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N04SLSemantics - Semantics of SL.pdf

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Department
Philosophy
Course
2250
Professor
James Hildebrand
Semester
Spring

Description
SEMANTICS FOR SENTENTIAL LOGIC Whereas the syntax for SL concerns what strings of vocabulary elements are well formed (are sentences of SL), the semantics concerns what meaning is to be ascribed to the vocabulary elements. These notes are divided into the following sections: Semantics for atomic sentences of SL Semantics for compound sentences of SL Determining the truth values of compound sentences Determining the truth values of atomic sentences Definitions of truth-functional properties and relations of sentences and groups of sentences of SL Truth functional truth, falsity, and indeterminacy Truth functional equivalence and non-equivalence Truth functional inconsistency and consistency Truth functional entailment and validity EXERCISES Comparison of truth functional with logical properties and relations SEMANTICS FOR ATOMIC SENTENCES OF SL English sentences, even English simple sentences, have a rich and complex meaning. This is not the case with sentences of SL. We will say that the meaning of an atomic sentence of SL is just its truth value. We further specify that there are only two truth values, T and F (or true and false), and that every atomic sentence of SL must have one or other but not both of these truth values. In English, the truth or falsity of a simple sentence is very often determined by the way the worl d is. The sentence “Grass is green,” for example, is taken to be true if and only if grass is in fact green. In SL the truth or falsity of atomic sentences is determined by truth value assignments. A truth value assignment assigns either a T or an F, but not both, to every atomic sentence of SL. Be careful about the difference between a truth value and a truth value assignment. T and F are truth values. A sentence of SL is assigned one or other of these two truth values by a truth value assignment. But a truth value assignment, as we are using that expression in this course, does not just assign one truth value to one atomic sentence of SL. A truth value assignment is the assignment of one or other of a T or an F to every atomic sentence of SL. The assignment of one truth value to one atomic sentence of SL is only a part of what a truth value assignment does. There are infinitely many truth value assignments. There is the truth value assignment that assigns a T to every atomic sentence. There is the truth value assignment that assigns a T to the alphabetically first atomic sentence, “A,” and an F to every other atomic sentence. There is the truth value assignment that assigns an F to every atomic sentence. There is the truth value assignment that assigns a T to “A,” “C,” and “H217 and an F to every other atomic sentence. And so on. In these notes (though not in the text), I will often refer to a truth value assignment by using the Greek letter “alpha” (α). Where I want to distinguish between different truth value assignments I will subscript this symbol (α1, α2, 3 , etc.). We are not concerned which of these infinitely many truth value assignments is the correct one. There is no correct one. Instead, we will standardly adopt the practice of considering all of them. There is an important consequence that must be noted and repeatedly stressed here. The fact that we are not concerned with which of the infinitely many truth value assignments is the correct one means that, in SL, truth is relative, not absolute. There is no absolute truth or falsity. Truth and falsity are relative to truth value assignments. We will never simply say that a sentence of SL is “true” or “false.” We will only ever say that it is “true on a truth value assignment,” “false on a truth value assignment,” “true on this truth value assignment but false on that one,” “true on all truth value assignments,” “true on no truth value assignment,” and so on. Whenever truth or falsity are ascribed to sentences of SL, it will have to be relative to one or more truth value assignments, and we will expect that those truth value assignments will have been mentioned. Though there are infinitely many truth value assignments, surveying them is not always as big a job as it might at first seem. If you are just considering one atomic sentence, “A,” then even though there are infinitely many truth value assignments, they are all of one or other of just two types, the ones that assign a T to “A” and the ones that assign an F to “A.” We do not need to worry about what assignments they make to other sentences of SL if we are only worried about the sentence, “A.” So, in effect, the infinitely many truth value assignments reduce to just two. Similarly, if