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 today (by midnight). HW #3 Assigned (due on 2/24). Natural deduction problems for LSL/Boolean Logic Adding 2-place (binary) predicates to the language (L2PL). L2PL semantics (Working with and constructing L2PL Interpretations) Constructing L2PL Interpretations An Example 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)
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 2 Constructing L2PL Interpretations An Example Establish the following claim, via the construction of a counterexample L2PL Interpretation. ( x) [ ( y)rxy ( z)( w)rwz ] ( x)( y)( z)(rxy Rzy)
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 3 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!
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 4 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.
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 5 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.
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 6 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.
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 7 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.
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 8 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
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 9 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
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 10 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
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 11 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
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 12 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.
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 13 The Rule of -Elimination ( E) Rule of -Elimination: For any formulae p and q, if p q occurs at a line j and p occurs at a line k, then q may be inferred at line m, labeling the line j, k E and writing on the left all numbers which appear on the left of line j and all numbers which appear on the left of line k. [Note: We may have either j < k or j > k.] a 1,..., a n (j) p q. b 1,..., b u (k) p. a 1,..., a n, b 1,..., b u (m) q j, k E Our example deduction concludes (we indicate the end of a proof with a ), in line (7), with a correct application of the -Elimination Rule: 1, 2, 3 (7) H 3, 6 E
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 14 How to Deduce a Conditional: I To deduce a conditional, we assume its antecedent and try to deduce its consequent from this assumption. If we are able to deduce the consequent from our assumption of the antecedent, then we discharge our assumption, and infer the conditional. To implement the I rule, we will first need a refined Rule of Assumptions that will allow us to assume arbitrary formulas for the sake of argument, later to be discharged after making desired deductions. Here s the refined rule of Assumptions: Rule of Assumptions (final version): At any line j in a proof, any formula p may be entered and labeled as an assumption (or premise, where appropriate). The number j should then be written on the left. Schematically: j (j) p Assumption (or: Premise)
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 15 How to Deduce a Conditional: II The I Rule Now, we need a formal Introduction Rule for the, which captures the intuitive idea sketched above (i.e., assuming the antecedent, etc.): Rule of -Introduction: For any formulae p and q, if q has been inferred at a line k in a proof and p is an assumption or premise occurring at line j, then at line m we may infer p q, labeling the line j, k I and writing on the left the same assumption numbers which appear on the left of line k, except that we delete j if it is one of these numbers. Note: we may have j < k, j > k, or j = k (why?). Schematically: j (j) p Assumption (or: Premise). a 1,..., a n (k) q. {a 1,..., a n }/j (m) p q j, k I
Branden Fitelson Philosophy 4515 (Advanced Logic) Notes 16 Using The I Rule: Another Example Let s do a deduction of: A (B C) (A B) (A C) 1 (1) A (B C) Premise 2 (2) A B Assumption 3 (3) A Assumption 2, 3 (4) B 2, 3 E 1, 3 (5) B C 1, 3 E 1, 2, 3 (6) C 4, 5 E 1, 2 (7) A C 3, 6 I 1 (8) (A B) (A C) 2, 7 I