The Soundness of SD
THE SOUNDNESS AND CONSISTENCY OF SD AND THE EXTENSIONALITY OF THE SENTENCES OF SL Up to now, we have treated the derivation system, SD, as an arbitrarily constructed system of rules to be used to derive sentences from other sentences. In fact, the rules were not arbitrarily constructed. They were designed to serve as a means for demonstrating the truth-functional validity of arguments. Each of the “straight” derivation rules was designed with the thought in mind that, if the sentences being appealed to as justifications for the rule application are true, the sentence derived in accord with that rule must be true as well. And the derivation system as a whole, with its restrictions on accessibility and closure, was designed to ensure that whatever sentences can be derived from a set of assumptions (i.e., occur under the scope just of those assumptions) must be truth-functionally entailed by those assumptions. It is not immediately obvious that this goal is in fact achieved by the rules of SD. Students who begin working with derivations under the disadvantage of already knowing something about the semantics of sentences of SL will often be suspicious about some of the rules. So a proof that SD really is sound — that if a sentence is derivable just under the scope of a set of assumptions using the rules of SD, then it is truth functionally entailed by that set of assumptions — is mandated. Soundness: If a sentence, P, of SL is derivable in SD from a set, Γ, of sentences of SL, then Γ truth-functionally entails P. We would also like to be assured that SD is consistent: that you cannot in fact (appearances notwithstanding) derive just anything using the rules of SD. Consistency: There is at least one sentence that is not a theorem of SD. As well as being assured that SD is sound, we would like to be assured that SD+ is sound. Proving this last point requires that we establish an important prior point concerning the sentences of SL: that they are extensional. The soundness of SD The strategy behind the soundness proof The soundness proof Proof of the basis clause Proof of the inductive step Case one: the assumption rule Case two: the reiteration rule Case three: ⊃I Cases four and five: ~E and ~I Cases six and seven: vE and ≡I Cases eight to twelve: the straight rules Corollary results: Theorems of SD are t-f true sentences of SL Sentences that are interderivable in SD are t-f equivalent Sets that are contradictory in SD are t-f inconsistent The consistency of SD
The soundness of SD+ Appendix: Proof of the extensionality of the sentences of SL Proofs of Metatheorems
STRATEGY FOR PROVING THE SOUNDNESS OF SD. If SD consisted just of the “straight” rules (⊃E, ≡E, &E, &I, vI, and the straight versions of vE, ~E, ~I and ≡I originally given in the SD notes), then proving its soundness would be easy. Each of the straight rules is truth preserving. Each rule specifies up to three sentences that you need to have in order to derive a further sentence, and it is easy to prove that, for each rule, if the sentences you need to have are all true on a truth value assignment, α, then the sentence the rule allows you to derive must be true on α as well. Consider, for example, the straight version of the vE rule. It specifies that from P v Q, P ⊃ R, and Q ⊃ R, you can derive R. And, as it turns out, if the three sentences the rule requires you to have are true on a tva, α, then there is no way that the sentence it allows you to derive, R, can be false on α. To see why, suppose P v Q, P ⊃ R, and Q ⊃ R are all true on a tva, α. Since α(P v Q)=T, either α(P)=T or α(Q)=T. But if α(P)=T then, since α(P ⊃ R)=T, α(R)=T. And if α(Q)=T then, since α(Q ⊃ R)=T, α(R)=T. So, either way, α(R)=T. The same case can be made for each of the straight rules. It follows that if you start off a straight derivation in SD with sentences that are all true on a truth value assignment, α, then this truth must get carried down to the all the sentences derived from them using the straight rules, to all the sentences derived from those sentences using the straight rules, and so on down to the sentence on the last line of the derivation (supposing the derivation has remained “straight” — i.e., no sub-derivations have been introduced). So, by mathematical induction, if all the initial assumptions of a straight derivation in SD are true on a tva, α, then the last line of that derivation must be true on α as well. So the set of initial assumptions truth functionally entails the sentence on the last line (because there is no way the set of initial assumptions could be true on a truth value assignment and the sentence on the last line be false on that assignment). So it follows that if a sentence, P, of SL is derivable using the straight rules of SD from a set, Γ, of sentences of SL, then Γ truth-functionally entails P. So the system of “straight” rules of SD is sound. Unfortunately, the system of straight rules is not also complete. While we can trust that any sentence derived using the straight rules must be t-f entailed by the set of initial assumptions, not all sentences that are truth functionally entailed by sets can be derived from those sets using just the straight rules. Most notably, no sentence that is t-f entailed by the empty set (no tf true sentence) is derivable using the straight rules. Each of the straight rules requires you to have some sentences to start with, so none can be applied when there are no initial assumptions. SD becomes complete when we add the sub-derivation rules to it. But the addition of these rules significantly complicates our effort to prove that SD is sound. We can no longer argue that if the sentences higher up in a derivation are true on a tva, α, then the sentences lower down must be true on α. This is because the sub-derivation rules allows us to assume sentences that bear no deductive relation to the sentences higher up on the tree and so cannot be shown to be true if those sentences are true.
This feature of the sub-derivation rules might lead us to wonder whether SD really is sound, or whether we might not have gone too far and made it unsound in our efforts to make it complete. We haven’t, but the addition of the sub-derivation rules does make it harder to prove that SD is sound. Since truth on α need not be carried down to the assumptions heading off sub-derivations, we must look for a different way to prove the soundness of SD. Fortunately, there is another such way. Though truth on a tva is not transmitted from line to line of a derivation in SD there is another property that is, and this other property serves just as well to establish the soundness of SD: the property of being truth functionally entailed by the set of open assumptions at that line. Recall that an open assumption is simply an assumption that has a continuing scope line. So whether an assumption is open or not depends on what line you are at in a derivation. All assumptions are open on the line where they are first made. And on each subsequent line, as long as their scope lines continue down to that line, they remain open. But beginning with the first line on which their scope lines terminate, they are considered to have been closed. So at each line, call it line k, in a derivation there is a (possibly empty) set, Γk, that we can call the set of open assumptions at that line. To determine the contents of this set you simply look up each scope line that is continuing at line k to see what assumptions are heading off that scope line. For example, in the following derivation scheme: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.
P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 P14 P15 P16
Γ1, the set of open assumptions at line 1, is {P1}, Γ2 is {P1, P2}, Γ3 is {P1, P2, P3}, Γ4 is {P1, P2, P3, P4}, Γ5 is {P1, P2, P3, P4}, Γ6 is {P1, P2, P3, P4}, Γ7 is {P1, P2, P3, P4, P7}, Γ8 is {P1, P2, P3, P4, P7}, Γ9 is {P1, P2, P3, P4, P7}, Γ10 is {P1, P2, P3, P4},
Γ11 is {P1, P2, P3,}, Γ12 is {P1, P2, P3, P12}, Γ13 is {P1, P2, P3, P12}, Γ14 is {P1, P2, P3, P12}, Γ15 is {P1, P2, P3}, and Γ16 is {P1, P2, P3}.
THE SOUNDNESS PROOF The crucial claim made by the soundness proof is that the rules of SD were designed in such a way as to ensure that if, at each previous line, the sentence on that line is truthfunctionally entailed by the set of assumptions that are open at that line, then any sentence that the rules permit us to write down on the next line must be such that it is truth-functionally entailed by the set of assumptions that remain open at that next line. If we combine this crucial claim with the claim that on the first line of any derivation in SD the sentence on that line must be truth functionally entailed by the set of open assumptions on that line, then we are in a position to argue, by mathematical induction on the lines in a derivation, that the sentence on the last line of any derivation in SD must be t-f entailed by the set of open assumptions on that line. To run this argument we consider the sentences in a derivation in SD to be ordered by their line numbers. So P1 is the sentence on line 1 of a derivation in SD, P2 is the sentence on line 2, P3 is the sentence on line 3, and so on. We also consider Γ1 to be the set of open assumptions at line 1, Γ2 the set of open assumptions at line 2, Γ3 to be the set of open assumptions at line 3, and so on. Here is the argument: Basis Clause: In any derivation in SD, Γ1 t- f entails P1. Inductive Step: If each of Γ1- Γk t- f entails each of P1- Pk respectively, then Γk+1 must t- f entail Pk+1. Conclusion: On the last line of any derivation in SD, line n, Γn must t- f entail Pn. Supposing that we can justify the basis clause and the inductive step, we need merely combine this result with the definition of derivability in SD to complete our soundness proof. According to the definition of derivability in SD, a sentence, P, of SL has been derived from a set, Γ, of sentences of SL if and only if P is derived on a line that has only a subset of the sentences in Γ in its set of open assumptions. (This is what it means to say that P occurs under the scope just of the sentences in Γ.) So if P has been derived from Γ in SD, then a subset of Γ must constitute the set of open assumptions at the line where P is derived. But then, by the conclusion just reached above, Γ must truth-functionally entail P. (If P is derived from a proper subset of Γ, that is, a set that does not contain all the sentences in Γ, then it will still be derived from Γ, because adding extra assumptions to the head of a derivation does nothing to change what can be derived from the prior assumptions.) So if P is derivable from a subset of Γ, Γ must t-f entail P. So SD is sound.