you are considering just two atomic sentences, “A” and “B,” then even though there are infinitely many truth value assignments, they are all of one or other of just four types, the ones that assign T to both “A” and “B,” the ones that assign T to “A” but F to “B,” the ones that assign F to “A” but T to “B,” and the ones that assign F to both. If you are considering just three atomic sentences there are just eight types of truth value assignments, if just four, sixteen, and so on. In general, the number of different types of truth value assignments you need to consider is 2 , where n is the number of sentences you are considering. These remarks exhaust the semantics of atomic sentences of SL. SEMANTICS FOR COMPOUND SENTENCES OF SL The meanings of compound sentences of SL are determined by the meanings of their component parts. But they are also determined by the meanings of the connectives that are attached to those component parts. The meanings of the five connectives of SL are defined in terms of truth values. However, the connectives do not themselves have truth values. Instead, they determine truth values for the compound sentences they are used to build, and it is the way they determine these values that constitutes their meaning. The connectives of SL are truth functional connectives. A truth functional connective creates a compound sentence that has a truth value that is completely determined by the truth values of its component parts. For each possible combination of truth values that the component parts might have, the connective must determine exactly one truth value for the compound. The meaning of the connective is just the rule it uses to determine the truth value of a compound sentence. Each connective has its own rule, which both tells us what the connective means and what truth values it assigns to a compound sentence. ~Rule: ~ creates a compound sentence with the opposite truth value of its immediate component. &Rule: & creates a compound sentence that is true iff both its immediate components are true and false iff at least one of its immediate components is false. vRule: v creates a compound sentence that is true iff at least one of its immediate components is true and false iff both immediate components are false. ⊃Rule: ⊃ creates a compound sentence that is true iff either its antecedent is false or its consequent is true and false iff its antecedent is true and its consequent is false. ≡Rule: ≡ creates a compound sentence that is true iff its immediate components have the same truth value and false iff its immediate components have different truth values. These rules ensure that every compound sentence of SL must also have one or the other, but not both of the two truth values, T and F on each possible truth value assignment to its components. So the meaning of compound sentences ends up being their truth values as well. The meaning of the connectives is given by the rules I have stated above. It is not given by what you might imagine are corresponding words of English. You should not think that “&” means “and” (or “but” or “moreover,” or “however,” etc.). You should not think that “v” means “or.” And so on. The meaning of the connectives is the rules given above. “&” means “true if and only if both immediate components are true.” “v” means “true if and only if at least one immediate component is true.” And so on. There are no words of English that behave, in all contexts, exactly as specified by these rules. Consequently, if you think of the connectives in terms of what you might imagine are the corresponding English names, you will only confuse yourself. Using the formal language, SL, to stand for or “translate” sentences of English or other natural languages is an art that we will reserve until much later. DETERMINING THE TRUTH VALUES OF COMPOUND SENTENCES If we are given a truth value assignment, α, to the atomic sentences of SL, we can use the rules for the connectives to determine the truth value of the compound ones on α. We do so by starting with the values α assigns to the atomic components and then, following the same constructive procedure we would follow were we considering how to construct the sentence using the syntactic rules, work up from atomic components to the compounds they build. Only now, we consider what truth value the connective assigns to the compound. Here is an example of what I mean. Consider the truth value assignment, α, that assigns T to “A” and “C” and F to “B” and every other atomic sentence of SL. (We represent this by writing: α(A)=T, α(B)=F, α(C)=T, not worrying about the other sentences since they aren’t part of the discussion.) Now let’s consider each of the sentences of SL in 2.1E#2 and calculate their truth values using the rules: 2a. 1. α(B)=F (given) 2. α(A & B)=F (from line 