunction Development, False (∧D,−x) A ∧ B,−x ↙         ↘ A,−x              B,−x

 Disjunction Development, True (∨D,+x) A ∨ B,+x ↙               ↘ A,+x      B,+x

 Disjunction Development, False (∨D,−x) A ∨ B,-x ↓ A,-x B,-x

 Negation Development, True (¬D,+x) ¬A,+x ↓ A,-x̄

 Negation Development, False (¬D,−x) ¬A,-x ↓ A,+x̄

(152, Note, names and abbreviations are my own and are not in the text.)

We test for validity first [as noted above] by setting every premise to true in the non-star world and the conclusion to false in the non-star world. We then apply all the rules possible, and if all the branches are closed [recall from above that closure occurs if we have a pair of the form A,+x and A,−x] then it is valid, and invalid otherwise [so it is invalid if any branches are open]. We then can make counter-models using completed open branches. On the basis of the world indicators in the branches, we assign to the formulas the values indicated by the true (+) and false (−) signs for the respective world (that is to say, “ if p,+x occurs on the branch, vwx(p) = 1, and if p,−x occurs on the branch, vwx(p) = 0.”) The equivalence between Routley Star semantics and FDE becomes apparent when we make the following translation: vw(p) = 1 iff 1; vw(p) = 0 iff 0.

Contents

8.5.1

[FDE as a Two-Valued, Possible World Semantics with Negation Being an Intensional Operator]

8.5.2

[Routley’s Star World w]

8.5.3

[Negation Defined. Validity.]

8.5.4

[Tableaux Rules for Routley Star Logic]

8.5.5

[Examples of Star Logic Tableaux]

8.5.6

[Star Counter-Models]

8.5.7

[Soundness and Completeness Proof  Postponement]

8.5.8

[The Equivalence of Routley Star semantics and FDE ]

Summary

8.5.1

[FDE as a Two-Valued, Possible World Semantics with Negation Being an Intensional Operator]

[We will now give an FDE two-valued, possible world semantics in which the negation is an intensional operator and thus is defined by means of related possible worlds.]

[In section 8.2 we gave the relational semantics for First Degree Entailment (FDE), and in section 8.4 we gave a many-valued semantics. We now will consider a third sort of semantics. It has just two values, but it also uses possible worlds where negation is an intensional operator, meaning that we define its truth conditions by referencing other worlds.]

We now have two equivalent semantics for FDE, a relational semantics and a many-valued semantics.5 For reasons to do with later chapters, we should have a third. This is a two-valued possible-world semantics, which treats negation as an intensional operator; that is, as an operator whose truth conditions require reference to worlds other than the world at which truth is being evaluated.

(151)

5. At least, they are equivalent given the standard set-theoretic reasoning employed in the reformulation. Such reasoning employs classical logic, however, and in a set theory based on a paraconsistent logic it may fail. See Priest (1993).

(151)

[contents]

8.5.2

[Routley’s Star World w]

[We can use star worlds to make such a two-valued FDE semantics. To define negation, we posit for any world w its companion star world w. A negated formula in a world w is valued true in that world not on the basis of it unnegated form being false in that world w, but rather on the basis of its unnegated form being false in the star world w.]

[Let us take a moment to quickly recall some things about possible world semantics. This comes from the brief summary of section 2.3:

An interpretation in our modal semantics takes the form ⟨W, R, v⟩, with W as the set of worlds, R as the accessibility relation, and v as the valuation function. ‘uRv’ can be understood as either, “world v is accessible from u,”  “in relation to u, situation v is possible,” or “world u access world v.” Negation, conjunction, and disjunction are evaluated (assigned 0 or 1) just as in classical propositional logic, except here we must specify in which world the valuation holds.

vwA) = 1 if vw(A) = 0, and 0 otherwise.

vw(AB) = 1 if vw(A) = vw (B) = 1, and 0 otherwise.

vw(AB) = 1 if vw(A) = 1 or vw (B) = 1, and 0 otherwise.

(21)

We are now going to talk about a world taking an operator called a Routley Star, which gives to a world its star world (also called its “reversal world”) symbolized with an asterisk, like w. It has significance for the evaluation of negated formulas. When the negation of a formula is true in a world, then we do not say that the unnegated form is also false in that world, like we would normally do especially for classical logic (see the above valuation rules). (It is hard to conceptualize. I will think of it for now as a world’s negation-affirmation companion world, and here we would have negation affirmation by other-worldly falsity (by reverse-worldly falsity). Negation in one world does not guarantee the falsity of the unnegated form in that same world. But the star world’s falsities affirm the negations in our world. As we continue, we will be able to characterize it further).]

Specifically, we assume that each world, w, comes with a mate, w, its star world, such that ¬A is true at w if A is false, not at w, but at w. If w = w (which may happen), then these conditions just collapse into the classical conditions for negation; but if not, they do not. The star operator is often described with a variety of metaphors; for example, it is sometimes described as a reversal operator; but it is hard to give it and its role in the truth conditions for negation a satisfying intuitive interpretation.

(151)

[contents]

8.5.3

[Negation Defined. Validity.]

[The evaluations for connectives are the same for conjunction and disjunction, but for negation, we say a negated formula is 1 in w if its unnegated form is 0 in w*. Validity is defined as truth preservation for all worlds and interpretations.]

[We now define star worlds and negation more formally. We define the star as being a function between worlds. It is simply the function that when applied twice for a given world (once to the world, then to its star world), it yields the first world. (It seems like an exclusive world exchanger. It establishes pairs of worlds that exchange one for the other by means of the function. Physically we might think of a singular tunnel that connects just two places. A simple tunnel of this sort will necessarily make it such that it takes you from place A to place B, and to no other place than that, and from place B it takes you back to A and not to any other place. We then have the normal evaluation rules for possible world semantics (see section 8.5.2 above), except only now we have that rule for negation which says a formula’s negation is true in a world if its unnegated form is false in that world’s star world. This also means that a negated formulation in the star world is true if it the unnegated form is true in the non-star world. (Consider the following situation: suppose in w* that A is 0. That means ¬A is 1 in w. But what is ¬A in w*? If would be 1 only if A is 0 in w. But we do not so far know what its value is.

w

vw(A)=?

vwA) = 1

w*

vw*(A) = 0

vw*A)=?