PROOF OF THE BASIS CLAUSE To fill out this argument sketch we need to prove the basis clause and the inductive step. The proof of the basis clause is straightforward. On the first line of any derivation in SD, the sentence on that line, P1, will have to be an assumption. (It cannot be a derived line since there is no rule of SD that allows you to derive a line without appealing to some earlier line as justification.) But if P1 is an assumption, then it is itself a member (in fact, the only member) of Γ1. (Γ1 is always {P1}.) And it is easy to prove that if a sentence, P, of SL is a member of a set, Γ, of sentences of SL then Γ truth functionally entails P. “A” metatheorem: if a sentence, P, of SL is a member of a set, Γ, of sentences of SL then Γ truth functionally entails P. (Click here for the proof.)
PROOF OF THE INDUCTIVE STEP It remains to prove the inductive step. Since any subsequent line of a derivation must either be a further assumption or must be justified by one of the 11 derivation rules, proving the inductive step means proving that, if all the previous lines of a derivation in SD, lines 1-k, have been truth functionally entailed by the assumptions that are open at those lines, then whether the next line, line k+1, is entered as an assumption or whether it is entered in accord with one of the 11 derivation rules, it must be truth functionally entailed by Γk+1, the set of assumptions that are open at that next line. The proof proceeds as follows: We first suppose the inductive hypothesis: Suppose that at each of lines 1-k of a derivation in SD the sentence on that line is t-f entailed by the set of open assumptions on that line. Now we prove that, under this supposition, whichever derivation rule we use to derive the sentence on line k+1, that sentence must be t-f entailed by the set of open assumptions at line k+1.
Case 1: Assumption If the sentence on line k+1, call it P, is an assumption then it is itself a member of the set of open assumptions at that line, call it Γk+1. So, by the argument used to prove the basis clause, Γk+1 must t-f entail P.
Case 2: Reiteration The following schema outlines the reiteration rule of SD:
i. P · · … … · k+1.
· · · P
i, R
For P to be derived at line k+1 by the R rule of SD, there must be some earlier occurrence of P on some earlier line, i. Moreover, the earlier occurrence of P at line i must be accessible at line k+1, which means that its scope line must continue unbroken down to line k+1. Since derivations cannot close without all the sub-derivations they contain being closed, all the scope lines to the left of P at line i must also continue unbroken down to line k+1. (If they didn’t the scope line immediately to the left of P couldn’t continue unbroken down to line k+1.) Consequently, no assumption that is open at line i can be closed at line k+1, which means that Γk+1 must be a superset of Γi. (A set equal to or larger than Γi.) By the inductive hypothesis, Γi t-f entails P. But then it follows by the R metatheorem of SL that Γk+1 must also t-f entail P. “R” metatheorem: if a set of sentences of SL truth functionally entails a sentence of SL then any superset of that set must also truth functionally entail that sentence. (Click here for the proof of this metatheorem.)
Case 3: ⊃I As a preliminary to the proof of this case, consider the following schema for any application of the ⊃I rule: θ
i. … k. k+1.
Q . . . R Q⊃R
For Q ⊃ R to be derived at line k+1 by the ⊃I rule of SD, there must be an earlier subderivation running from lines i-k that is headed off by the assumption of the antecedent, Q, at line i and that yields the consequent, R, at line k. Moreover, the sub-derivation that yields R at line k must be accessible at line k+1, which means that it must be closed at line k+1, and that the scope line immediately to its left (here labelled θ) must continue unbroken down to line k+1. Since derivations cannot close without all the sub-derivations they contain being closed, all the scope lines to the left of θ must also continue unbroken down to line k+1. (If they didn’t, θ couldn’t continue unbroken down to line k+1.) Consequently, no sub-derivation that is open at line k can be closed at line k+1, except for the sub-derivation headed off by Q at line i, which as we have said must be closed. So the set of open assumptions at line k, Γk, must be identical to Γk+1 ∪ {Q}. By the inductive hypothesis, Γk t-f entails R. Hence, since Γk and Γk+1 ∪ {Q} are identical, Γk+1 ∪ {Q} t-f entails R. But then it follows by the ⊃I metatheorem that Γk+1 t-f entails Q ⊃ R. ⊃I metatheorem: if Γ ∪ {Q} truth-functionally entails R, where Γ is a set of sentences of SL and Q and R are sentences of SL, then Γ truth-functionally entails Q ⊃ R. You were asked to prove this metatheorem previously as exercise #2b in Chapter 3.6. (Click here for the proof.)
Case 4: ~I To prove the ~I case, I am going to appeal to two metatheorems, which I will call the “X” metatheorem and the ~I metatheorem. X metatheorem: If a set, Γ, of sentences of SL truth-functionally entails both a sentence, P, of SL and ~P then Γ is truth functionally inconsistent. ~I metatheorem: If, for a set, Γ of sentences of SL and a sentence, P, of SL, Γ ∪ {P} is truth-functionally inconsistent, then Γ truth-functionally entails ~P. You can click here for the proof of these two metatheorems. (The first of them was proven in exercise 3.6 #3b.) I also want to appeal to the following schema for any application of the ~I rule:
i.-1 i.
P . . .
j.
Q . . . k. ~Q k+1. ~P Note that lines j and k in the above schema could have occurred in the reverse order. For ~P to be derived by ~I on line k+1, there must be an earlier sub-derivation running from lines i-k that is headed off by the assumption of P at line i, and that yields Q and ~Q immediately to the right of the scope line for this assumption at line k and some line, j, intermediate between i and k. Moreover, this sub-derivation must be accessible at line k+1, which means that it must be closed at line k+1, and that the scope line immediately to its left must continue unbroken down to line k+1. For reasons similar to those given when discussing case 3, this means that Γj and Γk are both identical to Γk+1 ∪ {P}. Since lines j and k are prior to line k+1, it follows by the inductive hypothesis that Γj and Γk and hence Γk+1 ∪ {P} t-f entail both Q and ~Q. But then, by the “X” metatheorem, Γk+1 ∪ {P} must be t-f inconsistent, and, by the ~I metatheorem, Γk+1 must t-f entail ~P.
Case 5: ~E The proof of case 5 is similar to the proof of case 4. Where the proof of case 4 appeals to the ~I metatheorem, the proof of case 5 appeals to the following ~E metatheorem: ~E metatheorem: If, for a set, Γ, of sentences of SL and a sentence, P, of SL, Γ ∪ {~P} is truth-functionally inconsistent, then Γ truth-functionally entails P. The proof of this metatheorem is almost identical to that of the ~I metatheorem.
Cases 6 and 7: vE and ≡I The proof of these cases is modeled on that of case 3 and invokes the following metatheorems:
vE metatheorem: If Γ t-f entails P v Q, Γ ∪ {P} t-f entails R, and Γ ∪ {Q} t-f entails R, then Γ t-f entails R. ≡I metatheorem: If Γ ∪ {P} t-f entails Q and Γ ∪ {Q} t-f entails P, then Γ t-f entails P ≡ Q. You should be able at this point to reconstruct the proof for yourself, or you can consult chapter 6.2 or the proofs at the end of this chapter.
Cases 8-12: The straight rules The proof of the straight rules is modeled on the proof of case 2 and invokes the following metatheorems, together with the “R” metatheorem: & metatheorem: Γ truth functionally entails (P & Q) if and only if Γ truth functionally entails both P and Q. vI metatheorem: If Γ truth functionally entails P then Γ truth functionally entails either one of (P v Q) and (Q v P). ⊃E metatheorem: If Γ truth functionally entails (P ⊃ Q) and Γ truth functionally entails P then Γ truth functionally entails Q. ≡E metatheorem: If Γ truth functionally entails (P ≡ Q) and Γ truth functionally entails either one of P or Q, then Γ truth functionally entails the other one as well. Supposing each of these metatheorems is true, we can argue along the following lines (taking the ⊃E case as an example): i. Q ⊃ P (or Q) · · · j. Q (or Q ⊃ P) · · · k+1. P i., j., ⊃E For P to be derived by ⊃E on line k+1, the derivation must contain at least two prior lines, call them lines i and j. One of these lines must have a sentence of the form Q ⊃ P on it, and the other must have a sentence of the form Q on it, though it does not matter which is which. Moreover, both of these lines must be accessible, which means that Γj must be a superset of Γi and Γk+1 must be a superset of Γj and hence of Γi. (That is because all scope lines that exist at line i must continue unbroken down to line j, and all scope lines that exist at lines i and j must continue unbroken down to line k+1.) Since lines i and j are prior to line k+1 it follows by the inductive hypothesis that Γi t-f entails Q ⊃ P (or Q as the case may be) and Γj t-f entails Q (or Q ⊃ P as the case may be). So, by the R metatheorem, and the fact that Γk+1 is a superset of both Γi and Γj, Γk+1 t-f entails both Q ⊃ P and Q. So, by the ⊃E metatheorem, Γk+1 t-f entails P.