1 by & rule) 2c. 1. α(A)=T (given) 2. α(A v C)=T (from line 1 by v rule) 2e. 1. α(A)=T (given) 2. α(~A)=F (from line 1 by ~ rule) 3. α(~A & ~C)=F (from line 2 by & rule) 2g. 1. α(B)=F (given) 2. α(B & (A v C))=F (from line 1 by & rule) 2i. 1. α(B)=F (given) 2. α((A & C) & B)=F (from line 1 by & rule) 2k. 1. α(C)=T (given) 2. α(B v C)=T (from line 1 by v rule) 3. α((B v C) v ~(B v C))=T (from line 2 by v rule) Note that in doing these exercises I have adopted the lazy expedient of not doing any more work than I absolutely have to. The & rule tells me that to prove that a conjunction is false on a truth value assignment (tva), all I need to do is determine that one of the conjuncts is false on that tva. So, once I have done that, I do not bother to look into what truth value the other conjunct has. It will make no difference; the conjunction will have to be false regardless. So I don’t bother with it and proceed on. You see me do this in 2a, 2e, 2g, and 2i. Indeed, in 2g and 2i you see me get crafty about it and zero in on the one conjunct I know I can show to be false even though it is not the first one in order or even though its partner is a compound that needs to be built up before the two can be conjoined. In 2c and 2k you see me adopt the same lazy method with the v rule. The v rule tells me that t o show that a disjunction is true on a tva I only need to prove that one of its disjuncts is true on that tva. Once I have done that, I can forget about the other disjunct, and that is what I do. In 2c, once I have proven that the first disjunct is true on α, I declare that the whole disjunction is true on α without bothering to look at the other disjunjct. And in 2k I get crafty about it and zero in on a path that will allow me to establish the truth value of the whole sentence in the fewest number of steps. Had I started with “B,” which is false on α, I would not have been in a position to draw any conclusion about the truth value of the disjunction, because a false disjunct is compatible with either a true or a false disjunction depending on whether the other disjunct is true or false. And had I set out to first determine the truth value of “~(B v C)” I would have wasted even more steps, since that disjunct ends up being false on α and again does not help me determine the truth value of the whole when the main connective is a “v.” So there was careful plotting behind my choice of approach. That having been said, all that starting the wrong place will do is cost you some extra work. Regardless of where you start, as long as you apply the rules correctly, you will get a result, and it will be the correct one. We can take the same crafty approach to the exercises in 3.1E#3. Here let us consider α to be the truth value assignment that assigns T to “B” and “C” and F to “A” and every other atomic sentence of SL. 3a. 1. α(A)=F (given) 2. α(~A)=T (from line 1 by ~ rule) 3. α(~A v (~C v ~B))=T (from line 2 by v rule) 4. α(~[~A v (~C v ~B)])=F (from line 3 by ~ rule) 3c. 1. α(A)=F (given) 2. α(A ⊃ B)=T (from line 1 by ⊃ rule) 3. α((A ⊃ B) v (B ⊃ C))=T (from line 2 by v rule) 3e. 1. α(B)=T (given) 2. α(C)=T (given) 3. α(B ≡ C)=T (from lines 1 and 2 by ≡ rule) 4. α((A ≡ B) v (B ≡ C))=T (from line 3 by v rule) 3g. 1. α(C)=T (given) 2. α(A v C)=T (from line 1 by v rule) 3. α(B ⊃ (A v C))=T (from line 2 by ⊃ rule) 4. α(~[B ⊃ (A v C)])=F (from line 3 by ~ rule) 5. α(~[B ⊃ (A v C)] & ~~B)=F (from line 4 by & rule) 3i. 1. α(A)=F (given) 2. α(B) =T (given) 3. α(~B)=F (from line 2 by ~ rule) 4. α(A ≡ ~B)=T (from lines 1 and 3 by ≡ rule) 5. α(~(A ≡ ~B))=F (from line 4 by ~ rule) 6. α(~A)=T (from line 1 by ~ rule) 7. α(~(A ≡ ~B) ≡ ~A)=F (from lines 5 and 6 by ≡ rule) 8. α(~[~(A ≡ ~B) ≡ ~A])=T (from line 7 by ~ rule) 9. α(B v C)=T (from line 2 by v rule) 10. α(~[~(A ≡ ~B) ≡ ~A] ≡ (B v C))=T (from lines 8 and 9 by ≡ rule) Note that in 3i there are almost no short cuts that can be taken, so we have no choice but to look at the truth value α assigns to every atomic component and work up from there. When it comes to proving that a conjunction is true, a disjunction false, a conditional false, or a biconditional either true or false there are no shortcuts. You can’t look at just one of the immediate components to get the answer. Doing the starred exercises in 2.1E#2 and 3.1E#3 after this same pattern is left to you as an assignment. DETERMINING THE TRUTH VALUES OF ATOMIC SENTENCES The rules for the connectives are stated in the form of “if and only if” sentences. This means that they hold in both directions. Given a truth value assignment to the immediate components of a sentence, the rules tell you what truth value the compound must have. But equally, given a truth value assignment to a compound sentence, the rules tell you what truth values its immediate components must