Suppose A is 0 in w, then:

w

vw(A)=0

vwA) = 1

w*

vw*(A) = 0

vw*A)=1

Or suppose A is 1 in w, then what?

w

vw(A)=1

vwA) = 1

w*

vw*(A) = 0

vw*A)=0

(Of course, the above situations can have all their values toggled for similar situations.) Here is what Priest says next:

Note that vw*A) = 1 iff vw**(A) = 0 iff vw(A) = 0. In other words, given a pair of worlds, w and w* each of A and ¬A is true exactly once.

(152)

I did not understand this, but I will guess it means the following. We are making the following pairings:

{1} vwA), vw*(A)

{2} vw*A), vw(A),

And we have two types of situations. Either

{a} in both worlds the formula and its negation have the same value, or

{b} in both worlds the formula and its negation have different values.

The we ask, for each type of situation and pairing, how many times is A true and how many times is ¬A true?

{1a}   vwA)=1, vw*(A)=0

{2a}  vw*A)=0, vw(A)=1

Here we see when looking horizontally, where we compare A from one world and ¬A from the other, we see that they are only true once between the two.

{1b}   vwA)=1, vw*(A)=0

{2b}  vw*A)=1, vw(A)=0

Here we see the same. Probably Priest means something entirely different, but this is my best guess for now.) Lastly, validity is defined as truth preservation over all worlds of all interpretations. (See section 2.3.11 for how this works in possible worlds semantics.)]

Formally, a Routley interpretation is a structure ⟨W, ∗, v⟩, where W is a set of worlds, is a function from worlds to worlds such that w∗∗ = w, and v assigns each propositional parameter either the value 1 or the value 0 at each world. v is extended to an assignment of truth values for all formulas by the conditions:

vw(AB) = 1 if vw(A) = vw (B) = 1, otherwise it is 0. .

vw(AB) = 1 if vw(A) = 1 or vw (B) = 1, otherwise it is 0.

vwA) = 1 if vw*(A) = 0, otherwise it is 0.

| Note that vw*A) = 1 iff vw**(A) = 0 iff vw(A) = 0. In other words, given a pair of worlds, w and w* each of A and ¬A is true exactly once. Validity is defined in terms of truth preservation over all worlds of all interpretations.

(151-152)

[contents]

8.5.4

[Tableaux Rules for Routley Star Logic]

[For the tableaux rules, we designate not just the truth value but also the world in which that formula has that value. This is especially important for negation, where the derived formulas are in the companion star world.]

[We will now learn the tableaux rules for this sort of Routley starred logic. Priest will be giving a general form that will work for us in forthcoming sections. But let me start with what seems to be the rendition more specifically suited for this section. Recall how in our FDE semantics, we place beside the formula either a plus or a minus sign, which intuitively means true and false. It seems now we have the following ideas for the tableaux. Recall from section 8.5.3 above the rule for negation:

vwA) = 1 if vw*(A) = 0, otherwise it is 0.

The determining factor here is whether or not the unnegated form of a formula is 0 in the star world. The other thing we noted in section 8.5.3 above was that the evaluation rule for negation works the other way, because  that vw*A) = 1 iff vw**(A) = 0 iff vw(A) = 0. It seems that the tableaux rules will not involve a specific designation of the star world and the non-star world, perhaps because of this symmetry we just noted, but I am not sure. Instead, the rules will give an overlined   to mean the companion star world. Here, the specific value that will be filled in for x or , is always 0, but in later chapters it will be some other natural number for reasons that we well learn about then. We might be tempted to think that 0 has something to do with falsity like it often does, but here it  does not seem to indicate a truth value, because we already have the plus and minus signs for that purpose. Instead, the 0 seems to just mean the given world, while 0# means its star world. (In other words, we normally would just write w and w*, but since we later will for some reason maybe need more such world distinctions, we for now will just use 0 and worry about the other options later. For right now, just see 0 as w, and see 0# as w*, in the tableaux.)  So I am guessing, but it seems that when we write for example B,+0#, we mean maybe formula B which is true in the star world. The rules for conjunction and disjunction are largely the same here, and we assume for them that the derivations will be for formulas in the same world. But for negation, the derived formulas will be in the star companion world. In our tableau development, a branch is closed when it contains “a pair of the form A,+x and A,−x,” in other words, that some is true and false in the same world (and not however that the formula and its negation are true in the same world.) (I assume that means the formulas will be on different lines, even though the word ‘pair’ is used here.) We set up our tableaux by stacking the premises and assigning them the positive value in world 0 and the conclusion is stacked below with the negative value in world o. ]

Appropriate tableaux for these semantics are easy to construct. Nodes are now of the form A,+x or A,−x, where x is either i or i#, i being a natural number. (In fact, i will always be 0, but we set things up in a slightly more general way for reasons to do with later chapters.) Intuitively, i# represents the star world of i. Closure occurs if we have a pair of the form A,+x and A,−x. The initial list comprises a node B,+0 for every premise, B, and A,−0, where A is the conclusion. The tableau rules are as follows, where x is either i or i#, and whichever of these it is, is the other.

 Conjunction Development, True (∧D,+x) A ∧ B,+x ↓ A,+x B,+x

 Conjunction Development, False (∧D,−x) A ∧ B,−x ↙         ↘ A,−x              B,−x

 Disjunction Development, True (∨D,+x) A ∨ B,+x ↙               ↘ A,+x      B,+x

 Disjunction Development, False (∨D,−x) A ∨ B,-x ↓ A,-x B,-x

 Negation Development, True (¬D,+x) ¬A,+x ↓ A,-x̄

 Negation Development, False (¬D,−x) ¬A,-x ↓ A,+x̄

(152, Note, names and abbreviations are my own and are not in the text. I add them for my own convenience, but they are based on the conventions in Agler’s text. Also: I changed the third one from a conjunction symbol to a disjunction, thinking it should be that way, but I very easily could be wrong about that. Please check the text!)

[contents]

8.5.5

[Examples of Star Logic Tableaux]

[We use these rules and procedures to work through an example of valid inference and an example of invalid inference.]