The proof of the remaining metatheorems and the application of each of them to prove its case is left to you as an exercise. (You can also consult the text or the end of this chapter.) Once it has been proven that, under the supposition of the inductive hypothesis, Γk+1 must t-f entail the sentence on line k+1 in each of these cases, the inductive step has been proven, and it then follows by mathematical induction that the last sentence of any derivation in SD must be t-f entailed by the set of open assumptions on that line. If that set is the empty set, then the sentence is t-f entailed by the empty set. If that set is the set of premises of an argument, then the set of those premises t-f entails that sentence, and the argument is t-f valid.
SOME COROLLARY RESULTS Now that I have proven that SD is sound, it is easy to prove that SD also gives us reliable proofs of the truth functional truth of sentences, the truth functional equivalence of pairs of sentences, and the truth functional inconsistency of sets. Corollary 1. Suppose that a sentence, P, is a theorem of SD. Then P derivable in SD from the empty set. Because SD is sound, it follows that the empty set t-f entails P. But it was proven in Chapter 3.6 exercise #2a that if the empty set t-f entails P, then P must be t-f true. So if P is a theorem of SD, then P is t-f true. Corollary 2. Suppose that P and Q are equivalent in SD. Then because SD is sound it follows that {P} t-f entails Q and {Q} t-f entails P. But then by the answer to exercise 3.5 #5b it follows that P and Q must be t-f equivalent. So if P and Q are equivalent in SD, P and Q must be t-f equivalent. Corollary 3. Suppose Γ is inconsistent in SD. Then there is some sentence, P, such that both P and ~P can both be derived from Γ in SD. Because SD is sound this means that Γ truth functionally entails both P and ~P. But then by the inconsistency metatheorem (discussed above in connection with case 4), Γ must be t-f inconsistent. So if Γ is inconsistent in SD, Γ must be t-f inconsistent.
THE CONSISTENCY OF SD As noted when discussing the consistency of the tree method, the term “consistency” is used in two senses. In one sense it names a property of sets of sentences. (A set of sentences is logically consistent if and only if there is no contradiction in affirming all of the sentences it contains; it is truth-functionally consistent if and only if there is a line on the truth table that assigns a “T” to every sentence in the set; and it is consistent in SD if and only if there is no sentence P such that both P and ~P are derivable from that set in SD.) In a second sense, consistency is a property of derivation systems, like the tree method and SD. A derivation system is said to be consistent if and only if for any sentence, P, both P and ~P are not theorems of system.
SD is consistent in this sense. To see why, let’s suppose the opposite and see what happens. Suppose that for some sentence, P, both P and ~P are theorems of SD. We just showed (Corollary 1 above) that any sentence that is a theorem of SD is t-f true. So it follows that P is t-f true. So it is true on every tva. But then, by ~rule, ~P must be false on every tva, and so false on at least one tva. So ~P can’t be t-f true. But we just supposed that ~P is also a theorem of SD, which means that it must be t-f true. So we have been led into a contradiction. Since we are led into an absurd result by supposing that for some sentence, P, both P and ~P are derivable from the empty set in SD, we can deny that supposition. So there is no sentence that is such that both it and its negation are theorems of SD. So SD is consistent.
THE SOUNDNESS OF SD+ The derivation system, SD+, consists of the rules of SD, three further derivation rules, MT, HS and DS, and the rules of replacement. We already know that the rules of SD are sound. And we do not need to do any further work to prove that MT, HS, and DS are sound. That is because these rules merely allow us to abbreviate a routine sequence of steps that we can otherwise perform using just the rules of SD. So wherever MT, HS, or DS are used to derive a sentence, P, we could instead run through the routine sequence of steps and derive P using just rules of SD. Since nothing can be derived using MT, HS, or DS that can’t be derived just using the rules of SD, and we know that the rules of SD are sound, it follows that MT, HS, and DS must be sound. However, the rules of replacement do not simply abbreviate routine sequences of steps in SD. When we use these rules we often replace one component of a sentence with a different sentential component. Now that we have proven that SD is sound, we can appeal to that result together with a further result, the extensionality metatheorem (proven in exercise 6.1#1e, or at the end of these notes) to prove that the rules of replacement are also sound. The proof appeals to a further fact that you should be able to verify for yourself (and that was set as an assignment in exercise 5.3#9): that each of the rules of replacement is an equivalence in SD, so that the sentence on the right side of the rule is derivable from the sentence on the left side, and the sentence on the left side derivable from the one on the right side. Here is the proof of the soundness of the rules of replacement. Proof: The rules of replacement hold between sentences that are equivalent in SD. So for each rule, P ⇔ Q, {P} yields Q in SD and {Q} yields P in SD. By Corollary 2 of the proof of the soundness of SD this means that P and Q must be t-f equivalent. But the sentences of SL are all extensional, that is, if P and Q are t-f equivalent, then R is t-f equivalent to [R] (P//Q), where “[R] (P//Q)” refers to the result of replacing one or more occurrences of Q in R with occurrences of P. It follows that if P ⇔ Q is a rule of replacement of SD+ and R is true on a tva, α, then α must assign a T to [R] (P//Q). Consequently, applying the rules of replacement cannot lead us from true sentences to false ones. So the rules of replacement are sound.
Appendix: PROOF OF THE EXTENSIONALITY OF THE SENTENCES OF SL In what follows, I prove that the sentences of SL are extensional.
That is, if P is t-f equivalent to Q, then replacing one or more occurrences of Q with occurrences of P in any longer sentence, R, containing Q will not change the truth value of the sentence. More precisely put, if P is t-f equivalent to Q, then any sentence, R, containing Q as one of its components, is t-f equivalent to [R](P//Q), where “[R](P//Q)” refers to the result of replacing one or more occurrences of Q in R with occurrences of P. (E.g., if “A” and “~~A” are tf equivalent, then “A & B” is t-f equivalent to “~~A & B.”) I give this proof by mathematical induction on the sentences of SL grouped by the number of occurrences of connectives in those sentences. So in the first group are all the sentences with 0 occurrences of connectives (the atomic sentences). In the second group are all the sentences with one connective, e.g., “~A” and “A & B.” In the third group are all the sentences with two occurrences of connectives, e.g., “~~A,” “~A v B.” And so on. The proof goes as follows: Basis Clause: All sentences of SL with 0 occurrences of connectives are extensional. Inductive Step: If all sentences of SL with 0-k occurrences of connectives are extensional, then all sentences with k+1 occurrences of connectives are extensional. Conclusion: All the sentences of SL are extensional. Proof of the Basis Clause: The sentences with 0 occurrences of connectives are the atomic sentences. But in the case of atomic sentences the following chain of consequences holds: 1. R must be the same sentence as Q. 2. [R](P//Q) must be the same sentence as P. 3. R must be t-f equivalent to [R](P//Q), that is, R is extensional. (1.) follows because atomic sentences have no components other than themselves. So the only way that a sentence, Q, can be a component of R when R is atomic is if Q are R are one and the same sentence. (2.) follows from (1.) because, given that R just is Q, the result, [R](P//Q), of putting P in the place of any occurrence of Q in R is equivalent to replacing R as a whole with P. (3.) follows directly from (1.) and (2.) because we are supposing that P and Q are t-f equivalent. So, if R has 0 occurrences of connectives, it is extensional. Proof of the Inductive Step: Suppose that the inductive hypothesis is true: All the sentences of SL with 0-k occurrences of connectives are extensional. Then there are two possibilities to consider: Q may just be R (as when, for example, Q and R are “~A” and P is “~~~A”), or Q may be a proper component of R. In the first of these cases, R must be extensional for the reason already given when proving the basis clause. In the second case, R must either be the result, ~S, of prefacing some proper component, S, with a “~,” or it must be the result of connecting two proper components, S and T, with a twoplace connective. Case 2a: If R is of the form ~S then, because S has only k connectives, it falls under the inductive hypothesis. So S must be extensional. This means that if P is t-f equivalent to Q, then S is t-f equivalent to [S](P//Q). But then on any truth value assignment either S and [S](P//Q) are both true, in which case ~S and ~[S](P//Q) are both false, or S and [S](P//Q) are both false, in which case ~S and ~[S](P//Q) are both true. So either way, ~S and ~[S](P//Q) must both have the same truth value. So they must be t-f equivalent. So if R is a sentence of the form ~S, then it must be extensional.
