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5 Propositional Logic: Consistency and Completeness 5 Propositional Logic: Consistency and completeness Reading: Metalogic Part II, 24, 15, 28-31 Contents 5.1 Soundness . 61 5.2 Consistency . 62 5.3 Completeness . 63 5.3.1 An Axiomatization of Propositional Logic . 63 5.3.2 Kalmar's Proof: Informal Exposition . 66 5.3.3 Kalmar's Proof . 68 5.4 Homework Exercises . 70 5.4.1 Questions . 70 5.4.2 Answers . 70 5.1 Soundness In this section, we establish the soundness of the system, i.e., Theorem 3 (Soundness). Every theorem is a tautology, i.e., If ` A then j= A. Proof The proof is by induction the length of the proof of A. For the Basis step, we show that each of the axioms is a tautology. For the induction step, we show that if A and A ⊃ B are tautologies, then B is a tautology. Case 1 (PS1): AB B ⊃ A (A ⊃ (B ⊃ A)) TT TT TF TT FT FT FF TT Case 2 (PS2) 62 5 Propositional Logic: Consistency and completeness XYZ ABC B ⊃ C A ⊃ (B ⊃ C) A ⊃ B A ⊃ C Y ⊃ ZX ⊃ (Y ⊃ Z) TTT TTTTTT TTF FFTFFT TFT TTFTTT TFF TTFFTT FTT TTTTTT FTF FTTTTT FFT TTTTTT FFF TTTTTT Case 3 (PS3) AB » B » A » B ⊃∼ A A ⊃ B (» B ⊃∼ A) ⊃ (A ⊃ B) TT FFTTT TF TFFFT FT FTTTT FF TTTTT Case 4 (MP). If A is a tautology, i.e., true for every assignment of truth values to the atomic letters, and if A ⊃ B is a tautology, then there is no assignment which makes A T and B F. Since every assignment makes A T, every assignment must also make B T; so B must also be a tautology. ¤ 5.2 Consistency From Church (1956): The notion of consistency is semantically motivated in that one wants one's system to be free from absurdity or contradiction. It is, however, a syntactical notion, and there are several ways of de¯ning it. De¯nition 29 A logical system is Consistent with Respect to a partic- ular transformation by which each sentence or propositional form A is trans- formed into a sentence or propositional form A0, if there is no sentence or propositional form A such that ` A and ` A0. Lemma 5.2.1 The Propositional Calculus with respect to the transformation of A into » A. Proof. By the de¯nition of tautology, it is not the case both that A and » A are tautologies. In fact, if A is a tautology, » A is a contradiction. Since every theorem is a tautology, it cannot be both that ` A and `» A. QED. De¯nition 30 A logical system is Absolutely Consistent if not all its sen- tences and propositional forms are theorems. Lemma 5.2.2 The Propositional Calculus is absolutely consistent. 5.3 Completeness 63 Proof. The w® A^ » A is not a tautology, and since every theorem is a tautology, 6` (A^ » A). So there is at least one w® that is not provable. QED. De¯nition 31 A logical system is Consistent in the sense of Post (with respect to a certain category of primitive symbols designated as `propositional variables') if a w® consisting of a propositional variable alone is not a theorem. Lemma 5.2.3 The Propositional Calculus is consistent in the sense of Post Proof. A w® consisting of a propositional variable alone is not a tautology because its value for the value F of the variable is F. Since every theorem is a tautology, a propositional variable is not a theorem. QED. 5.3 Completeness Historical Note. Before the notion of a structure or a model had become clear, logicians had attempted to de¯ne the notion of completeness syntactically, although, as we all know, its motivation was semantical. Here (following Church) are some of the de¯nitions pro®ered. De¯nition 32 A Logical system is complete with respect to a given transforma- tion by which each sentence or propositional form A is transformed into a sentence