Priest will now show a valid inference and an invalid one using the tableaux rules. [Recall from section 8.3.3 and section 8.5.4 above that to test for validity, we first set up the tableaux by stacking the premises and conclusion, with the premises assigned the positive symbol and the conclusion assigned the negative symbol. We need to recall some other terminology. This comes from section section 1.4: “A tableau is complete iff every rule that can be applied has been applied” (8). “A branch is closed iff there are formulas of the form A and ¬A on two of its nodes; otherwise it is open. A closed branch is indicated by writing an × at the bottom. A tableau itself is closed iff every branch is closed; otherwise it is open” (8). Furthermore: “A is a proof-theoretic consequence of the set of formulas Σ (Σ ⊢ A) iff there is a complete tree whose initial list comprises the members of Σ and the negation of A, and which is closed. We write ⊢ A to mean that φ ⊢ A, that is, where the initial list of the tableau comprises just ¬A. ‘Σ ⊬ A’ means that it is not the case that Σ ⊢ A(8-9). So to test for validity, we need to see if the set-up we mentioned above in its complete tree form is a closed tree. It is a complete tree if we have applied all applicable rules to it. It is also a closed tree if it has no open branches (see Agler’s Symbolic Logic section 4.2.2). And a branch is closed, as we said above, if we have a pair of the form A,+x and A,−x. So in other words, after we set up our tableau by stacking premises and conclusion, setting premises to true in world 0 and the conclusion to false in world 0, we then fully develop the tree, looking to see if all the branches are complete or not. If they all are, it is valid, because that means it cannot be that the premises are true and the conclusion false. But if there are any open branches, then we can produce a counter-example by means of them, and so the inference is invalid. In the following I will follow Agler’s style of tree development. See Agler’s Symbolic Logic section 4.1.1.]

Here are tableaux demonstrating that ¬(B ∧ ¬C) ∧ A BC) ∨ D and p ∧ (q∨¬q) ⊬ r

 ¬(B ∧ ¬C) ∧ A ⊢ (¬B ∨ C) ∨ D 1. . 2. . 3. . 4. . 5. . 6. . 7. . 8. . 9. . 11. . 12. . 13. . 14. . ¬(B∧¬C) ∧ A,+0 . (¬B ∨ C) ∨ D,−0   . (¬B ∨ C),−0   . D,−0 .  ¬B,−0 .  C,−0   . B,+0# .  ¬(B∧¬C),+0 .  A,+0 .  B∧¬C,−0#  ↙             ↘  # B,−0#           ¬C,−0# .  ×                  C,+0                      × P . P . 2∨D,−x . 2∨D,−x .  3∨D,−x .  3∨D,−x   . 5¬D,−x   . 1∧D,+x . 1∧D,+x . 8¬D,+x . 11∧D,−x . 12¬D,−x

[based on p.152-153]

Line two is pursued as far as possible. Then line one is pursued to produce closure.

 p ∧ (q ∨ ¬q) ⊬ r 1. . 2. . 3. . 4. . 5. . 6. . p ∧ (q ∨ ¬q),+0 . r,−0   . p,+0   . q ∨ ¬q,+0 ↙             ↘ q,+0            ¬q,+0 .                   q,−0#    . P . P . 1∧D,+x . 1∧D,+x .  4∨D,+x .  5¬D,+x   .

[based on p.153]

(quoting with my table modifications from pp.152-153)

[contents]

8.5.6

[Star Counter-Models]

[We make counter-models using completed open branches. On the basis of the world indicators in the branches, we assign to the formulas the values indicated by the true (+) and false (−) signs for the respective world. When there is negation, however, we need to use values in the star-companion world.]

Priest will now explain how we fashion counter-models. [This is quite complicated, so let go through its parts carefully. I will be making some guesses here, but we will test them. Possibly it can be simply understood that you assign 0 to − valued formulas and you assign 1 to + valued formulas, all the while designating which world that value for the formula holds in. (It seems we only make these assignments to fully decomposed formulas, so we would not designate values to a negated formula it seems, but I am not sure.) In the above formula, we have two open branches. Let us start with the right branch, like Priest does. In the table there is

p,+0

So we assign

vwo(p) = 1

And there is

q,−0#

So we assign

vwo#(q) = 0

And we also have

r, −0

So we assign it:

vwo(r) = o

The original formula was:

p ∧ (q∨¬q) ⊬ r

Before we move on, let us recall from section 8.5.3 above how we define validity:

Validity is defined in terms of truth preservation over all worlds of all interpretations.

(151-152)

I am not certain about the following, but suppose for some inference, its truth is preserved in a world but not in its star world. The fact that it is not preserved in one world would seem to be enough for it to not have been preserved over all worlds of all interpretations. In other words, in the following, our concern will be with showing that in the non-star world, the evaluation we gave above will make the premises true and the conclusion false. Let us look at what we have so far:

p ∧ (q∨¬q) ⊬ r

vwo(p) = 1

vwo#(q) = 0

vwo(r) = o

We have mostly values for the non-star world (by non-star I mean that we started with one, the non-star, and with one star we designated the star world, even though with two-stars we also get the first world.) We have values mostly for the non-star world. How do we get the non-star world value for ¬q? We are saying that q is false in the star world. Recall from section 8.5.3 above our semantics for negation:

vwA) = 1 if vw*(A) = 0, otherwise it is 0.

So since q is 0 in the star world, that makes ¬q 1 in the non-star world. But we do not know the value of q in the non-star world. Maybe it will not matter. The other connectives all take their criteria in the same world, so let us do a primitive visual evaluation by playing with the truth values:

vwo(p) = 1

vwo(q) = ?

vwoq) = 1

vwo(r) = o

p ∧ (q ∨ ¬q) ⊬ r

1 ∧ (? ∨ 1) ⊬ 0

1 ∧ (1) ⊬ 0

1 ⊬ 0

Let us test this with the left branch, which is also open.

vwo(p) = 1

vwo(q) = 1

vwoq) = ?

vwo(r) = o

p ∧ (q∨¬q) ⊬ r

1 ∧ (1 ∨ ?) ⊬ 0

1 ∧ (1) ⊬ 0

1 ⊬ 0

I may not understand this right, so here is the text:]

To read off a counter-model from an open branch: W = {w0,w0# } (there are only ever two worlds);  w*0 = w0# and (w0#)* = w0. (W and ∗ are always the same, no matter what the tableau.) v is such that if p,+x occurs on the branch, vwx(p) = 1, and if p,−x occurs on the branch, vwx(p) = 0. Thus, the counter-model defined by the righthand open branch of the second tableau of 8.5.5 has vwo(p) = 1, vwo(r) = 0 and vwo#(q) = 0. It is easy to check directly that this interpretation does the job. Since q is false at w0# , ¬q is true at w0, as, therefore, is q∨¬q; but p is true at w0, hence p ∧ (q ∨ ¬q) is true at w0. But r is false at w0, as required.