Case 2b: If R is the result of connecting two proper components, S and T, with a twoplace connective, then there are two sub-cases to consider: the proper component, Q, in R that gets replaced by P could be either a) one or both of the immediate components of R (i.e., S or T), or b) one or more proper components of R other than R itself or its immediate components. Case 2b(i): In this case [R](P//Q) is either [R](P//S) or [R](P//T) or both. But the truth value of S & T, S v T, S ⊃ T, and S ≡ T is in each case determined exclusively by the truth value of its immediate components. Consequently if P is t-f equivalent to S or to T, any substitution of P for either of S or T will not change the truth value of the resulting sentence on any truth value assignment, since putting P for S or T does not result in a change to the truth value of the component. So R must be t-f equivalent to [R](P//Q). So if the component, Q, in R that gets replaced by P is either or both of its immediate components, then R is extensional. Case 2b(ii): In this case since R has k+1 occurrences of connectives, each of its immediate components, S and T must contain k or fewer occurrences of connectives (so that the sum of the occurrences of connectives in both is exactly k). But then, by the inductive hypothesis, each of S and T is extensional. By definition, that means that if P is t-f equivalent to Q, S must be t-f equivalent to [S](P//Q) and T must be t-f equivalent to [T](P//Q). Consequently, regardless of whether P is put for Q in S, T, or both, the result will be a sentence with a left immediate component that is t-f equivalent to S (if only S itself, in the case where no substitution is made) and a sentence with a right immediate component that is t-f equivalent to T (if onlyT itself in the case where no substitution is made). But then, since the truth value of S & T, S v T, S ⊃ T, or S ≡ T is in each case determined exclusively by the truth value of its immediate components, R must be t-f equivalent to [R](P//Q). So if the component, Q, in R that gets replaced by P is one of the components of R other than R itself or one of its immediate components, then R is extensional. So, If R is the result of connecting two proper components, S and T, with a two-place connective, then R must be extensional. So, either way, if Q is a proper component of R, then R must be extensional. So, either way, if all the sentences of SL with 0-k occurrences of connectives are extensional, and R is any sentence with k+1 occurrences of connectives it must be extensional.
PROOFS OF METATHEOREMS 1. “Assumption” metatheorem: if a sentence, P, of SL is a member of a set, Γ, of sentences of SL then Γ truth functionally entails P. Proof: If P is a member of Γ then any truth value assignment that makes all the sentences in Γ true must make P true. So there is no truth value assignment that makes all the sentences in Γ true and P false. So Γ truth functionally entails P. Back 2. “R” metatheorem: if Γ is a set of sentences of SL that truth functionally entails a sentence, P, of SL and Γ′ is a superset of Γ (a set that contains all the sentences in Γ and possibly other sentences of SL as well), then Γ′ must also truth functionally entail P. Proof [This is a corollary of 3.6E #4c]:
Suppose that some set, Γ, of sentences of SL truth functionally entails a sentence, P, of SL. Now consider a superset, Γ′, of Γ. Since Γ′ includes all the sentences in Γ, any truth value assignment that makes all the sentences in Γ′ true must make all the sentences in Γ true. But, since we are supposing that Γ t-f entails P, no truth value assignment that makes all the sentences in Γ true can make P false. So, no truth value assignment that makes all the sentences in Γ′ true can make P false. So Γ′ must truth-functionally entail P. Back 3. ⊃I metatheorem: if Γ ∪ {P} truth-functionally entails Q, where Γ is a set of sentences of SL and P and Q are sentences of SL, then Γ truth-functionally entails P ⊃ Q. Proof [Proven as part of 3.6E #2b]: Suppose that Γ ∪ {P} truth-functionally entails Q. Then there is no truth value assignment on which all the sentences in Γ ∪ {P} are true and Q is false. So there is no truth value assignment on which all the sentences in Γ are true, P is true, and Q is false. So on any truth value assignment on which all the sentences in Γ are true, P cannot be true and Q false. So, on any truth value assignment on which all the sentences in Γ are true, P ⊃ Q must be true. So Γ truthfunctionally entails P ⊃ Q. Back 4. X metatheorem: If a set, Γ, of sentences of SL truth-functionally entails both a sentence, P, of SL and a sentence, ~P, of SL then Γ is truth functionally inconsistent. Proof [Proven as 3.6E #3b]: Suppose that Γ t-f entails both P and ~P. We know from the definition of t-f entailment that if a set t-f entails a sentence, then on any truth value assignment on which that sentence is false, the members in the set cannot all be true. But exactly one of P or ~P must be false on every tva (by ~ rule). It follows that at least one of the sentences in Γ must be false on every tva. So Γ must be t-f inconsistent. Back 5. ~I metatheorem: If, for a set, Γ of sentences of SL and a sentence, P, of SL, Γ ∪ {P} is truth-functionally inconsistent, then Γ truth-functionally entails ~P. Proof: Suppose that Γ ∪ {P} is truth-functionally inconsistent. Then on any truth value assignment on which all the sentences in Γ are true, P must be false. So, on any truth value assignment on which all the sentences in Γ are true, ~P must be true. So Γ truth-functionally entails ~P. Back 6. ~E metatheorem: If, for a set, Γ of sentences of SL and a sentence, P, of SL, Γ ∪ {~P} is truth-functionally inconsistent, then Γ truth-functionally entails P. Proof [Proven as 3.6E #1c]: The proof is identical to that of the ~I metatheorem replacing occurrences of P with ~P, and occurrences of ~P with P. Back 7. ≡I metatheorem: If Γ ∪ {P} t-f entails Q and Γ ∪ {Q} t-f entails P, then Γ t-f entails P ≡ Q. Proof: Suppose that Γ ∪ {P} t-f entails Q and Γ ∪ {Q} t-f entails P. Then on any truth value assignment on which all the sentences in Γ are true, it cannot be the case that all the sentences in {P} are
true and Q is false, and it cannot be the case that all the sentences in {Q} are true and P is false. But P is the only sentence in {P} and Q is the only sentence in {Q}. So, on any truth value assignment on which all the sentences in Γ are true, it cannot be the case that P is true and Q is false, and it cannot be the case that Q is true and P is false. So, on any truth value assignment on which all the sentences in Γ are true, P and Q cannot have different truth values, which means that P ≡ Q must be true. But in that case, Γ t-f entails P ≡ Q. 8. v metatheorem: If Γ t-f entails P v Q, Γ ∪ {P} t-f entails R, and Γ ∪ {Q} t-f entails R, then Γ tf entails R. Proof: Suppose Γ t-f entails P v Q, Γ ∪ {P} t-f entails R, and Γ ∪ {Q} t-f entails R. Then on any truth value assignment, α, on which all the sentences in Γ are true, P v Q must be true, which means either P must be true or Q must be true. Suppose P is true on α. Then, because P is the only sentence in {P}, all the sentences in Γ ∪ {P} are true on α. But then R must be true on α. Now suppose it is instead Q that is true on α. Then, because Q is the only sentence in {Q}, all the sentences in Γ ∪ {Q} are true on α. But then R must again be true on α. So either way, R is true on α. So, on any truth value assignment, α, on which all the sentences in Γ are true, R is true. So, on our assumptions, Γ t-f entails R. That is, if Γ t-f entails P v Q, Γ ∪ {P} t-f entails R, and Γ ∪ {Q} t-f entails R, then Γ t-f entails R. 9. & metatheorem: Γ truth functionally entails (P & Q) if and only if Γ truth functionally entails both P and Q. Proof: Γ t-f entails P & Q if and only if, on any truth value assignment on which all the sentences in Γ are true, P & Q is true. But P & Q is true if and only if both P and Q are true. So Γ t-f entails P & Q if and only if, on any truth value assignment on which all the sentences in Γ are true, both P and Q are true. So Γ t-f entails P & Q if and only if it t-f entails both P and Q. 10. vI metatheorem: If Γ truth functionally entails P then Γ truth functionally entails both (P v Q) and (Q v P). Proof: Suppose Γ t-f entails P. Then on any truth value assignment on which all the sentences in Γ are true P must be true. But in this case, both P v Q and Q v P must be true as well. So if Γ t-f entails P it must t-f entail both P v Q and Q v P. 11. ⊃ metatheorem: If Γ truth functionally entails (P ⊃ Q) and Γ truth functionally entails P then Γ truth functionally entails Q. Proof: Suppose Γ t-f entails both P ⊃ Q and P. Then on any tva on which all the sentences in Γ are true, both P ⊃ Q and P must be true. But on any tva on which both P ⊃ Q and P are true, Q is true. So, if Γ t-f entails both P ⊃ Q and P, then on any tva on which all the sentences in Γ are true, Q is true. So if Γ truth functionally entails both P ⊃ Q and P then Γ truth functionally entails Q. 12. ≡E metatheorem: If Γ truth functionally entails (P ≡ Q) and Γ truth functionally entails either one of P or Q, then Γ truth functionally entails the other one as well. Proof: Suppose Γ t-f entails both P ≡ Q and one of its immediate components. Then on any truth value assignment on which all the sentences in Γ are true, both P ≡ Q and one of its immediate components must be true. But on any truth value assignment on which both P ≡ Q and one of
its immediate components are true, the other must be true as well. So, if Γ t-f entails both P ≡ Q and one of its immediate components, then on any truth value assignment on which all the sentences in Γ are true, the other immediate component must be true as well. So, if Γ t-f entails both P ≡ Q and one of its immediate components, it must t-f entail the other immediate component as well.
