Overview of Today s Lecture

Transcription

1 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 1 Overview of Today s Lecture Administrative Stuff HW #1 grades and solutions have been posted Please make sure to work through the solutions HW #2 due on Friday (2/10). Adding 2-place (binary) predicates to the language (L2PL). Symbolizations in L2PL (Rewind) Definition of a Limit (Take Three!) L2PL semantics (Working with and constructing L2PL Interpretations ) Can no longer think in terms of & and for and Infinite domains are sometimes required for L2PL invalidity Next Topic: Proof Theory (Natural Deduction) for LSL (Chapter 4)

2 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 2 L2PL Symbolization (Rewind) The Definition of a Limit (Correction). Let f be a real-valued function (i.e., suppose f : R R). Then, a limiting value of f is defined as: lim f (x) = L ( y > 0)( z > 0)( x) [ 0 < x c < z ] f (x) L < y x c In order to symbolize this definition in L2PL, we ll need (i) one monadic predicate, (ii) two binary relations, and (iii) a domain D = R. The monadic predicate we ll need is: (i) Gv v > 0. The 2 binary relations we ll need are: Dxz 0 < x c < z, and Lxy f (x) L < y. With these ingredients, we can now write the definition of a limit in our L2PL notation. Let s do this in two stages. First, Logish, then L2PL. Last time, I skipped some steps... For every y, if Gy then there is a z such that Gz and for all x Dxz implies Lxy. ( y) { Gy ( z) [ Gz & ( x) ( Dxz Lxy )]} I had been too hasty in my previous L2PL symbolization, which explains why Mathematica wasn t giving the expected answers. This works.

3 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 3 L2PL Interpretations III (I 1 ) Let D be the set consisting of George W. Bush (α) and Jeb Bush (β). And, let Bxy: x is a brother of y. Determine I 1 -truth-values for: 1. ( x)( y)bxy 2. ( y)( x)bxy α β (1) is on I 1, since both of its D-instances are on I 1. ( y)bay is on I 1 because its instance Bab is on I 1. That is, α, β Ext(B). Note: Ext(B) = { α, β, β, α }. ( y)bby is on I 1 because its instance Bba is on I 1. (2) is on I 1, since both of its D-instances are on I 1. ( x)bxa is on I 1 because its instance Baa is on I 1. That is, α, α Ext(B). ( x)bxb is on I 1 because its instance Bbb is on I 1.

4 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 4 L2PL Interpretations IV Just as with LMPL, L2PL interpretations can be used as counterexamples to validity claims. Establishing claims works just as you d expect. We have just seen an L2PL interpretation that shows the following: ( x)( y)rxy ( x)( y)rxy Interpretation I 1 on the previous slide is a counterexample. Why? ( x)( y)bxy is on I 1, since both of its instances are on I 1. ( x)( y)rxy is on I 1, since both of its instances are on I 1. Here is a very important L2PL invalidity: ( x)( y)rxy, ( x)( y)( z)[(rxy & Ryz) Rxz] ( x)rxx ( ) ( ) reveals a surprising difference between LMPL (and LSL) and L2PL sometimes infinite interpretations are needed to prove in L2PL!

5 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 5 Why ( ) is So Important L2PL vs LMPL: Infinite Domains In LMPL, if p is true on any interpretation I, then it is true on a finite interpretation. Indeed, p will be true on an interpretation of size no greater than 2 k, where k is the # of monadic predicate letters in p. In L2PL, some statements are true only on infinite interpretations. It is for this reason that there is no general decision procedure for validity (or logical truth) in L2PL. ( ) on the last slide is a good example of this. ( ) ( x)( y)rxy, ( x)( y)( z)[(rxy & Ryz) Rxz] ( x)rxx Fact. Only infinite interpretations I can be counterexamples to the validity in ( ). To see why, try to construct such an interpretation. We start by showing that no interpretation I 1 with a 1-element domain can be an interpretation on which the premises of ( ) are and its conclusion is. Then, we will repeat this argument for I 2 and I 3. This reasoning can, in fact, be shown correct for all (finite) n. So, only I s with infinite domains will work [e.g., D = N, Rxy: x < y]. Begin with a 1-element domain {α}. For the conclusion of (4) to be, no