have, and so what truth values the components of those components must have down to atomic components. For example, if I am told that “~(A ⊃ B)” is true on a truth value assignment, α, I can reason “backwards” using the rules as follows: 1. α(~(A ⊃ B))=T (given) 2. α(A ⊃ B)=F (from line 1 by ~ rule) 3. α(A)=T and α(B)=F (from line 2 by ⊃ rule) Just as the ~ rule tells me that ~P must be true if P is false, so it tells me that ~P is true only if P is false, from which it follows that just as I can infer the truth of ~P from the falsity of its immediate component, P, so I can infer the falsity of P from the truth of its negation, ~P. And just as the ⊃ rule tells me that P ⊃ Q must be false if P is true and Q is false, so it tells me that P ⊃ Q is false only if P is true and Q is false, from which it follows that just as I can infer the falsity of P ⊃ Q from the truth of its antecedent, P, and falsity of its consequent, Q, so I can infer the truth of P and falsity of Q from the fact that a conditional that has P as its antecedent and Q as its consequent is false. There is just one wrinkle. When we reason “backwards” from truth values of compounds to truth values of components like this, we sometimes run into two or even three or more alternatives. For example, if I am told that “~(A ≡ B)” is true on α, and I reason back, I run into two alternatives: 1. α(~(A ≡ B))=T (given) 2. α(A ≡ B)=F (from line 1 by ~ rule) 3. α(A)≠α(B) (from line 2 by ≡ rule) 4. Either α(A)=T and α(B)=F, or α(A)=F and α(B)=T (from line 3) The same follows if I am told that “~(A ≡ B)” is false on α, only now my two alternatives are either that α assigns a T to both “A” and “B” or that it assigns an F to both of them. Things are even worse whenever I run into the claim that a conditional or a disjunction is true or a conjunction is false on a tva. That opens three different possibilities for the way truth values to the immediate components could go on that tva. This can get messy if the different possibilities are themselves compound sentences that open further possibilities. All the same, it is worth looking at how our reasoning would proceed in simpler cases. Let’s consider exercise 2.1#2 again, only this time, let’s assume that each sentence is true on α and attempt to reason back as far as we can to the truth values of the atomic components. Here is how the unstarred exercises turn out: 2a. 1. α(A & B)=T (given) 2. α(A)=α(B)=T (from line 1 by & rule) 2c. 1. α(A v C)=T (given) 2. At least one of “A” and “C” is true on α (from line 1 by v rule) 3. Either α(A)=α(C)=T; or α(A)=T and α(B)=F; or α(A)=F and α(B)=T (from line 2) 2e. 1. α(~A & ~C)=T (given) 2. α(~A)=α(~C)=T (from line 1 by & rule) 3. α(A)=α(C)=F (from line 2 by ~rule) 2g. 1. α(B & (A v C))=T (given) 2. α(B)=α(A v C)=T (from line 1 by & rule) 3. α(B)=T and at least one of “A” and “C” is true on α (from line 2 by v rule) 4. Either α (B)=α(A)=α(C)=T; or α (B)=α(A)=T and α (C)=F; or α (B)=α(C)=T and α (A)=F (from line 3) 2i. 1. α((A & C) & B)=T (given) 2. α((A & C)=α(B)=T (from line 1 by & rule) 3. α(A)=α(C)=α(B)=T (from line 2 by & rule) 2k. 1. α((B v C) v ~(B v C))=T (given) 2. At least one of α(B v C) and α(~(B v C)) is true (from line 1 by v rule) 3. Either α(B v C)=α(~(B v C))=T; or α(B v C)=T and α(~(B v C))=F or α(B v C)=F and α(~(B v C))=T (from line 2) 4. Either α(B v C)=T and α(B v C))=F; or α(B v C)=T and α(B v C))=T; or α(B v C)=F and α(B v C)=F (from line 3 by ~ rule) 5. α(B v C)=T or α(B v C)=F (from line 4 because by definition of a truth value assignment, the first of the three alternatives at line 4 is unacceptable — no sentence can be both true and false on any truth value assignment — and the other two include redundancies that have simply been eliminated here) 6. Either at least one of “B” and “C” is true on α or α(B)=α(C)=F (from line 5 by v rule) 7. Either α(B)=α(C)=T; or α(B)=T and α(C)=F; or α(B)=F and α(C)=T; or α(B)=α(C)=F (from line 6) Exercise 2k above is interesting for a couple of reasons. You see that what can sometimes happen as you reason back from a truth value assignment to a compound sentence is not just that multiple alternatives open up, but that contradictions can emerge that “close” certain alternatives off. No truth value assignment can assign both a T and an F to the same sentence (atomic or compound) so when your reasoning leads you to that result, it closes off the alternative. This raises an even more interesting possibility. Could it happen that there is no good alternative? The answer is that it could. To take a simple case, there is no way that “A & ~A” could be true on any truth value assignment. Any truth value assignment that assigns T to “A” must assign F to “~A” (by ~rule) and so F to “A & ~A” (by & rule). Any truth value assignment that assigns F to “A” must assign