The soundness of SD+ Appendix: Proof of the extensionality of the sentences of SL Proofs of Metatheorems
STRATEGY FOR PROVING THE SOUNDNESS OF SD. If SD consisted just of the “straight” rules (⊃E, ≡E, &E, &I, vI, and the straight versions of vE, ~E, ~I and ≡I originally given in the SD notes), then proving its soundness would be easy. Each of the straight rules is truth preserving. Each rule specifies up to three sentences that you need to have in order to derive a further sentence, and it is easy to prove that, for each rule, if the sentences you need to have are all true on a truth value assignment, α, then the sentence the rule allows you to derive must be true on α as well. Consider, for example, the straight version of the vE rule. It specifies that from P v Q, P ⊃ R, and Q ⊃ R, you can derive R. And, as it turns out, if the three sentences the rule requires you to have are true on a tva, α, then there is no way that the sentence it allows you to derive, R, can be false on α. To see why, suppose P v Q, P ⊃ R, and Q ⊃ R are all true on a tva, α. Since α(P v Q)=T, either α(P)=T or α(Q)=T. But if α(P)=T then, since α(P ⊃ R)=T, α(R)=T. And if α(Q)=T then, since α(Q ⊃ R)=T, α(R)=T. So, either way, α(R)=T. The same case can be made for each of the straight rules. It follows that if you start off a straight derivation in SD with sentences that are all true on a truth value assignment, α, then this truth must get carried down to the all the sentences derived from them using the straight rules, to all the sentences derived from those sentences using the straight rules, and so on down to the sentence on the last line of the derivation (supposing the derivation has remained “straight” — i.e., no sub-derivations have been introduced). So, by mathematical induction, if all the initial assumptions of a straight derivation in SD are true on a tva, α, then the last line of that derivation must be true on α as well. So the set of initial assumptions truth functionally entails the sentence on the last line (because there is no way the set of initial assumptions could be true on a truth value assignment and the sentence on the last line be false on that assignment). So it follows that if a sentence, P, of SL is derivable using the straight rules of SD from a set, Γ, of sentences of SL, then Γ truth-functionally entails P. So the system of “straight” rules of SD is sound. Unfortunately, the system of straight rules is not also complete. While we can trust that any sentence derived using the straight rules must be t-f entailed by the set of initial assumptions, not all sentences that are truth functionally entailed by sets can be derived from those sets using just the straight rules. Most notably, no sentence that is t-f entailed by the empty set (no tf true sentence) is derivable using the straight rules. Each of the straight rules requires you to have some sentences to start with, so none can be applied when there are no initial assumptions. SD becomes complete when we add the sub-derivation rules to it. But the addition of these rules significantly complicates our effort to prove that SD is sound. We can no longer argue that if the sentences higher up in a derivation are true on a tva, α, then the sentences lower down must be true on α. This is because the sub-derivation rules allows us to assume sentences that bear no deductive relation to the sentences higher up on the tree and so cannot be shown to be true if those sentences are true.
This feature of the sub-derivation rules might lead us to wonder whether SD really is sound, or whether we might not have gone too far and made it unsound in our efforts to make it complete. We haven’t, but the addition of the sub-derivation rules does make it harder to prove that SD is sound. Since truth on α need not be carried down to the assumptions heading off sub-derivations, we must look for a different way to prove the soundness of SD. Fortunately, there is another such way. Though truth on a tva is not transmitted from line to line of a derivation in SD there is another property that is, and this other property serves just as well to establish the soundness of SD: the property of being truth functionally entailed by the set of open assumptions at that line. Recall that an open assumption is simply an assumption that has a continuing scope line. So whether an assumption is open or not depends on what line you are at in a derivation. All assumptions are open on the line where they are first made. And on each subsequent line, as long as their scope lines continue down to that line, they remain open. But beginning with the first line on which their scope lines terminate, they are considered to have been closed. So at each line, call it line k, in a derivation there is a (possibly empty) set, Γk, that we can call the set of open assumptions at that line. To determine the contents of this set you simply look up each scope line that is continuing at line k to see what assumptions are heading off that scope line. For example, in the following derivation scheme: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.
P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 P14 P15 P16
Γ1, the set of open assumptions at line 1, is {P1}, Γ2 is {P1, P2}, Γ3 is {P1, P2, P3}, Γ4 is {P1, P2, P3, P4}, Γ5 is {P1, P2, P3, P4}, Γ6 is {P1, P2, P3, P4}, Γ7 is {P1, P2, P3, P4, P7}, Γ8 is {P1, P2, P3, P4, P7}, Γ9 is {P1, P2, P3, P4, P7}, Γ10 is {P1, P2, P3, P4},
Γ11 is {P1, P2, P3,}, Γ12 is {P1, P2, P3, P12}, Γ13 is {P1, P2, P3, P12}, Γ14 is {P1, P2, P3, P12}, Γ15 is {P1, P2, P3}, and Γ16 is {P1, P2, P3}.
THE SOUNDNESS PROOF The crucial claim made by the soundness proof is that the rules of SD were designed in such a way as to ensure that if, at each previous line, the sentence on that line is truthfunctionally entailed by the set of assumptions that are open at that line, then any sentence that the rules permit us to write down on the next line must be such that it is truth-functionally entailed by the set of assumptions that remain open at that next line. If we combine this crucial claim with the claim that on the first line of any derivation in SD the sentence on that line must be truth functionally entailed by the set of open assumptions on that line, then we are in a position to argue, by mathematical induction on the lines in a derivation, that the sentence on the last line of any derivation in SD must be t-f entailed by the set of open assumptions on that line. To run this argument we consider the sentences in a derivation in SD to be ordered by their line numbers. So P1 is the sentence on line 1 of a derivation in SD, P2 is the sentence on line 2, P3 is the sentence on line 3, and so on. We also consider Γ1 to be the set of open assumptions at line 1, Γ2 the set of open assumptions at line 2, Γ3 to be the set of open assumptions at line 3, and so on. Here is the argument: Basis Clause: In any derivation in SD, Γ1 t- f entails P1. Inductive Step: If each of Γ1- Γk t- f entails each of P1- Pk respectively, then Γk+1 must t- f entail Pk+1. Conclusion: On the last line of any derivation in SD, line n, Γn must t- f entail Pn. Supposing that we can justify the basis clause and the inductive step, we need merely combine this result with the definition of derivability in SD to complete our soundness proof. According to the definition of derivability in SD, a sentence, P, of SL has been derived from a set, Γ, of sentences of SL if and only if P is derived on a line that has only a subset of the sentences in Γ in its set of open assumptions. (This is what it means to say that P occurs under the scope just of the sentences in Γ.) So if P has been derived from Γ in SD, then a subset of Γ must constitute the set of open assumptions at the line where P is derived. But then, by the conclusion just reached above, Γ must truth-functionally entail P. (If P is derived from a proper subset of Γ, that is, a set that does not contain all the sentences in Γ, then it will still be derived from Γ, because adding extra assumptions to the head of a derivation does nothing to change what can be derived from the prior assumptions.) So if P is derivable from a subset of Γ, Γ must t-f entail P. So SD is sound.
PROOF OF THE BASIS CLAUSE To fill out this argument sketch we need to prove the basis clause and the inductive step. The proof of the basis clause is straightforward. On the first line of any derivation in SD, the sentence on that line, P1, will have to be an assumption. (It cannot be a derived line since there is no rule of SD that allows you to derive a line without appealing to some earlier line as justification.) But if P1 is an assumption, then it is itself a member (in fact, the only member) of Γ1. (Γ1 is always {P1}.) And it is easy to prove that if a sentence, P, of SL is a member of a set, Γ, of sentences of SL then Γ truth functionally entails P. “A” metatheorem: if a sentence, P, of SL is a member of a set, Γ, of sentences of SL then Γ truth functionally entails P. (Click here for the proof.)
