On some Metatheorems about FOL

Transcription

1 On some Metatheorems about FOL February 25, 2014 Here I sketch a number of results and their proofs as a kind of abstract of the same items that are scattered in chapters 5 and 6 in the textbook. You notice that part 1 is nothing but exactly the same things about FOL as those given about propositional logic previously. Remember that propositional logic is a part (sublogic) of FOL. Also remember that FOL yields the accepted explication of provability and explicitness (see Hilbert's Thesis). Therefore, the results listed are of fundamental importance for provability and computability and thus for all of computer science. 1 Soundness and Completeness 1. Soundness of the semantic tableau method ST : The semantic tableau method for investigating validity of sentences A in FOL is sound with respect to the semantics of FOL. What does this mean?... It means that for every closed fml (sentence) A in a language of FOL, if the semantic tableau method applied to it yields that the tableau of A closes, then A is valid according to the semantics (i.e. v I (A) = T for all interpretations I).That is: = ST A = A. How do you prove it?... Prove that all rules α, β, γ and δ preserve truth in the following sense. For an arbitrary interpretation I it holds that truth on one level implies truth on next level below (for β at least in one of the two branches) and truth is preserved along each individual branch when you go in the opposite direction upwards to the root.. Consequently, if there are only contradictions in the leaves (when the tableau is closed), then the fml on top, at the root, must be a contradiction, too. 2. Soundness of Gentzen's deductive calculus G w.r.t ST : Gentzen's deductive calculus G for proving theorems in G is sound with respect to the semantic tableau method. That is: G A = ST A. How is it proved?... Carry out the duality of all fmls in a closed semantic tableau of a fml A and turn the tableau upside-down. This yields a deduction of A at the root (at the bottom after the tableau has been turned). This operation on the semantic tableau is pure routine. And it works in both ways! Therefore we also get: 1

2 3. Completeness of Gentzen's deductive calculus G w.r.t ST : Gentzen's deductive calculus G for proving theorems in G is complete with respect to the semantic tableau method. What does this mean?... It means that every fml that can be shown to be valid using the semantic tableau method can be proved as a theorem in G. That is: = ST A G A. 4. Soundness of Hilbert's deductive calculus H: Hilbert's deductive calculus H for proving theorems in H is sound with respect to the semantics of FOL. What does this mean?... It means that for every closed fml (sentence) A in a language of FOL, if A is proved in H as a theorem, then A is valid (in the semantics). That is: H A = A. How is it proved?... Prove that all axioms, Ax 1,..., Ax 5 in H are valid and that the rules MP, Ass and GEN preserve validity. This is rather easy to do. 5. Completeness of Hilbert's deductive calculus H w.r.t. G: Hilbert's deductive calculus H for proving theorems in H is complete with respect to Gentzen's deductive calculus G for proving theorems in G. What does this mean?... It means that for every closed fml A, if A is provable in G, then A is also provable in H. G A H A. How is it proved?... Prove that all rules α, β, γ and δ in G correspond to equivalent derivable rules of deduction in H. 6. Completeness of Hilbert's deductive calculus H: Hilbert's deductive calculus H for proving theorems in H is complete with respect to the semantics of FOL. What does this mean?... It means that every valid fml A is provable in H. = A H A. How is it proved?... Prove it by rst proving Completeness of the semantic tableau method ((7) below). Because, if 7 holds, then also G is complete w.r.t. to the semantics, due to 1, 2 and 3. Then 5 yields 6, i.e. Completeness of H w.r.t. the semantics of FOL. 7. Completeness of the semantic tableau method ST : The semantic tableau method for investigating validity of sentences A in FOL is complete with respect to the semantics of FOL. What does this mean?... It means that for every valid fml A in a language of FOL the semantic tableau method applied to A yields that the tableau of A closes. = A = ST A. How do you prove it?... Prove instead the equivalent contrapositive statement: If the semantic tableau for a fml A is not closed, then you can construct an interpretation in which A is true by examining and using the leave of an open branch in the tableau of A. 2

3 FINAL RESULT: The semantic tableau method ST and the deductive calculi G and H are all sound and complete w.r.t. the semantics of FOL. That is: = ST A = A and G A = A and H A = A. 2 Compactness, Undecidability and Incompleteness results 8. Compactness theorem: A set U of sentences in a language of FOL is unsatisable if and only if there exists a nite subset U 0 U that is unsatisable. Proof: The if-part is trivial, because extending an unsatisable set keeps it unsatisable. For the only if-part you can use the soundness and completeness of FOL and notice that if U is unsatisable, then U is inconsistent. Therefore you can prove a contradiction from U as the set of assumptions. However, every proof is nite in length and uses only a nite number of assumptions from U. Let U 0 be the set of these assumptions actually used. Thus you have proved a contradiction from a nite subset U 0 U. Therefore, due to the the soundness and completeness of FOL, U 0 must be unsatisable. The compactness theorem can also be given in its contrapositive form: A set U of sentences in a language of FOL is satisable if and only if every nite subset U 0 U is satisable. 9. FOL is undecidable. What does this mean?... It means that there exists no method by which one can decide (in a nite number of steps) for every fml A in any language of FOL whether it is a logical theorem, and thus valid, or not. Note, if it is valid, then you can show it (in a nite number of steps) using e.g. the semantic tableau method, because the tableau has to close in a nite number of steps. However, if for some fml when you try to establish that it is valid and its tableau has not closed, yet, it still may close... or not. Therefore, the validity for an arbitrary fml is not decidable. How is it proved?... The exact proof is somewhat involved. However, the fact that semantic tableaux for fmls in FOL may become innite hints that there could be fmls such that we never know for sure whether the tableau closes or not. As a matter of fact such fmls exist. Comment. If the language of FOL is very simple, e.g. having only unary relations (and a nite number of non-constants, of course), then one can decide for each fml in the language whether it is a logical theorem / valid or not. 10. Not every theory in FOL is axiomatizable. What does it mean?... Remember that a theory is set T of fmls that is closed under logical consequence (i.e. for every fml A in the FOL-language in question T = A if and only if A T ). 3