(153)

8.5.7

[Soundness and Completeness Proof  Postponement]

[Later we prove soundness and completeness proofs for this system.]

We prove the soundness and completeness of this star tableau system in sections  8.7.10–8.7.16 (153).

[contents]

8.5.8

[The Equivalence of Routley Star semantics and FDE ]

[Routley Star semantics and FDE are equivalent when we make the following translation: vw(p) = 1 iff 1; vw(p) = 0 iff 0.]

[Recall our relational semantics from section 8.2. There we had two truth-values, 1 and 0, but the semantics allowed for a formula to relate also to both truth values or to have no relation to a truth-value. We note now in our star semantics that we have not spoken yet of the both and neither possibilities. But in fact the star semantics and the relational semantics are equivalent. We later look at the proof for that, so perhaps then we will understand in greater depth now that is. But for now, we note that

a relational interpretation, ρ, is equivalent to a pair of worlds, w and w.  Specifically, the relation and the worlds do exactly the same job when they are related by the condition:

vw(p) = 1 iff 1

vw(p) = 0 iff 0

(I do not yet entirely understand how this works, but let us work through this. I am assuming that we are talking about the FDE semantics as the relational semantics. First note that an example from that section 8.3.7 is the same formula that we just worked on in section 8.5.6 above, which was also shown to be invalid:

p ∧ (q∨¬q) ⊬ r

But since there are three values to work with, it could get very laborious laying out every possible relational situation. But I still would like something visual so I can better grasp what makes these systems equivalent. In the next chapter, in section 8.6.1, Priest writes:

As we have seen (8.4.8 and 8.4.11), both of the following are false in FDE: p q ∨ ¬q, p ∧ ¬p q. This is essentially because there are truth-value gaps (for the former) and truth-value gluts (for the latter). In particular, then, FDE does not suffer from the problem of explosion (4.8).

(154)

In those sections, we saw how they are valid in more restricted version of FDE, but invalid in FDE. But we did not make the tableaux for them in FDE, so let us now do it for both FDE and Routley star semantics. For FDE, recall from section 8.3.2 the meanings of the + and − signs, from section 8.3.3 the way to set up the table for testing validity, from section 8.3.4 the tableaux development rules, from section 8.3.5 the rule for closing branches, and recall again the following from section 1.4 on how to test for validity: “A tableau is complete iff every rule that can be applied has been applied” (8). “A branch is closed iff there are formulas of the form A and ¬A on two of its nodes; otherwise it is open. A closed branch is indicated by writing an × at the bottom. A tableau itself is closed iff every branch is closed; otherwise it is open” (8). Furthermore: “A is a proof-theoretic consequence of the set of formulas Σ (Σ ⊢ A) iff there is a complete tree whose initial list comprises the members of Σ and the negation of A, and which is closed. We write ⊢ A to mean that φ ⊢ A, that is, where the initial list of the tableau comprises just ¬A. ‘Σ ⊬ A’ means that it is not the case that Σ ⊢ A(8-9). For testing for validity in Routley star semantics, see the tableaux rules above in section 8.5.4 and example proofs above in section 8.5.5. As a reminder, in FDE branches close when they have:

A,+ and A,−

and in Routley star semantics, they close when they have:

A,+x and A,−x

So, if the branch instead has both something and its negation (in the same world), then that is not enough to close the branch. Also recall how we find counter models from open branches. In FDE:

Counter-models can be read off from open branches in a simple way. For every parameter, p, if there is a node of the form p,+, set 1; if there is a node of the form ¬p,+, set 0. No other facts about ρ obtain.

(146. See section 8.3.8)

In Routley Star semantics:

To read off a counter-model from an open branch: W = {w0,w0# } (there are only ever two worlds);  w*0 = w0# and (w0#)* = w0. (W and ∗ are always the same, no matter what the tableau.) v is such that if p,+x occurs on the branch, vwx(p) = 1, and if p,−x occurs on the branch, vwx(p) = 0.

(153, see section 8.5.6 above)

FDE

 p ∧ ¬p ⊬FDE q 1. . 2. . 3. . 4. p ∧ ¬p,+ . q,−   . p,+   . ¬p,+   . [completed open] P . P . 1∧D,+ . 1∧D,+ .  Invalid

Counter-model: 1, 0

Routley Star (here as R*)

 p ∧ ¬p ⊬R* q 1. . 2. . 3. . 4. . 5. p ∧ ¬p,+0 . q,−0   . p,+0 . ¬p,+0   . p,−0#   . [completed open] P . P . 1∧D,+ . 1∧D,+ .  4¬D,+x .  Invalid

Counter-model: vw(p) = 1,  vw(q) = 0,  vw*(p) = 0

Now let us compare semantic evaluations. Recall from section 8.2.8 that for or the FDE relational semantics, validity is defined as:

Semantic consequence is defined, in the usual way, in terms of truth preservation, thus:

Σ ⊨ A iff for every interpretation, ρ, if 1 for all B ∈ Σ then 1

(144)

Doing all the evaluations will be a bit laborious, so bear with me (or scroll past this):

Σ ⊨ A iff for every interpretation, ρ, if 1 for all B ∈ Σ then 1.

p ∧ ¬p q iff for every interpretation, ρ, if (p ∧ ¬p)ρ1 then 1.

¬1 iff 0

¬0 iff 1

A1 iff 1 and 1

A0 iff 0 or 0

A1 iff 1 or 1

A0 iff 0 and 0

Configuration one (1,1):

1

1

¬0

(p ∧ ¬p)ρ0

1

0 ⊨ 1

[In configuration one, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration two (1,b):

1

1

0

¬0

(p ∧ ¬p)ρ0

1

0

0 ⊨ 1

0 ⊨ 0

[In configuration two, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration three (1,n):

1

¬0

(p ∧ ¬p)ρ0

[q has no value relation]

0 ⊨

[In configuration three, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration four (1,0):

1

0

¬0

(p ∧ ¬p)ρ0

0

0 ⊨ 0

[In configuration four, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration five (b,1):

1

0

1

¬0

¬1

(p ∧ ¬p)ρ1

(p ∧ ¬p)ρ0

1

1 ⊨ 1

0 ⊨ 1

[In configuration five, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration six (b, b):

1

0

1

0

¬1

¬0

(p ∧ ¬p)ρ1

(p ∧ ¬p)ρ0

1

0

1 ⊨ 1

1 ⊨ 0

0 ⊨ 1

0 ⊨ 0

[In configuration six, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration seven (b,n):

1

0

¬0

¬1

(p ∧ ¬p)ρ0

(p ∧ ¬p)ρ1

[q has no value relation]

0 ⊨

1 ⊨

[In configuration seven, it is not the case that if the premises are 1 than the conclusion is 1.]