PROOF OF THE INDUCTIVE STEP It remains to prove the inductive step. Since any subsequent line of a derivation must either be a further assumption or must be justified by one of the 11 derivation rules, proving the inductive step means proving that, if all the previous lines of a derivation in SD, lines 1-k, have been truth functionally entailed by the assumptions that are open at those lines, then whether the next line, line k+1, is entered as an assumption or whether it is entered in accord with one of the 11 derivation rules, it must be truth functionally entailed by Γk+1, the set of assumptions that are open at that next line. The proof proceeds as follows: We first suppose the inductive hypothesis: Suppose that at each of lines 1-k of a derivation in SD the sentence on that line is t-f entailed by the set of open assumptions on that line. Now we prove that, under this supposition, whichever derivation rule we use to derive the sentence on line k+1, that sentence must be t-f entailed by the set of open assumptions at line k+1.
Case 1: Assumption If the sentence on line k+1, call it P, is an assumption then it is itself a member of the set of open assumptions at that line, call it Γk+1. So, by the argument used to prove the basis clause, Γk+1 must t-f entail P.
Case 2: Reiteration The following schema outlines the reiteration rule of SD:
i. P · · … … · k+1.
· · · P
i, R
For P to be derived at line k+1 by the R rule of SD, there must be some earlier occurrence of P on some earlier line, i. Moreover, the earlier occurrence of P at line i must be accessible at line k+1, which means that its scope line must continue unbroken down to line k+1. Since derivations cannot close without all the sub-derivations they contain being closed, all the scope lines to the left of P at line i must also continue unbroken down to line k+1. (If they didn’t the scope line immediately to the left of P couldn’t continue unbroken down to line k+1.) Consequently, no assumption that is open at line i can be closed at line k+1, which means that Γk+1 must be a superset of Γi. (A set equal to or larger than Γi.) By the inductive hypothesis, Γi t-f entails P. But then it follows by the R metatheorem of SL that Γk+1 must also t-f entail P. “R” metatheorem: if a set of sentences of SL truth functionally entails a sentence of SL then any superset of that set must also truth functionally entail that sentence. (Click here for the proof of this metatheorem.)
Case 3: ⊃I As a preliminary to the proof of this case, consider the following schema for any application of the ⊃I rule: θ
i. … k. k+1.
Q . . . R Q⊃R
For Q ⊃ R to be derived at line k+1 by the ⊃I rule of SD, there must be an earlier subderivation running from lines i-k that is headed off by the assumption of the antecedent, Q, at line i and that yields the consequent, R, at line k. Moreover, the sub-derivation that yields R at line k must be accessible at line k+1, which means that it must be closed at line k+1, and that the scope line immediately to its left (here labelled θ) must continue unbroken down to line k+1. Since derivations cannot close without all the sub-derivations they contain being closed, all the scope lines to the left of θ must also continue unbroken down to line k+1. (If they didn’t, θ couldn’t continue unbroken down to line k+1.) Consequently, no sub-derivation that is open at line k can be closed at line k+1, except for the sub-derivation headed off by Q at line i, which as we have said must be closed. So the set of open assumptions at line k, Γk, must be identical to Γk+1 ∪ {Q}. By the inductive hypothesis, Γk t-f entails R. Hence, since Γk and Γk+1 ∪ {Q} are identical, Γk+1 ∪ {Q} t-f entails R. But then it follows by the ⊃I metatheorem that Γk+1 t-f entails Q ⊃ R. ⊃I metatheorem: if Γ ∪ {Q} truth-functionally entails R, where Γ is a set of sentences of SL and Q and R are sentences of SL, then Γ truth-functionally entails Q ⊃ R. You were asked to prove this metatheorem previously as exercise #2b in Chapter 3.6. (Click here for the proof.)
Case 4: ~I To prove the ~I case, I am going to appeal to two metatheorems, which I will call the “X” metatheorem and the ~I metatheorem. X metatheorem: If a set, Γ, of sentences of SL truth-functionally entails both a sentence, P, of SL and ~P then Γ is truth functionally inconsistent. ~I metatheorem: If, for a set, Γ of sentences of SL and a sentence, P, of SL, Γ ∪ {P} is truth-functionally inconsistent, then Γ truth-functionally entails ~P. You can click here for the proof of these two metatheorems. (The first of them was proven in exercise 3.6 #3b.) I also want to appeal to the following schema for any application of the ~I rule:
i.-1 i.
P . . .
j.
Q . . . k. ~Q k+1. ~P Note that lines j and k in the above schema could have occurred in the reverse order. For ~P to be derived by ~I on line k+1, there must be an earlier sub-derivation running from lines i-k that is headed off by the assumption of P at line i, and that yields Q and ~Q immediately to the right of the scope line for this assumption at line k and some line, j, intermediate between i and k. Moreover, this sub-derivation must be accessible at line k+1, which means that it must be closed at line k+1, and that the scope line immediately to its left must continue unbroken down to line k+1. For reasons similar to those given when discussing case 3, this means that Γj and Γk are both identical to Γk+1 ∪ {P}. Since lines j and k are prior to line k+1, it follows by the inductive hypothesis that Γj and Γk and hence Γk+1 ∪ {P} t-f entail both Q and ~Q. But then, by the “X” metatheorem, Γk+1 ∪ {P} must be t-f inconsistent, and, by the ~I metatheorem, Γk+1 must t-f entail ~P.
Case 5: ~E The proof of case 5 is similar to the proof of case 4. Where the proof of case 4 appeals to the ~I metatheorem, the proof of case 5 appeals to the following ~E metatheorem: ~E metatheorem: If, for a set, Γ, of sentences of SL and a sentence, P, of SL, Γ ∪ {~P} is truth-functionally inconsistent, then Γ truth-functionally entails P. The proof of this metatheorem is almost identical to that of the ~I metatheorem.
Cases 6 and 7: vE and ≡I The proof of these cases is modeled on that of case 3 and invokes the following metatheorems:
vE metatheorem: If Γ t-f entails P v Q, Γ ∪ {P} t-f entails R, and Γ ∪ {Q} t-f entails R, then Γ t-f entails R. ≡I metatheorem: If Γ ∪ {P} t-f entails Q and Γ ∪ {Q} t-f entails P, then Γ t-f entails P ≡ Q. You should be able at this point to reconstruct the proof for yourself, or you can consult chapter 6.2 or the proofs at the end of this chapter.
Cases 8-12: The straight rules The proof of the straight rules is modeled on the proof of case 2 and invokes the following metatheorems, together with the “R” metatheorem: & metatheorem: Γ truth functionally entails (P & Q) if and only if Γ truth functionally entails both P and Q. vI metatheorem: If Γ truth functionally entails P then Γ truth functionally entails either one of (P v Q) and (Q v P). ⊃E metatheorem: If Γ truth functionally entails (P ⊃ Q) and Γ truth functionally entails P then Γ truth functionally entails Q. ≡E metatheorem: If Γ truth functionally entails (P ≡ Q) and Γ truth functionally entails either one of P or Q, then Γ truth functionally entails the other one as well. Supposing each of these metatheorems is true, we can argue along the following lines (taking the ⊃E case as an example): i. Q ⊃ P (or Q) · · · j. Q (or Q ⊃ P) · · · k+1. P i., j., ⊃E For P to be derived by ⊃E on line k+1, the derivation must contain at least two prior lines, call them lines i and j. One of these lines must have a sentence of the form Q ⊃ P on it, and the other must have a sentence of the form Q on it, though it does not matter which is which. Moreover, both of these lines must be accessible, which means that Γj must be a superset of Γi and Γk+1 must be a superset of Γj and hence of Γi. (That is because all scope lines that exist at line i must continue unbroken down to line j, and all scope lines that exist at lines i and j must continue unbroken down to line k+1.) Since lines i and j are prior to line k+1 it follows by the inductive hypothesis that Γi t-f entails Q ⊃ P (or Q as the case may be) and Γj t-f entails Q (or Q ⊃ P as the case may be). So, by the R metatheorem, and the fact that Γk+1 is a superset of both Γi and Γj, Γk+1 t-f entails both Q ⊃ P and Q. So, by the ⊃E metatheorem, Γk+1 t-f entails P.
The proof of the remaining metatheorems and the application of each of them to prove its case is left to you as an exercise. (You can also consult the text or the end of this chapter.) Once it has been proven that, under the supposition of the inductive hypothesis, Γk+1 must t-f entail the sentence on line k+1 in each of these cases, the inductive step has been proven, and it then follows by mathematical induction that the last sentence of any derivation in SD must be t-f entailed by the set of open assumptions on that line. If that set is the empty set, then the sentence is t-f entailed by the empty set. If that set is the set of premises of an argument, then the set of those premises t-f entails that sentence, and the argument is t-f valid.