4 Theory T is axiomatizable if there exists a nite (or decidable) subset U T such that if A T, then U A (and thus also U = A).Thus, 10 means that there are theories T for which no such axioms U exists. The result is not surprising and it can be proved simply by noticing that (if the language is not very simple) the set of all theories has a cardinality ℵ 1 = 2 ℵ0 whereas cardinality of the set of all nite or decidable sets of axioms is ℵ 0 < ℵ 1. That is, there are not enough of dierent axiom systems for all the theories there are. This holds as soon as the language is rich enough to allow for determining an innite number of equivalence classes w.r.t. to logical equivalence. Kurt Gödel proved 1930 the following seminal metatheorem: 11. Gödel's incompleteness theorem: Elementary arithmetic (number theory) is unaxiomatizable. What does it mean?... The theory of natural numbers with addition and multiplication is not axiomatizable and consequently there are problems (and equations) in elementary arithmetic that cannot be solved. This means also that any set of axioms for elementary arithmetic is doomed to be incomplete or inconsistent. How is it proved?... The proof uses self-reference, so-called, as its most crucial step. Herein every expression in the language of formal arithmetic is given a code in the set of natural numbers (themselves). Then one studies the codes of all theorems of formal arithmetic, in particular, and nds out that they do not constitute a decidable set of natural numbers, which actually settles the question of axiomatizability in the negative. In all, the proof is rather long and involved. We shall need some set theory, here, something about innite sets. 3 Lindenbaum's lemma and LST-theorems 12. Lindenbaum's lemma: Every satisable (and thus consistent) set U of sentences (closed fmls) in a language of FOL can be extended to a maximal (i.e. complete) satisable (consistent) set of sentences T (i.e. to a complete theory). This means also that every satisable (consistent) set U of sentences is a subset of some satisable (consistent) maximal theory T. This result may seem rather trivial. Yet it is a manifestation in a special context of the Axiom of Choice (AC) in set theory, one of the most debated axioms in mathematics. The proof runs as follows: (i) enumerate recursively (i.e. generate in some systematic way all sentences in the language in question), A 0, A 1,..., A n,...; (ii) start with U, then as step 0 form U {A 0 } and U { A 0 } and choose any one of them that is consistent (and satisable) and call it U 0 (iii) extend U 0 in the same way by either A 1 or A 1, etc.... step n form U n 1 {A n } and U n 1 { A n } and choose any one of them that is consistent (and satisable) and call it U n. (iv) run through all A 0, A 1,..., A n,..., i.e. all sentences and 4

5 construct the limit lim n ω0 U n = n N U n. This set is maximal and consistent (satisable) by construction and can serve as theoryt in the lemma. Logicians insist calling this important theorem a lemma (subsidiary proposition), because it is so useful in the proofs of many other important theorems. 13. The Löwenheim-Skolem-Tarski theorems (LST): (Upward-LST) If a set U of sentences in FOL has models of arbitrary high nite cardinality (interpretations with domains of arbitrary high cardinality), then U has models of every innite cardinality. (Downward-LST) If a set U of sentences in FOL has models of some innite cardinality, then U has models of cardinality ℵ 0 = the lowest innite cardinality (= the cardinality of the set natural numbers). In other words, one cannot express and distinguish dierent innite cardinalities in any language of FOL, nor can one make the distinction between potential (nite but arbitrarily large) and actual innity. Therefore, one cannot even express the general concept of niteness in any FOL-language. This result is related to the halting problem for Turing machines. Because, if one could dene niteness in a FOL-language, then one could program the halting condition for a Turing machine. The proof of the downward LST is very simple, because if the semantic tableau yields an innite model for a set U of sentences, then this model is of cardinality ℵ 0. For the proof of the upward LST you make the observation that in dening the assignments σ I : Var D, for an interpretation I there are no restrictions with regards to the cardinality of the domain D of the interpretation I. Thus, all restrictions with regards to the cardinality stem from the set U of sentences that one interprets. If U has innite models, then the model can be of any innite cardinality (you can extend any model by new elements such that you still get a model of U). 5