Configuration eight (b,0):

1

0

0

¬0

¬1

(p ∧ ¬p)ρ0

(p ∧ ¬p)ρ1

0

0 ⊨ 0

1 ⊨ 0

[In configuration eight, it is not the case that if the premises are 1 than the conclusion is 1.]

Configuration nine (n, 1):

1

p has no value relation]

[p ∧ ¬p has no value relation]

1

⊨ 1

[In configuration nine, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration ten (n, b):

1

0

p has no value relation]

[p ∧ ¬p has no value relation]

1

0

⊨ 1

⊨ 0

[In configuration ten, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration eleven (n, n):

[neither p nor q has a value relation]

p has no value relation]

[p ∧ ¬p has no value relation]

[q has no value relation]

[In configuration eleven, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration twelve (n, 0):

0

p has no value relation]

[p ∧ ¬p has no value relation]

0

⊨ 0

[In configuration twelve, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration thirteen (0, 1):

0

1

¬1

(p ∧ ¬p)ρ0

1

0 ⊨ 1

[In configuration thirteen, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration fourteen (0, b):

0

1

0

¬1

(p ∧ ¬p)ρ0

1

0

0 ⊨ 1

0 ⊨ 0

[In configuration fourteen, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration fifteen (0, n):

0

¬1

(p ∧ ¬p)ρ0

[q has no value]

0 ⊨

[In configuration fifteen, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration sixteen (0,0):

0

0

¬1

(p ∧ ¬p)ρ0

0 ⊨ 0

[In configuration fifteen, it is the case that if the premises are 1 than the conclusion is 1.]

So in configuration seven (b, n): 1, 0,  and in configuration eight (b, 0): 1, 0, 0 we have the premises true but the conclusion false. Thus p ∧ ¬p q is invalid in FDE.

p ∧ ¬p FDE q

[Recall that our counter-model that we assessed from the tableau above was: 1, 0. The wording is “No other facts about ρ obtain.” So since this works for the situation (b, n): 1, 0, then I suppose the tableaux counter-model method indicates just one of the possible counter-examples and not all of them (if not, then of course I did my evaluations wrong.)] Let us now evaluate it in Routley star semantics. Let us begin with the semantical rules and evaluate it under all possible assignments.

??? p ∧ ¬p R* q ???

vwA) = 1 if vw*(A) = 0, otherwise it is 0.

vw(AB) = 1 if vw(A) = vw (B) = 1, otherwise it is 0.

[Validity is defined in terms of truth preservation over all worlds of all interpretations.]

World-Truth situation 1:

vw(p) = 1

vw(q) = 1

vw(p) = 1

vw(q) = 1

vw(¬p) = 0

vw(¬p) = 0

vw(p ¬p) = 0

vw(p ¬p) = 0

vw(q) = 1

vw(q) = 1

vw

0 ⊨ 1

vw

0 ⊨ 1

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 2:

vw(p) = 1

vw(q) = 1

vw(p) = 1

vw(q) = 0

vw(¬p) = 0

vw(¬p) = 0

vw(p ¬p) = 0

vw(p ¬p) = 0

vw(q) = 1

vw(q) = 0

vw

0 ⊨ 1

vw

0 ⊨ 0

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 3:

vw(p) = 1

vw(q) = 1

vw(p) = 0

vw(q) = 1

vw(¬p) = 1

vw(¬p) = 0

vw(p ¬p) = 1

vw(p ¬p) = 0

vw(q) = 1

vw(q) = 1

vw

1 ⊨ 1

vw

0 ⊨ 1

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 4:

vw(p) = 1

vw(q) = 0

vw(p) = 1

vw(q) = 1

vw(¬p) = 0

vw(¬p) = 0

vw(p ¬p) = 0

vw(p ¬p) = 0

vw(q) = 0

vw(q) = 1

vw

0 ⊨ 0

vw

0 ⊨ 1

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 5:

vw(p) = 0

vw(q) = 1

vw(p) = 1

vw(q) = 1

vw(¬p) = 0

vw(¬p) = 1

vw(p ¬p) = 0

vw(p ¬p) = 1

vw(q) = 1

vw(q) = 1

vw

0 ⊨ 1

vw

1 ⊨ 1

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 6:

vw(p) = 1

vw(q) = 1

vw(p) = 0

vw(q) = 0

vw(¬p) = 1

vw(¬p) = 0

vw(p ¬p) = 1

vw(p ¬p) = 0

vw(q) = 1

vw(q) = 0

vw

1 ⊨ 1

vw

0 ⊨ 0

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 7:

vw(p) = 1

vw(q) = 0

vw(p) = 0

vw(q) = 1

vw(¬p) = 1

vw(¬p) = 0

vw(p ¬p) = 1

vw(p ¬p) = 0

vw(q) = 0

vw(q) = 1

vw

1 ⊨ 0

vw

0 ⊨ 1

[Here it is not the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 8:

vw(p) = 0

vw(q) = 0

vw(p) = 1

vw(q) = 1

vw(¬p) = 0

vw(¬p) = 1

vw(p ¬p) = 0

vw(p ¬p) = 1

vw(q) = 0

vw(q) = 1

vw

0 ⊨ 0

vw

1 ⊨ 1

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 9:

vw(p) = 0

vw(q) = 1

vw(p) = 0

vw(q) = 1

vw(¬p) = 1

vw(¬p) = 1

vw(p ¬p) = 0

vw(p ¬p) = 0

vw(q) = 1

vw(q) = 1

vw

0 ⊨ 1

vw

0 ⊨ 1

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 10:

vw(p) = 1

vw(q) = 0

vw(p) = 1

vw(q) = 0

vw(¬p) = 0

vw(¬p) = 0

vw(p ¬p) = 0

vw(p ¬p) = 0

vw(q) = 0

vw(q) = 0

vw

0 ⊨ 0

vw

0 ⊨ 0

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 11:

vw(p) = 1

vw(q) = 0

vw(p) = 0

vw(q) = 0

vw(¬p) = 1

vw(¬p) = 0

vw(p ¬p) = 1

vw(p ¬p) = 0

vw(q) = 0

vw(q) = 0

vw

1 ⊨ 0

vw

0 ⊨ 0

[Here it is not the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 12:

vw(p) = 0

vw(q) = 1

vw(p) = 1

vw(q) = 0

vw(¬p) = 0

vw(¬p) = 1

vw(p ¬p) = 0

vw(p ¬p) = 1

vw(q) = 0

vw(q) = 0

vw

0 ⊨ 0

vw

1 ⊨ 0

[Here it is not the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 13:

vw(p) = 0

vw(q) = 1

vw(p) = 0

vw(q) = 0

vw(¬p) = 1

vw(¬p) = 1

vw(p ¬p) = 0

vw(p ¬p) = 0

vw(q) = 1

vw(q) = 0

vw

0 ⊨ 1

vw

0 ⊨ 0

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 14:

vw(p) = 0

vw(q) = 0

vw(p) = 1

vw(q) = 0

vw(¬p) = 0

vw(¬p) = 1

vw(p ¬p) = 0

vw(p ¬p) = 1

vw(q) = 0

vw(q) = 0

vw

0 ⊨ 0

vw

1 ⊨ 0

[Here it is not the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 15:

vw(p) = 0

vw(q) = 0

vw(p) = 0

vw(q) = 1

vw(¬p) = 1

vw(¬p) = 1

vw(p ¬p) = 0

vw(p ¬p) = 0

vw(q) = 0

vw(q) = 1

vw

0 ⊨ 0

vw

0 ⊨ 1

[Here it is the case that if the premises are 1 than the conclusion is 1.]

World-Truth situation 16:

vw(p) = 0

vw(q) = 0

vw(p) = 0

vw(q) = 0

vw(¬p) = 1

vw(¬p) = 1

vw(p ¬p) = 0

vw(p ¬p) = 0

vw(q) = 0

vw(q) = 0

vw

0 ⊨ 0

vw

0 ⊨ 0

[Here it is the case that if the premises are 1 than the conclusion is 1.]

Thus under the following evaluations, the premises are true and the conclusion false: World-Truth situation 7, vw(p) = 1, vw(q) = 0, vw(p) = 0, vw(q) = 1; World-Truth situation 11, vw(p) = 1, vw(q) = 0, vw(p) = 0, vw(q) = 0; World-Truth situation 12: vw(p) = 0, vw(q) = 1, vw(p) = 1, vw(q) = 0, and World-Truth situation 14: vw(p) = 0, vw(q) = 0, vw(p) = 1, vw(q) = 0.  So

p ∧ ¬p R* q

[Please be aware that I probably made many mistakes in the work above, and I will correct them as I find them. I also realize that in fact I may be doing this all entirely wrong anyway, but I am working it out fully to try to learn how the machinery works.] [Recall that the counter-model that we assessed from the tableau above was: vw(p) = 1,  vw(q) = 0,  vw*(p) = 0. This seems to apply to the situations 7 and 11 above, but not to12 and 14. Probably I did 12 and 14 wrong, or maybe the tableau counter-model method does not indicate all possible counter-examples.] Now we have the other formula to check:

p q ∨ ¬q

p q ∨ ¬q

Let us do so first using tableau rules and then using semantic evaluations.

FDE

 p ⊬FDE q  ∨ ¬q 1. . 2. . 3. . 4. p,+ . q  ∨ ¬q,−   . q,−   . ¬q,−   . [completed open] P . P . 2∨D,− . 2∨D,− .  Invalid

Counter-model: 1

Routley Star (here as R*)

 p ⊬R* q  ∨ ¬q 1. . 2. . 3. . 4. . 5. p,+0 . q  ∨ ¬q,−0   . q,−0   . ¬q,−0   . q,+0*   . [completed open] P . P . 2∨D,−x . 2∨D,−x .  4¬D,−x . Invalid

Counter-model: vw(p) =1, vw(q) =0, vw*(q) =1

Now let us do a semantic evaluation in FDE:

??? p FDE q ∨ ¬q ???

Σ ⊨ A iff for every interpretation, ρ, if 1 for all B ∈ Σ then 1.

p FDE q ∨ ¬q iff for every interpretation, ρ, if pρ1 then (q ¬q)ρ1.

¬1 iff 0

¬0 iff 1

A1 iff 1 and 1

A0 iff 0 or 0

A1 iff 1 or 1

A0 iff 0 and 0

Configuration one (1,1):

1

1

¬qρ0

(q ¬q)ρ1

1

1 ⊨ 1

[In configuration one, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration two (1,b):

1

1

0

¬qρ0

¬qρ1

(q ¬q)ρ1

(q ¬q)ρ0

1

1 ⊨ 1

1 ⊨ 0

[In configuration two, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration three (1,n):

1

[¬q has no value relation]

[(q ¬q) has no value relation]

1

1 ⊨

[In configuration three, it is not the case that if the premises are 1 than the conclusion is 1.]

Configuration four (1,0):

1

0

¬qρ1

(q ¬q)ρ1

1

1 ⊨ 1

[In configuration four, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration five (b,1):

1

0

1

¬qρ0

(q ¬q)ρ1

1

0

1 ⊨ 1

0 ⊨ 1

[In configuration five, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration six (b, b):

1

0

1

0

¬qρ0

¬qρ1

(q ¬q)ρ1

(q ¬q)ρ0

1

0

1 ⊨ 1

1 ⊨ 0

0 ⊨ 1

0 ⊨ 0

[In configuration six, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration seven (b,n):

1

0

[¬q has no value relation]

[(q ¬q) has no value relation]

1

0

0 ⊨

1 ⊨

[In configuration seven, it is not the case that if the premises are 1 than the conclusion is 1.]

Configuration eight (b,0):

1

0

0

¬qρ1

(q ¬q)ρ1

1

0

1 ⊨ 1

0 ⊨ 1

[In configuration eight, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration nine (n, 1):

1

¬qρ0

(q ¬q)ρ1

[p has no value relation]

⊨ 1

[In configuration nine, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration ten (n, b):

1

0

¬qρ0

¬qρ1

(q ¬q)ρ1

(q ¬q)ρ0

[p has no value relation]

⊨ 1

⊨ 0

[In configuration ten, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration eleven (n, n):

[neither p nor q has a value relation]

[¬q has no value relation]

[q ¬q has no value relation]

[p has no value relation]

[In configuration eleven, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration twelve (n, 0):

0

¬qρ1

(q ¬q)ρ1

[p has no value relation]

⊨ 0

[In configuration twelve, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration thirteen (0, 1):

0

1

¬qρ0

(q ¬q)ρ1

0

0 ⊨ 1

[In configuration thirteen, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration fourteen (0, b):

0

1

0

¬qρ0

¬qρ1

(q ¬q)ρ1

(q ¬q)ρ0

0

0 ⊨ 1

0 ⊨ 0

[In configuration fourteen, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration fifteen (0, n):

0

[¬q has no value relation]

[q ¬q has no value relation]

0

0 ⊨

[In configuration fifteen, it is the case that if the premises are 1 than the conclusion is 1.]