F to “A & ~A.” And all truth value assignments must be of one or other of these two types. So no tva can make “A & ~A” true. So what do we say about those cases? If every alternative way of assign ing a T to a compound sentence closes off, we know the sentence could not be true on any truth value assignment whatsoever (not just on α, the particular one we have been considering). If every alternative way of assigning an F to a compound sentence closes off, we know the sentence could not be false on any truth value assignment whatsoever. (Note that these are only possibilities when reasoning back from truth value assignments to compound sentences. By definition of a tva, any atomic sentence is true on infinitely many tva’s and false on infinitely many others.) As for the alternative of every way of assigning a T closing off and every way of assigning an F closing off, that won’t happen. No truth value assignment can leave any atomic sentences without exactly one truth value, and the connective rules do not permit any compound sentence to have anything other than exactly one truth value, given the assignments made to the components. Consequently, when all the truth values to the atomic components are determined (as they must be by definition on any tva), all the truth values to the compounds made from those atoms are determined (as a consequence of the way the rules work), all the compounds made from those atoms and compounds are determined, and so on up. The possibility of ever arriving at a compound sentence that does not have exactly one of the two truth values is foreclosed. This means that if all the ways of assigning a T close off, every possible way of assigning an F must open up. And when all the ways of assigning an F close off, every possible way of assigning a T opens up. This is the other interesting thing about exercise 2k. The sentence, “(B v C) v ~(B v C)” consists of two atomic components, and we know when there are just two atomic sentences we are concerned with, there are only 2 =4 possible variants on truth value assignments available to consider: the ones that make both “B” and “C” true, the ones that make both false, a nd the ones that make one or the other true and its partner false. And, as you see, all four are represented at line 7 of the exercise. What this tells us is that there is no way that this sentence could be false, because every possible truth value assignment has been found to make it true. And in fact, if we tried to make it false, we would very quickly find that we run into a contradiction at every step, illustrating that there can be no truth value assignment that does this. (The following demonstration adopts the conventions of using “⊥” to designate contradiction, of indenting lines of demonstration that follow from a supposition, and of “outdenting” consequences of a supposition — in this case, consequences of having demonstrated that a supposition leads to an absurdity.) 2k (for false) 1. Suppose: α((B v C) v ~(B v C))=F 2. : α(B v C)=α(~(B v C))=F (from line 1 by v rule) 3. : α(B v C)=F and α(B v C)=T (from line 2 by ~ rule) 4. : ⊥ (from line 3 and the definition of a truth value assignment) 5. α((B v C) v ~(B v C)) cannot =F (from lines 1-4 which demonstrate that the alternative supposition generates a contradiction) 6. α((B v C) v ~(B v C))=T (from line 5 by definition of a truth value assignment) Do the starred exercises in 2.1#2 on your own by considering the assigned sentence to be true on a truth value assignment, α, and then working back to identify all the alternative possible ways α might make assignments to the atomic components. This will help you develop your familiarity with what has just been said. This will not be a huge task. Keep in mind that none of the assigned sentences contains more than three atomic components, so at most 8 possibilities will open up. Usually, there will be considerably fewer. TRUTH FUNCTIONAL PROPERTIES AND RELATIONS Truth functional truth, falsity, and indeterminacy Recall that I said that there are infinitely many truth value assignments. I also said that we do not care which of these truth value assignments is the correct one. We don’t consider there to be any one that is more correct than any other. As we have just seen, many sentences of SL are such that they are true on some truth value assignments and false on others. We call such sentences truth functionally indeterminate. To call such sentences “indeterminate” does not mean that their truth value is not determined. It rather means that it is determined in different ways by different truth value assignments. So, “indeterminate” is perhaps not the best name for what we mean. But there we go. It is the name used in the text, so we will follow that practice here. The atomic sentences of SL are all truth functionally