SOME COROLLARY RESULTS Now that I have proven that SD is sound, it is easy to prove that SD also gives us reliable proofs of the truth functional truth of sentences, the truth functional equivalence of pairs of sentences, and the truth functional inconsistency of sets. Corollary 1. Suppose that a sentence, P, is a theorem of SD. Then P derivable in SD from the empty set. Because SD is sound, it follows that the empty set t-f entails P. But it was proven in Chapter 3.6 exercise #2a that if the empty set t-f entails P, then P must be t-f true. So if P is a theorem of SD, then P is t-f true. Corollary 2. Suppose that P and Q are equivalent in SD. Then because SD is sound it follows that {P} t-f entails Q and {Q} t-f entails P. But then by the answer to exercise 3.5 #5b it follows that P and Q must be t-f equivalent. So if P and Q are equivalent in SD, P and Q must be t-f equivalent. Corollary 3. Suppose Γ is inconsistent in SD. Then there is some sentence, P, such that both P and ~P can both be derived from Γ in SD. Because SD is sound this means that Γ truth functionally entails both P and ~P. But then by the inconsistency metatheorem (discussed above in connection with case 4), Γ must be t-f inconsistent. So if Γ is inconsistent in SD, Γ must be t-f inconsistent.
THE CONSISTENCY OF SD As noted when discussing the consistency of the tree method, the term “consistency” is used in two senses. In one sense it names a property of sets of sentences. (A set of sentences is logically consistent if and only if there is no contradiction in affirming all of the sentences it contains; it is truth-functionally consistent if and only if there is a line on the truth table that assigns a “T” to every sentence in the set; and it is consistent in SD if and only if there is no sentence P such that both P and ~P are derivable from that set in SD.) In a second sense, consistency is a property of derivation systems, like the tree method and SD. A derivation system is said to be consistent if and only if for any sentence, P, both P and ~P are not theorems of system.
SD is consistent in this sense. To see why, let’s suppose the opposite and see what happens. Suppose that for some sentence, P, both P and ~P are theorems of SD. We just showed (Corollary 1 above) that any sentence that is a theorem of SD is t-f true. So it follows that P is t-f true. So it is true on every tva. But then, by ~rule, ~P must be false on every tva, and so false on at least one tva. So ~P can’t be t-f true. But we just supposed that ~P is also a theorem of SD, which means that it must be t-f true. So we have been led into a contradiction. Since we are led into an absurd result by supposing that for some sentence, P, both P and ~P are derivable from the empty set in SD, we can deny that supposition. So there is no sentence that is such that both it and its negation are theorems of SD. So SD is consistent.
THE SOUNDNESS OF SD+ The derivation system, SD+, consists of the rules of SD, three further derivation rules, MT, HS and DS, and the rules of replacement. We already know that the rules of SD are sound. And we do not need to do any further work to prove that MT, HS, and DS are sound. That is because these rules merely allow us to abbreviate a routine sequence of steps that we can otherwise perform using just the rules of SD. So wherever MT, HS, or DS are used to derive a sentence, P, we could instead run through the routine sequence of steps and derive P using just rules of SD. Since nothing can be derived using MT, HS, or DS that can’t be derived just using the rules of SD, and we know that the rules of SD are sound, it follows that MT, HS, and DS must be sound. However, the rules of replacement do not simply abbreviate routine sequences of steps in SD. When we use these rules we often replace one component of a sentence with a different sentential component. Now that we have proven that SD is sound, we can appeal to that result together with a further result, the extensionality metatheorem (proven in exercise 6.1#1e, or at the end of these notes) to prove that the rules of replacement are also sound. The proof appeals to a further fact that you should be able to verify for yourself (and that was set as an assignment in exercise 5.3#9): that each of the rules of replacement is an equivalence in SD, so that the sentence on the right side of the rule is derivable from the sentence on the left side, and the sentence on the left side derivable from the one on the right side. Here is the proof of the soundness of the rules of replacement. Proof: The rules of replacement hold between sentences that are equivalent in SD. So for each rule, P ⇔ Q, {P} yields Q in SD and {Q} yields P in SD. By Corollary 2 of the proof of the soundness of SD this means that P and Q must be t-f equivalent. But the sentences of SL are all extensional, that is, if P and Q are t-f equivalent, then R is t-f equivalent to [R] (P//Q), where “[R] (P//Q)” refers to the result of replacing one or more occurrences of Q in R with occurrences of P. It follows that if P ⇔ Q is a rule of replacement of SD+ and R is true on a tva, α, then α must assign a T to [R] (P//Q). Consequently, applying the rules of replacement cannot lead us from true sentences to false ones. So the rules of replacement are sound.
Appendix: PROOF OF THE EXTENSIONALITY OF THE SENTENCES OF SL In what follows, I prove that the sentences of SL are extensional.
That is, if P is t-f equivalent to Q, then replacing one or more occurrences of Q with occurrences of P in any longer sentence, R, containing Q will not change the truth value of the sentence. More precisely put, if P is t-f equivalent to Q, then any sentence, R, containing Q as one of its components, is t-f equivalent to [R](P//Q), where “[R](P//Q)” refers to the result of replacing one or more occurrences of Q in R with occurrences of P. (E.g., if “A” and “~~A” are tf equivalent, then “A & B” is t-f equivalent to “~~A & B.”) I give this proof by mathematical induction on the sentences of SL grouped by the number of occurrences of connectives in those sentences. So in the first group are all the sentences with 0 occurrences of connectives (the atomic sentences). In the second group are all the sentences with one connective, e.g., “~A” and “A & B.” In the third group are all the sentences with two occurrences of connectives, e.g., “~~A,” “~A v B.” And so on. The proof goes as follows: Basis Clause: All sentences of SL with 0 occurrences of connectives are extensional. Inductive Step: If all sentences of SL with 0-k occurrences of connectives are extensional, then all sentences with k+1 occurrences of connectives are extensional. Conclusion: All the sentences of SL are extensional. Proof of the Basis Clause: The sentences with 0 occurrences of connectives are the atomic sentences. But in the case of atomic sentences the following chain of consequences holds: 1. R must be the same sentence as Q. 2. [R](P//Q) must be the same sentence as P. 3. R must be t-f equivalent to [R](P//Q), that is, R is extensional. (1.) follows because atomic sentences have no components other than themselves. So the only way that a sentence, Q, can be a component of R when R is atomic is if Q are R are one and the same sentence. (2.) follows from (1.) because, given that R just is Q, the result, [R](P//Q), of putting P in the place of any occurrence of Q in R is equivalent to replacing R as a whole with P. (3.) follows directly from (1.) and (2.) because we are supposing that P and Q are t-f equivalent. So, if R has 0 occurrences of connectives, it is extensional. Proof of the Inductive Step: Suppose that the inductive hypothesis is true: All the sentences of SL with 0-k occurrences of connectives are extensional. Then there are two possibilities to consider: Q may just be R (as when, for example, Q and R are “~A” and P is “~~~A”), or Q may be a proper component of R. In the first of these cases, R must be extensional for the reason already given when proving the basis clause. In the second case, R must either be the result, ~S, of prefacing some proper component, S, with a “~,” or it must be the result of connecting two proper components, S and T, with a twoplace connective. Case 2a: If R is of the form ~S then, because S has only k connectives, it falls under the inductive hypothesis. So S must be extensional. This means that if P is t-f equivalent to Q, then S is t-f equivalent to [S](P//Q). But then on any truth value assignment either S and [S](P//Q) are both true, in which case ~S and ~[S](P//Q) are both false, or S and [S](P//Q) are both false, in which case ~S and ~[S](P//Q) are both true. So either way, ~S and ~[S](P//Q) must both have the same truth value. So they must be t-f equivalent. So if R is a sentence of the form ~S, then it must be extensional.