6 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 6 object can be related to itself: ( x) Rxx. Thus, we must have Raa: But, to make the first premise, we need there to be some y such that Ray is. That means we need another object β to allow Rab. Thus: α α β Now, because we need the conclusion to remain, we must have Rbb. And, because we need the first premise to remain, we need there to be some y such that Rby is. We could try to make Rba, as follows: α β

7 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 7 But, this picture is not consistent with the second premise being and (at the same time) the conclusion being. If R is transitive, then Rab & Rba (as pictured) entails Raa, which makes the conclusion. Transitivity of R and entails: α β α β Thus, the only way to consistently ensure that there is some y such that Rby is to introduce yet another object γ (such that Rbc), which yields: α β Again, in order to make the conclusion, we must have Rcc, and in order to make the first premise, there must be some y such that Rcy. We could try to make either Rca or Rcb true. But, both of these choices will end-up with the same sort of inconsistency we just saw with β.

8 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 8 In other words, no finite interpretation will give us what we want here. However, if we let D = N and Rxy: x < y, then we get what we want That is, the relation Rxy: x < y on the natural numbers N is such that: For all x, there exists a y such that x < y. [seriality] For all x, y, z, if x < y and y < z, then x < z. [transitivity] For all x, x x. [irreflexivity] It is crucial that the set N of all natural numbers is infinite. The relation < cannot satisfy all three of these properties on any finite domain. I.e., no finite subset of N will suffice to show that the invalidity in (4) holds. Equivalently, the following sentence of L2PL is on all finite I s: p ( x)( y)rxy & ( x)( y)( z)[(rxy & Ryz) Rxz] & ( x) Rxx This sort of thing cannot happen in LMPL. In this sense, the introduction of a single 2-place predicate involves a quantum leap in complexity.

9 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 9 Some Further Remarks on Validity in L2PL As I just explained, there is no general decision procedure for claims in L2PL. This is because we can t always establish claims in finite time. However, there is a method for proving claims natural deduction. And, L2PL s natural deduction system is exactly the same as LMPL s! Before we get to proofs, however, I want to look at the alternating quantifier example that I said separates LMPL and L2PL. As we have seen, ( x)( y)rxy ( y)( x)rxy. But, the converse entailment does hold. That is, ( y)( x)rxy ( x)( y)rxy. We will prove i.e., deduce ( y)( x)rxy ( x)( y)rxy shortly. Before we do that, let s think about ( y)( x)rxy ( x)( y)rxy using our definitions, and our informal method of thinking of as & and as. This is interesting for both directions of the entailment. But, we need to be much more careful here than with LMPL!

10 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 10 First, consider what ( y)( x)rxy says on a domain of size n: ( y)( x)rxy n ( x)rxa ( x)rxb ( x)rxn n (Raa & & Rna) (Rab & & Rnb) (Ran & & Rnn) Next, consider what ( x)( y)rxy says on a domain of size n: ( x)( y)rxy n ( y)ray & ( y)rby & & ( y)rny n (Raa Ran) & (Rba Rbn) & & (Rna Rnn) Then, we notice that these two sentential forms are intimately related. Specifically, we note that ( y)( x)rxy has the following n-form: X n = (p 1 & p 2 & & p n ) (q 1 & q 2 & & q n ) (r 1 & r 2 & & r n ) And, we notice that ( x)( y)rxy has the following n-form: Y n = (p 1 q 1 r 1 ) & (p 2 q 2 r 2 ) & & (p n q n r n ) Fact. X n Y n, for any n. Each disjunct of X n entails every conjunct of Y n. Caution! This doesn t show that ( y)( x)rxy ( x)( y)rxy! Fact. Y n X n, for all n > 1. This can be shown (next slide) using only LSL reasoning. This does show that ( x)( y)rxy ( y)( x)rxy.

11 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 11 The moral is that our informal semantical approach to the quantifiers works for LMPL, since no infinite domains are required for in LMPL. However, our informal semantical approach breaks down for L2PL, since we sometimes need an infinite domain to establish in L2PL. In L2PL, if the informal method above reveals p n q n for some finite n, then it does follow that p q. For instance, Y 2 X 2 on the last slide: (Raa Rab) & (Rba Rbb) (Raa & Rba) (Rab & Rbb) This is just an LSL problem with 4-atoms [A = Raa, B = Rab, C = Rba, D = Rbb]. Truth-tables will generate a counterexample. On the other hand, if (in L2PL) our informal method indicates (as above) that p n q n for all finite n, this does not guarantee p q. E.g.: p = ( x)( y)rxy & ( x)( y)( z)[(rxy & Ryz) Rxz]. q = ( x)rxx. We showed above (informally) that p n q n for all finite n. But, we also saw that there are infinite interpretations on which p is but q is.