indeterminate. This is because every possible way of assigning one or other but not both of a T and an F to all the atomic sentences of SL is a truth value assignment. That means that, for any given atomic sentence of SL, there is at least one truth value assignment (in fact, there are infinitely many) that makes that sentence true, and at least one (in fact infinitely many) that makes it false. If nothing else, the two truth value assignments that make every atomic sentence of SL true and that make every atomic sentence of SL false will do this. But, as we have also just seen, while every atomic sentence of SL is true on at least one truth value assignment and false on at least one other truth value assignment (in fact, infinitely many of each), there are some compound sentences of SL that are not like this. There are some that are true on every truth value assignment and others that are false on every truth value assignment. We will say that a sentence of SL is truth functionally true if and only if it is true on every truth value assignment. And we will also say that a sentence of SL is truth functionally false if and only if it is false on every truth value assignment. The sentence, A ⊃ A is truth functionally true. We can prove this as we proved the truth functional truth of sentence 2k above: 1. Suppose: there is a truth value assignment, α, such that α(A ⊃ A)=F 2. : α(A)=T and α(A)=F (from line 1 by ⊃ rule) 3. : ⊥ (from line 2 and the definition of a truth value assignment) 4. It is not the case that there is a truth value assignment, α, such that α(A ⊃ A)=F (from lines 1-3 which demonstrate that the alternative supposition generates a contradiction) 5. “A ⊃ A” must be true on every tva, that is, it must be t-f true (from line 4 by definition of t-f truth). Note that it would not do to instead suppose that there is a tva, α, such that α(A ⊃ A)=T — at least not without some careful tweaking. The reason tweaking would be necessary is that we want to show is that “A ⊃ A” is truth functionally true, that is, true on every tva. But α is only one tva. So even if we prove that “A ⊃ A” is true on a tva, α, that is not good enough. T-F indeterminate sentences are also true on a tva (just not on every tva). The only way to prove that a sentence is T-F true is to prove that it is true on every tva. Since there are infinitely many truth value assignments, that would mean either surveying them all, or showing that the opposite, having even one tva that assigns F, is impossible. That is what we did above. Admittedly, in the case where a sentence is built up from just one atomic component, all the infinitely many tva’s reduce to just two: the one’s that assign a T to that component and the ones that assign an F. This opens the possibility of mounting an argument that “A ⊃ A” must be truth functionally true because every possible type of truth value assignment assigns a T to it. Such an argument might run as follows: 1. On any truth value assignment, α, either α(A)=T or α(A)=F (by definition of a tva) 2. Suppose: α(A)=T 3. : then α(A ⊃ A)=T (from line 2 by ⊃ rule — true consequent) 4. Now suppose instead that α(A)=F 5. :then α(A ⊃ A) still =T (from line 2 by ⊃ rule — false antecedent) 6. So, either way, α(A ⊃ A)=T (from lines 1, 2-3 and 4-5) 7. So on any truth value assignment, α, α(A ⊃ A)=T (from lines 1 and 5) 8. “A ⊃ A” must be true on every tva, that is, it must be t-f true (from line 7 by definition of t-f truth). While this argument also works, it would have been considerably longer had we been considering a sentence with two even three (let alone more) atomic components, since then there would have been four or eight (or exponentially more) different cases to consider. That is why showing that there is a contradiction in the opposite assumption (in this case, the assumption that there is a tva that assigns an F) is generally a much more elegant strategy to adopt. In the remainder of this chapter I will consistently adopt this more elegant strategy, however “indirect” it may seem. Rather than argue that a sentence is a certain way, I will argue that if you suppose that it is the opposite way you will run into a contradiction, from which it follows that it cannot be the opposite way. This argumentative strategy is in fact called “indirect proof.” We can use a variant on indirect proof to prove that sentences are T-F false as well as T-F true. For example, the sentence, A ≡ ~A is T-F false. Here is an argument that shows this: 1. Suppose: that there is a tva, α, such that α(A ≡ ~A)=T 2. : Then either α(A)=T and α(~A)=T, or α(A)=F and α(~A)=F (from line 1 by ≡ rule) 3. : But then either α(A)=T and α(A)=F, or α(A)=F and α(A)=T (from line 2 by ~ rule) 4. : So, either way, α must assign two different truth values to A (from line 3) 5. : ⊥ (from line 