Case 2b: If R is the result of connecting two proper components, S and T, with a twoplace connective, then there are two sub-cases to consider: the proper component, Q, in R that gets replaced by P could be either a) one or both of the immediate components of R (i.e., S or T), or b) one or more proper components of R other than R itself or its immediate components. Case 2b(i): In this case [R](P//Q) is either [R](P//S) or [R](P//T) or both. But the truth value of S & T, S v T, S ⊃ T, and S ≡ T is in each case determined exclusively by the truth value of its immediate components. Consequently if P is t-f equivalent to S or to T, any substitution of P for either of S or T will not change the truth value of the resulting sentence on any truth value assignment, since putting P for S or T does not result in a change to the truth value of the component. So R must be t-f equivalent to [R](P//Q). So if the component, Q, in R that gets replaced by P is either or both of its immediate components, then R is extensional. Case 2b(ii): In this case since R has k+1 occurrences of connectives, each of its immediate components, S and T must contain k or fewer occurrences of connectives (so that the sum of the occurrences of connectives in both is exactly k). But then, by the inductive hypothesis, each of S and T is extensional. By definition, that means that if P is t-f equivalent to Q, S must be t-f equivalent to [S](P//Q) and T must be t-f equivalent to [T](P//Q). Consequently, regardless of whether P is put for Q in S, T, or both, the result will be a sentence with a left immediate component that is t-f equivalent to S (if only S itself, in the case where no substitution is made) and a sentence with a right immediate component that is t-f equivalent to T (if onlyT itself in the case where no substitution is made). But then, since the truth value of S & T, S v T, S ⊃ T, or S ≡ T is in each case determined exclusively by the truth value of its immediate components, R must be t-f equivalent to [R](P//Q). So if the component, Q, in R that gets replaced by P is one of the components of R other than R itself or one of its immediate components, then R is extensional. So, If R is the result of connecting two proper components, S and T, with a two-place connective, then R must be extensional. So, either way, if Q is a proper component of R, then R must be extensional. So, either way, if all the sentences of SL with 0-k occurrences of connectives are extensional, and R is any sentence with k+1 occurrences of connectives it must be extensional.
PROOFS OF METATHEOREMS 1. “Assumption” metatheorem: if a sentence, P, of SL is a member of a set, Γ, of sentences of SL then Γ truth functionally entails P. Proof: If P is a member of Γ then any truth value assignment that makes all the sentences in Γ true must make P true. So there is no truth value assignment that makes all the sentences in Γ true and P false. So Γ truth functionally entails P. Back 2. “R” metatheorem: if Γ is a set of sentences of SL that truth functionally entails a sentence, P, of SL and Γ′ is a superset of Γ (a set that contains all the sentences in Γ and possibly other sentences of SL as well), then Γ′ must also truth functionally entail P. Proof [This is a corollary of 3.6E #4c]:
Suppose that some set, Γ, of sentences of SL truth functionally entails a sentence, P, of SL. Now consider a superset, Γ′, of Γ. Since Γ′ includes all the sentences in Γ, any truth value assignment that makes all the sentences in Γ′ true must make all the sentences in Γ true. But, since we are supposing that Γ t-f entails P, no truth value assignment that makes all the sentences in Γ true can make P false. So, no truth value assignment that makes all the sentences in Γ′ true can make P false. So Γ′ must truth-functionally entail P. Back 3. ⊃I metatheorem: if Γ ∪ {P} truth-functionally entails Q, where Γ is a set of sentences of SL and P and Q are sentences of SL, then Γ truth-functionally entails P ⊃ Q. Proof [Proven as part of 3.6E #2b]: Suppose that Γ ∪ {P} truth-functionally entails Q. Then there is no truth value assignment on which all the sentences in Γ ∪ {P} are true and Q is false. So there is no truth value assignment on which all the sentences in Γ are true, P is true, and Q is false. So on any truth value assignment on which all the sentences in Γ are true, P cannot be true and Q false. So, on any truth value assignment on which all the sentences in Γ are true, P ⊃ Q must be true. So Γ truthfunctionally entails P ⊃ Q. Back 4. X metatheorem: If a set, Γ, of sentences of SL truth-functionally entails both a sentence, P, of SL and a sentence, ~P, of SL then Γ is truth functionally inconsistent. Proof [Proven as 3.6E #3b]: Suppose that Γ t-f entails both P and ~P. We know from the definition of t-f entailment that if a set t-f entails a sentence, then on any truth value assignment on which that sentence is false, the members in the set cannot all be true. But exactly one of P or ~P must be false on every tva (by ~ rule). It follows that at least one of the sentences in Γ must be false on every tva. So Γ must be t-f inconsistent. Back 5. ~I metatheorem: If, for a set, Γ of sentences of SL and a sentence, P, of SL, Γ ∪ {P} is truth-functionally inconsistent, then Γ truth-functionally entails ~P. Proof: Suppose that Γ ∪ {P} is truth-functionally inconsistent. Then on any truth value assignment on which all the sentences in Γ are true, P must be false. So, on any truth value assignment on which all the sentences in Γ are true, ~P must be true. So Γ truth-functionally entails ~P. Back 6. ~E metatheorem: If, for a set, Γ of sentences of SL and a sentence, P, of SL, Γ ∪ {~P} is truth-functionally inconsistent, then Γ truth-functionally entails P. Proof [Proven as 3.6E #1c]: The proof is identical to that of the ~I metatheorem replacing occurrences of P with ~P, and occurrences of ~P with P. Back 7. ≡I metatheorem: If Γ ∪ {P} t-f entails Q and Γ ∪ {Q} t-f entails P, then Γ t-f entails P ≡ Q. Proof: Suppose that Γ ∪ {P} t-f entails Q and Γ ∪ {Q} t-f entails P. Then on any truth value assignment on which all the sentences in Γ are true, it cannot be the case that all the sentences in {P} are
true and Q is false, and it cannot be the case that all the sentences in {Q} are true and P is false. But P is the only sentence in {P} and Q is the only sentence in {Q}. So, on any truth value assignment on which all the sentences in Γ are true, it cannot be the case that P is true and Q is false, and it cannot be the case that Q is true and P is false. So, on any truth value assignment on which all the sentences in Γ are true, P and Q cannot have different truth values, which means that P ≡ Q must be true. But in that case, Γ t-f entails P ≡ Q. 8. v metatheorem: If Γ t-f entails P v Q, Γ ∪ {P} t-f entails R, and Γ ∪ {Q} t-f entails R, then Γ tf entails R. Proof: Suppose Γ t-f entails P v Q, Γ ∪ {P} t-f entails R, and Γ ∪ {Q} t-f entails R. Then on any truth value assignment, α, on which all the sentences in Γ are true, P v Q must be true, which means either P must be true or Q must be true. Suppose P is true on α. Then, because P is the only sentence in {P}, all the sentences in Γ ∪ {P} are true on α. But then R must be true on α. Now suppose it is instead Q that is true on α. Then, because Q is the only sentence in {Q}, all the sentences in Γ ∪ {Q} are true on α. But then R must again be true on α. So either way, R is true on α. So, on any truth value assignment, α, on which all the sentences in Γ are true, R is true. So, on our assumptions, Γ t-f entails R. That is, if Γ t-f entails P v Q, Γ ∪ {P} t-f entails R, and Γ ∪ {Q} t-f entails R, then Γ t-f entails R. 9. & metatheorem: Γ truth functionally entails (P & Q) if and only if Γ truth functionally entails both P and Q. Proof: Γ t-f entails P & Q if and only if, on any truth value assignment on which all the sentences in Γ are true, P & Q is true. But P & Q is true if and only if both P and Q are true. So Γ t-f entails P & Q if and only if, on any truth value assignment on which all the sentences in Γ are true, both P and Q are true. So Γ t-f entails P & Q if and only if it t-f entails both P and Q. 10. vI metatheorem: If Γ truth functionally entails P then Γ truth functionally entails both (P v Q) and (Q v P). Proof: Suppose Γ t-f entails P. Then on any truth value assignment on which all the sentences in Γ are true P must be true. But in this case, both P v Q and Q v P must be true as well. So if Γ t-f entails P it must t-f entail both P v Q and Q v P. 11. ⊃ metatheorem: If Γ truth functionally entails (P ⊃ Q) and Γ truth functionally entails P then Γ truth functionally entails Q. Proof: Suppose Γ t-f entails both P ⊃ Q and P. Then on any tva on which all the sentences in Γ are true, both P ⊃ Q and P must be true. But on any tva on which both P ⊃ Q and P are true, Q is true. So, if Γ t-f entails both P ⊃ Q and P, then on any tva on which all the sentences in Γ are true, Q is true. So if Γ truth functionally entails both P ⊃ Q and P then Γ truth functionally entails Q. 12. ≡E metatheorem: If Γ truth functionally entails (P ≡ Q) and Γ truth functionally entails either one of P or Q, then Γ truth functionally entails the other one as well. Proof: Suppose Γ t-f entails both P ≡ Q and one of its immediate components. Then on any truth value assignment on which all the sentences in Γ are true, both P ≡ Q and one of its immediate components must be true. But on any truth value assignment on which both P ≡ Q and one of
its immediate components are true, the other must be true as well. So, if Γ t-f entails both P ≡ Q and one of its immediate components, then on any truth value assignment on which all the sentences in Γ are true, the other immediate component must be true as well. So, if Γ t-f entails both P ≡ Q and one of its immediate components, it must t-f entail the other immediate component as well.