12 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 12 Truth vs Proof ( vs ) Recall: p q iff it is impossible for p to be true while q is false. We have methods (truth-tables) for establishing and claims. These methods are especially good for claims, but they get very complex for claims. Is there another more natural way to prove s? Yes! In Chapter 4, we will learn a natural deduction system for LSL. This is a system of rules of inference that will allow us to prove all valid LSL arguments in a purely syntactical way (no appeal to semantics). The notation p q means that there exists a natural deduction proof of q from p in our natural deduction system for sentential logic. p q is short for p deductively entails q. While has to do with truth, does not. has only to do with what can be deduced, using a fixed set of formal, natural deduction rules.

13 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 13 Happily, our system of natural deduction rules is sound and complete: Soundness. If p q, then p q. [no proofs of invalidities!] Completeness. If p q, then p q. [proofs of all validities!] We will not prove the soundness and completeness of our system of natural deduction rules. I will say a few things about soundness as we go along, but completeness is much harder to establish (140A!). We ll have rules that permit the elimination or introduction of each of the connectives &,,,, within natural deductions. These rules will make sense, from the point of view of the semantics. A proof of q from p is a sequence of LSL formulas, beginning with p and ending with q, where each formula in the sequence is deduced from previous lines, via a correct application of one of the rules. Generally, we will be talking about deductions of formulas q from sets of premises p 1,..., p n. We call these p 1,..., p n q s sequents.

14 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 14 An Example of a Natural Deduction Involving & and The following is a valid LSL argument form: A & B C & D (A & D) H H Here s a (7-line) natural deduction proof of the sequent corresponding to this argument: A & B, C & D, (A & D) H H. 1 (1) A & B Premise 2 (2) C & D Premise 3 (3) (A & D) H Premise 1 (4) A 1 &E 2 (5) D 2 &E 1, 2 (6) A & D 4, 5 &I 1, 2, 3 (7) H 3, 6 E

15 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 15 The Rule of Assumptions (Preliminary Version) Rule of Assumptions (preliminary version): The premises of an argument-form are listed at the start of a proof in the order in which they are given, each labeled Premise on the right and numbered with its own line number on the left. Schematically: j (j) p Premise We can see that our example proof begins, as it should, with the three premises of the argument-form, written as follows: 1 (1) A & B Premise 2 (2) C & D Premise 3 (3) (A & D) H Premise

16 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 16 The Rule of &-Elimination (&E) Rule of &-Elimination: If a conjunction p & q occurs at line j, then at any later line k one may infer either conjunct, labeling the line j &E and writing on the left all the numbers which appear on the left of line j. Schematically: a 1,..., a n (j) p & q a 1,..., a n (j) p & q. OR. a 1,..., a n (k) p j &E a 1,..., a n (k) q j &E We can see that our example deduction continues, in lines (4) and (5), with two correct applications of the &-Elimination Rule: 1 (4) A 1 &E 2 (5) D 2 &E

17 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 17 The Rule of &-Introduction (&I) Rule of &-Introduction: For any formulae p and q, if p occurs at line j and q occurs at line k then the formula p & q may be inferred at line m, labeling the line j, k &I and writing on the left all numbers which appear on the left of line j and all which appear on the left of line k. [Note: we may have j < k, j > k, or j = k. Why?] a 1,..., a n (j) p. b 1,..., b u (k) q. a 1,..., a n, b 1,..., b u (m) p & q j, k &I We can see that our example deduction continues, in lines (6), with a correct application of the &-Introduction Rule: 1, 2 (6) A & D 4, 5 &I

18 Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 18 Deduction #2 Using the Rules &E and &I Consider the valid LSL argument form: A & (B & C) C & (B & A) Let s do a deduction of this argument form: 1 (1) A & (B & C) Premise 1 (2) A 1 &E 1 (3) B & C 1 &E 1 (4) B 3 &E 1 (5) C 3 &E 1 (6) B & A 4, 2 &I 1 (7) C & (B & A) 5, 6 &I NOTE: &E can only be applied to formulas whose main connective is &, and &E must be applied to that particular connective.