4 and the definition of a truth value assignment) 6. So there can be no tva, α, such that α(A ≡ ~A)=T (from lines 1-5) 7. So“A ≡ ~A” must be t-f false (from line 6 by definition. of t-f falsity) Proving that a sentence is t-f indeterminate requires an entirely different approach. Rather than start with a truth value assignment to the whole sentence and reason backwards, we try to think up two truth value assignments to the atomic parts: one that will make the compound sentence true and one that will make it false. For example, the sentence, A ⊃ ~A is t-f indeterminate. To prove it, I think up two truth value assignments, one that makes it true, and one that makes it false. I claim that the truth value assignment, α , th1t assigns a T to A, makes the sentence false, and the truth value assignment, α , that2assigns an F to A makes the sentence true. Here are my demonstrations: 1. α (1)=T (given) 2. α (1A)=F (1, ~rule) 3. α (1 ⊃ ~A)=F (1, 2, ⊃ rule: true antecedent but false consequent) 4. α (2)=F (given) 5. α (2 ⊃ ~A)=T (4, ⊃ rule: false antecedent) 6. “A ⊃ ~A” has a different truth value on α tha1 it does on α (fro2 lines 3 and 5) 7. “A ⊃ ~A” is t-f indeterminate (from line 6 by def. of t-f indeterminacy) This was a simple case. In other cases it may be more difficult to think up the truth value assignments that will do each job. In these cases, what we do is start off by assuming that the whole sentence is true and reasoning backwards, just as we do when trying to prove a sentence is t-f false. If the sentence really is t-f indeterminate and not t-f false, then we won’t run into a contradiction. Instead, we will end up figuring out one or more truth value assignments to the atomic components that will make the sentence true. Then we can assume the sentence is false and reason backwards. If the sentence really is t-f indeterminate and not t-f true, we will find one or more tva’s that will make it false. I illustrate one side of this procedure below, in connection with proving t-f non-equivalence. Truth functional equivalence and non-equivalence We can proceed in the same way to define and determine a truth-functional relation that holds between sentences of SL, that of truth functional equivalence. We will say that a pair of sentences of SL is truth functionally equivalent if and only if there is no truth value assignment on which they have different truth values. A pair of sentences of SL is not truth functionally equivalent if and only if there is at least one truth value assignment on which they have different truth values. There is a common misunderstanding of the notion of t-f equivalence that needs to be cleared up right at the outset. When we say that there is no truth value assignment on which two sentences of SL, P and Q, receive different truth values, we just mean that there is no tva on which P and Q have different truth values from one another. We do not mean that there is no tva on which P and Q have different truth values from any other tva. We do not require that P must individually receive the same truth value on every truth value assignment (that it must be either t-f true or t-f false) and that Q must receive this same value on every tva. All that we require is that, on any truth value assignment you happen to pick, P and Q are both the same on that assignment. They might both be true on one, and both false on another. As long as they are both the same on each, the definition has been satisfied. Put in a picture, this suffices for t-f equivalence, P Q α F F 1 α2 F F α3 T T α4 F F α5 T T α6 F F • • • • • • • • • supposing that on each of the infinitely many lines that follow, there are either two T’s or two F’s — supposing there is no line with a T and an F or an F and a T on it. We do not insist on this: P Q P Q α T T α F F 1 1 α 2 T T α 2 F F α 3 T T α 3 F F α 4 T T otrhis: α 4 F F α 5 T T α 5 F F α 6 T T α 6 F F • • • • • • • • • • • • • • • • • • As long as on each row they are both the same, its is good. They do not have to be the same on one row that they are on another. To prove that a pair of sentences is t-f equivalent there are two alternatives that must be ruled out. We first suppose that the first sentence is true and the second one is false and show that this leads to a contradiction. Then we suppose that the second one is false and the first one is true and show that this leads to a contradiction. From this it follows that there is no way the two sentences can have different truth values. For example, the pair of sentences, A ⊃ B ~B ⊃ ~A are t-f equivalent. Here is the proof: 1. Suppose: there is a tva, α, s.t. α(A ⊃ B)=T and α(~B ⊃ ~A)=F 2. : α(~B)=T and α(~A)=F (1, ⊃ rule applied to false ~B ⊃ ~A) 3. : α(B)=F and α(A)=T (2, ~ rule) 4. : α(A ⊃ B)=F (3, ⊃ rule) 5. : ⊥ (from 1 clause of line 1 and line 4 by