Configuration sixteen (0,0):

0

0

¬qρ1

(q ¬q)ρ1

0

0 ⊨ 1

[In configuration fifteen, it is the case that if the premises are 1 than the conclusion is 1.]

So in configuration three (1, n): 1; and in configuration seven (b, n): 1, 0; we have the premises at least true but the conclusion not at least true. Thus p q ∨ ¬q is invalid in FDE.

p FDE q ∨ ¬q

[Recall that our tableau counter-model was 1, which gives us configuration three but not seven.] Let us now evaluate it in Routley Star semantics. And let us begin with the semantical rules and evaluate it under all possible assignments.

??? p R* q ∨ ¬q ???

vwA) = 1 if vw*(A) = 0, otherwise it is 0.

vw(AB) = 1 if vw(A) = 1 or vw (B) = 1, otherwise it is 0.

[Validity is defined in terms of truth preservation over all worlds of all interpretations.]

World-Truth situation 1:

vw(p) = 1

vw(q) = 1

vw(p) = 1

vw(q) = 1

vw(¬q) = 0

vw(¬q) = 0

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 1

vw(p) = 1

vw(p) = 1

vw

1 ⊨ 1

vw

1 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 2:

vw(p) = 1

vw(q) = 1

vw(p) = 1

vw(q) = 0

vw(¬q) = 1

vw(¬q) = 0

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 1

vw(p) = 1

vw(p) = 1

vw

1 ⊨ 1

vw

1 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 3:

vw(p) = 1

vw(q) = 1

vw(p) = 0

vw(q) = 1

vw(¬q) = 0

vw(¬q) = 0

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 1

vw(p) = 1

vw(p) = 0

vw

1 ⊨ 1

vw

0 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 4:

vw(p) = 1

vw(q) = 0

vw(p) = 1

vw(q) = 1

vw(¬q) = 0

vw(¬q) = 1

vw(q ∨ ¬q) = 0

vw(q ∨ ¬q) = 1

vw(p) = 1

vw(p) = 1

vw

1 ⊨ 0

vw

1 ⊨ 1

[Here it is not the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 5:

vw(p) = 0

vw(q) = 1

vw(p) = 1

vw(q) = 1

vw(¬q) = 0

vw(¬q) = 0

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 1

vw(p) = 0

vw(p) = 1

vw

0 ⊨ 1

vw

1 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 6:

vw(p) = 1

vw(q) = 1

vw(p) = 0

vw(q) = 0

vw(¬q) = 1

vw(¬q) = 0

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 0

vw(p) = 1

vw(p) = 0

vw

1 ⊨ 1

vw

0 ⊨ 0

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 7:

vw(p) = 1

vw(q) = 0

vw(p) = 0

vw(q) = 1

vw(¬q) = 0

vw(¬q) = 1

vw(q ∨ ¬q) = 0

vw(q ∨ ¬q) = 1

vw(p) = 1

vw(p) = 0

vw

1 ⊨ 0

vw

0 ⊨ 1

[Here it is not the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 8:

vw(p) = 0

vw(q) = 0

vw(p) = 1

vw(q) = 1

vw(¬q) = 0

vw(¬q) = 1

vw(q ∨ ¬q) = 0

vw(q ∨ ¬q) = 1

vw(p) = 0

vw(p) = 1

vw

0 ⊨ 0

vw

1 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 9:

vw(p) = 0

vw(q) = 1

vw(p) = 0

vw(q) = 1

vw(¬q) = 0

vw(¬q) = 0

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 1

vw(p) = 0

vw(p) = 0

vw

0 ⊨ 1

vw

0 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 10:

vw(p) = 1

vw(q) = 0

vw(p) = 1

vw(q) = 0

vw(¬q) = 1

vw(¬q) = 1

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 1

vw(p) = 1

vw(p) = 1

vw

1 ⊨ 1

vw

1 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 11:

vw(p) = 1

vw(q) = 0

vw(p) = 0

vw(q) = 0

vw(¬q) = 1

vw(¬q) = 1

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 1

vw(p) = 1

vw(p) = 0

vw

1 ⊨ 1

vw

0 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 12:

vw(p) = 0

vw(q) = 1

vw(p) = 1

vw(q) = 0

vw(¬q) = 1

vw(¬q) = 0

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 0

vw(p) = 0

vw(p) = 0

vw

0 ⊨ 1

vw

0 ⊨ 0

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 13:

vw(p) = 0

vw(q) = 1

vw(p) = 0

vw(q) = 0

vw(¬q) = 1

vw(¬q) = 0

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 0

vw(p) = 0

vw(p) = 0

vw

0 ⊨ 1

vw

0 ⊨ 0

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 14:

vw(p) = 0

vw(q) = 0

vw(p) = 1

vw(q) = 0

vw(¬q) = 1

vw(¬q) = 1

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 1

vw(p) = 0

vw(p) = 1

vw

0 ⊨ 1

vw

1 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 15:

vw(p) = 0

vw(q) = 0

vw(p) = 0

vw(q) = 1

vw(¬q) = 0

vw(¬q) = 1

vw(q ∨ ¬q) = 0

vw(q ∨ ¬q) = 1

vw(p) = 0

vw(p) = 0

vw

0 ⊨ 0

vw

0 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

World-Truth situation 16:

vw(p) = 0

vw(q) = 0

vw(p) = 0

vw(q) = 0

vw(¬q) = 1

vw(¬q) = 1

vw(q ∨ ¬q) = 1

vw(q ∨ ¬q) = 1

vw(p) = 0

vw(p) = 0

vw

0 ⊨ 1

vw

0 ⊨ 1

[Here it is the case that if the premises are 1 then the conclusion is 1.]

So as we can see, in World-Truth situation 4: vw(p) = 1, vw(q) = 0, vw(p) = 1, vw(q) = 1, and in situation 7: vw(p) = 1, vw(q) = 0, vw(p) = 0, vw(q) = 1, the premises in the non-star world are true but the conclusion is not true. So:

p R* q ∨ ¬q

[Our tableau counter-model gave us: vw(p) =1, vw(q) =0, vw*(q) =1, which seems to apply to both situations above.] Now recall that Priest writes:

a relational interpretation, ρ, is equivalent to a pair of worlds, w and w. Specifically, the relation and the worlds do exactly the same job when they are related by the condition:

vw(p) = 1 iff 1

vw(p) = 0 iff 0

[I wonder if for each “configuration” in FDE there is a corresponding “world-truth situation” in Routley star semantics. So let us see. Take for example

configuration three (1, n): 1

I think we would make as the corresponding Routley arrangement:

vw(p) = 1, vw(q) = 0, vw(p) = 1, vw(q) = 1

This corresponds with world-truth situation 4, which also had premises true and conclusion false. So it seems promising so far. But let us further conceive this equivalence. Imagine this, in one reality (I want do avoid the word “world” but that is what I mean), there is a fact that is true and another fact that has no truth value (like a future contingent perhaps.) In another reality, there is a pair of worlds. In one world, the first fact is true, and it is also true in the companion world, and the second fact is false, but it is true in the companion world. What conceptually makes these equivalent? Let us try this for all the pairings.

Configuration one (1,1):

1

1

vw(p) = 1, vw(q) = 1, vw(p) = 1, vw(q) = 1

World-Truth situation 1:

vw(p) = 1

vw(q) = 1

vw(p) = 1

vw(q) = 1

Configuration two (1,b):

1

1

0

vw(p) = 1, vw(q) = 1, vw(p) = 1, vw(q) = 0

World-Truth situation 2:

vw(p) = 1

vw(q) = 1

vw(p) = 1

vw(q) = 0

Configuration three (1,n):

1

vw(p) = 1, vw(q) = 0, vw(p) = 1, vw(q) = 1

World-Truth situation 4:

vw(p) = 1

vw(q) = 0

vw(p) = 1

vw(q) = 1

Configuration four (1,0):

1

0

vw(p) = 1, vw(q) = 0, vw(p) = 1, vw(q) = 0

World-Truth situation 10:

vw(p) = 1

vw(q) = 0

vw(p) = 1

vw(q) = 0

Configuration five (b,1):

1

0

1

vw(p) = 1, vw(q) = 1, vw(p) = 0, vw(q) = 1

World-Truth situation 3:

vw(p) = 1

vw(q) = 1

vw(p) = 0

vw(q) = 1

Configuration six (b, b):

1

0

1

0

vw(p) = 1, vw(q) = 1, vw(p) = 0, vw(q) = 0

World-Truth situation 6:

vw(p) = 1

vw(q) = 1

vw(p) = 0

vw(q) = 0

Configuration seven (b,n):

1

0

vw(p) = 1, vw(q) = 0, vw(p) = 0, vw(q) = 1

World-Truth situation 7:

vw(p) = 1

vw(q) = 0

vw(p) = 0

vw(q) = 1

Configuration eight (b,0):

1

0

0

vw(p) = 1, vw(q) = 0, vw(p) = 0, vw(q) = 0

World-Truth situation 11:

vw(p) = 1

vw(q) = 0

vw(p) = 0

vw(q) = 0

Configuration nine (n, 1):

1

vw(p) = 0, vw(q) = 1, vw(p) = 1, vw(q) = 1

World-Truth situation 5:

vw(p) = 0

vw(q) = 1

vw(p) = 1

vw(q) = 1

Configuration ten (n, b):

1

0

vw(p) = 0, vw(q) = 1, vw(p) = 1, vw(q) = 0

World-Truth situation 12:

vw(p) = 0

vw(q) = 1

vw(p) = 1

vw(q) = 0

Configuration eleven (n, n):

[neither p nor q has a value relation]

vw(p) = 0, vw(q) = 0, vw(p) = 1, vw(q) = 1

World-Truth situation 8:

vw(p) = 0

vw(q) = 0

vw(p) = 1

vw(q) = 1

Configuration twelve (n, 0):

0

vw(p) = 0, vw(q) = 0, vw(p) = 1, vw(q) = 0

World-Truth situation 14:

vw(p) = 0

vw(q) = 0

vw(p) = 1

vw(q) = 0

Configuration thirteen (0, 1):

0

1

vw(p) = 0, vw(q) = 1, vw(p) = 0, vw(q) = 1

World-Truth situation 9:

vw(p) = 0

vw(q) = 1

vw(p) = 0

vw(q) = 1

Configuration fourteen (0, b):

0

1

0

vw(p) = 0, vw(q) = 1, vw(p) = 0, vw(q) = 0

World-Truth situation 13:

vw(p) = 0

vw(q) = 1

vw(p) = 0

vw(q) = 0

Configuration fifteen (0, n):

0

vw(p) = 0, vw(q) = 0, vw(p) = 0, vw(q) = 1

World-Truth situation 15:

vw(p) = 0

vw(q) = 0

vw(p) = 0

vw(q) = 1

Configuration sixteen (0,0):

0

0

vw(p) = 0, vw(q) = 0, vw(p) = 0, vw(q) = 0

World-Truth situation 16:

vw(p) = 0

vw(q) = 0

vw(p) = 0

vw(q) = 0

I will need to think more about how to understand this equivalence, and I will add more later if I figure anything out. I note for now that whenever a variable has the value  of both in FDD, then in the equivalent star situation, it is 1 in w and 0 in w*. And whenever a variable has the value of neither, it is  0 in w and 1 in w*.  And when it is just 1 in FDE, then it is 1 in both w and w*. And when it is just 0 in FDE, then it is 0 in both w and w*. As I said, I will come back to say more if I find a good way to think about the equivalence. Here is the quotation:]

It is not at all obvious that the ∗ semantics are equivalent to the relational semantics, but it is not too difficult to establish this. Essentially, it is because a relational interpretation, ρ, is equivalent to a pair of worlds, w and w. Specifically, the relation and the worlds do exactly the same job when they are related by the condition:

vw(p) = 1 iff 1

vw(p) = 0 iff 0

| for all parameters, p. The proof of the equivalence is given in 8.7.17 and 8.7.18.

(153-154)

[contents]

From:

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

Sources cited by Priest:

(1993), ‘Can Contradictions be True?, II’, Proceedings of the Aristotelian Society, Supplementary Volume 68: 35–54.

.