definition of a tva) 6. So there cannot be a tva, α, s.t. α(A ⊃ B)=T and α(~B ⊃ ~A)=F (1-5) 7. Suppose: there is a tva, α, s.t. α(A ⊃ B)=F and α(~B ⊃ ~A)=T 8. : α(A)=T and α(B)=F (7 ⊃ rule applied to false A ⊃ B) 9. : α(~A)=F and α(~B)=T (8, ~ rule) 10. : α(~B ⊃ ~A)=F (9, ⊃ rule) 11. : ⊥ (2nd clause of 7, 4, def. of tva) 12. So there cannot be a tva, α, s.t. α(A ⊃ B)=F and α(~B ⊃ ~A)=T (7-11) 13. So no tva can assign different truth values to “A ⊃ B” and “~B ⊃ ~A” (from 6 and 12) 14. So they are t-f equivalent (13, def. of t-f equivalence). If a pair is not t-f equivalent, one or other or both of these attempts will fail to lead to a contradiction, and we will end up discovering a truth value assignment on which the two sentences have different truth values. For example, the sentences, ~(A v B) ~A v ~B are not t-f equivalent. If we try to prove that they are equivalent, we won’t find the contradictions we are looking for. In this case, if we suppose that the first sentence is true and the second sentence is false, we will run into a contradiction. 1. Suppose: there is a tva, α, s.t. α(~(A v B))=T and α(~A v ~B)=F 2. : α(~A)=F and α(~B)=F (1, v rule applied to false ~A v ~B) 3. : α(A)=T and α(B)=T (2, ~ rule) 4. : α(A v B)=T (3, v rule) 5. : α(~(A v B)=F (4, ~rule) 6. : ⊥ (from 1 clause of line 1 and line 4 by definition of a tva) But, interestingly, we don’t find a contradiction if we suppose that the first sentence is false and the second sentence is true. 1. Suppose: there is a tva, α, sst. α(~(A v B))=F and α(~A v ~B)=T 2. : α(A v B)=T (1 clause of line 1, ~ rule) 3. : α assigns a T to at least one of “A” and “B” (2, v rule) 4. Suppose: α(A)=T but α(B)=F 5. : α(~B)=T (4, ~ rule) 6. : α(~A v ~B)=T (5, v rule) This argument is incomplete. But as far as it goes, it has uncovered an assignment that α could make to all the atomic components of the two sentences that makes one false and the other true without running into the contradiction we were looking for. This suggests that any truth value assignment that assigns T to “A” and F “B” will also assign different truth values to “~(A v B)” and “~A v ~B.” Since there is at least one truth value assignment on which these sentences receive different truth values, and they are not t-f equivalent. To rigorously prove this, I take the truth value assignment my earlier, failed attempt at proving equivalence led me to discover, and show by appeal to the semantic rules that it do es indeed lead, without producing any contradictions, to an assignment of F to “~(A v B)” and T to “~A v ~B.” Here is that demonstration: 1. α(A)=T (given) 2. α(A v B)=T (1, v rule) 3. α(~(A v B))=F (3, ~ rule) 4. α(B)=F (given) 5. α(~B)=T (4, ~rule) 6. α(~A v ~B)=T (5, v rule) 7. While “~(A v B)” is false on α “~A v ~B” is true on α (from lines 3 and 6) 8. “~(A v B)” and “~A v ~B” are not t-f equivalent (from line 7 by def. of t-f non- equivalence) Truth functional inconsistency and consistency A set of sentences of SL is truth functionally inconsistent if and only if there is no truth value assignment that assigns a T to every sentence in that set. A set of sentences of SL is truth functionally consistent if and only if there is at least one truth value assignment that assigns a T to every sentence in that set. Once again, there are common misunderstandings that need to be forestalled at the outset. For a set to be t-f inconsistent we only require that there be no tva that assigns a T to every sentence in the set. We do not require that there be a tva that assigns an F to each set member. That is a far stronger demand than what we are looking for. We only require that each tva assign an F to at least one set member. Moreover, we do not require that it be the same set member for each tva that gets the F. A set does not have to contain a t-f false sentence in order to be t-f inconsistent (though that certainly suffices). It just has to be the case that one or other of the set members is false on each tva. In determining truth-functional inconsistency or consistency we are concerned with one and only one thing: the presence or absence of a truth value assignment that assigns a T to every sentence in the set. Other truth value assignments are of absolutely no interest to us. In particular, truth value assignments that assign an F to every sentence in the set are of absolutely no interest to us. So, given the set, {P, Q, R} of sentences of SL, and the following truth value assignments, α -1 t7 set the members, P Q R α 1 F F F α T F T 2 α 3 F T T α 4 F T F α T T T 5 α 6 T F F α 7 T T F • • • • • • • • • • • • we declare the set to be t-f consistent. The fact that there is a truth value assign1ent, α , that makes all the sentences in the set false is absolutely irrelevant to our decision.
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