or propositional form A0 if, for every sentence or propositional form B, either ` B or the system, upon addition of B to it as an axiom, becomes inconsistent with respect to the given transformation. De¯nition 33 A logical system is absolutely complete if, for every sentence or propositional form B, either ` B or the system, upon addition of B to it as an axiom, becomes absolutely inconsistent. De¯nition 34 A logical system is complete in the sense of Post if, for every sen- tence or propositional form B, either ` B or the system, upon addition of B to it as an axiom, becomes inconsistent in the sense of Post. 5.3.1 Kalmar's Proof: Informal Exposition So, we want to develop rules of inference for each of our connectives. For example, suppose our w® is A ⊃ B, wh ere A and are each as complicated as one wants. And suppose that we have our basic statement Xi. Here are the possibilities: Xi A B A ⊃ B Xi TTT Xi TFF Xi FTT Xi FFT 64 5 Propositional Logic: Consistency and completeness So, we have the following rules of inference, corresponding to the four lines on the table: Xi ⊃ A;Xi ⊃ B ` Xi ⊃ (A ⊃ B) Xi ⊃ A;Xi ⊃∼ B ` Xi ⊃∼ (A ⊃ B) Xi ⊃∼ A;Xi ⊃ B ` Xi ⊃ (A ⊃ B) Xi ⊃∼ A;Xi ⊃∼ B ` Xi ⊃ (A ⊃ B) Our job is to provide such rules for each of the connectives: CONJUNC- TION Xi A B A ^ B Xi TTT Xi TFF Xi FTF Xi FFF So, we have the following rules of inference, corresponding to the four lines on the table: Xi ⊃ A;Xi ⊃ B ` Xi ⊃ (A ^ B) Xi ⊃ A;Xi ⊃∼ B ` Xi ⊃∼ (A ^ B) Xi ⊃∼ A;Xi ⊃ B ` Xi ⊃∼ (A ^ B) Xi ⊃∼ A;Xi ⊃∼ B ` Xi ⊃∼ (A ^ B) DISJUNCTION Xi ABA_B Xi TTT Xi TFT Xi FTT Xi FFF So, we have the following rules of inference, corresponding to the four lines on the table: Xi ⊃ A;Xi ⊃ B ` Xi ⊃ (A_B) Xi ⊃ A;Xi ⊃∼ B ` Xi ⊃ (A_B) Xi ⊃∼ A;Xi ⊃ B ` Xi ⊃ (A_B) Xi ⊃∼ A;Xi ⊃∼ B ` Xi ⊃∼ (A_B) BICONDITIONAL Xi A B A ´ B Xi TTT Xi TFF Xi FTF Xi FFT 5.3 Completeness 65 So, we have the following rules of inference, corresponding to the four lines on the table: Xi ⊃ A;Xi ⊃ B ` Xi ⊃ (A ´ B) Xi ⊃ A;Xi ⊃∼ B ` Xi ⊃∼ (A ´ B) Xi ⊃∼ A;Xi ⊃ B ` Xi ⊃∼ (A ´ B) Xi ⊃∼ A;Xi ⊃∼ B ` Xi ⊃ (A ´ B) NEGATION Xi A » A »» A Xi TFT So, we have the following rules of inference: Xi ⊃ A ` Xi ⊃∼∼ A Next, we put in the following axiom schema: (X1 ^ X2 ^ ¢ ¢ ¢ ^ Xn) ⊃ Xi where 1 · i · n. Next, to work out tautologies, the second part of our requirement, we include the following rules: (A ^ B) ⊃ C; (A^ » B) ⊃ C ` A ⊃ C B ⊃ C; » B ⊃ C ` C This system is relatively cumbersome, but it can be simpli¯ed and then shown to be sound and complete. That is the heart of Kalmar's pro of. 5.3.2 Kalmar's Proof In this section, we are going to describe a completeness theorem due to Kalmar. This theorem is interesting because it shows how the notion of proof is intimately tied to the truth table idea. Let A1;:::; An be the atomic letters in a w® A, and let v be a truth value assignment to A1;:::; An. If v(Ai) = T then ®i = Ai; if v(Ai) = F then ®i =»Ai Then, we show ®1; : : : ; ®n ` ® For example, consider the w® A ⊃ B. We have the following truth table AB A ⊃ B TT T TF F FT T FF T 66 5 Propositional Logic: Consistency and completeness So, we have to show the following, corresponding to each line of the table: A; B ` A ⊃ B A; » B `» (A ⊃ B) »A; B ` A ⊃ B »A; » B ` A ⊃ B Supposing that we have this truth table lemma, we can prove completeness as follows: Claim (Completeness). If a w® A is a tautology, then it is a theorem. Proof. Let A be a tautology, and let A1;:::; An be the statement letters in it. For a given assignment v, we know that ®1; : : : ; ®n ` ® And since A is a tautology, we have ®1; : : : ; ®n ` A Now, since the assignment to each letter is independent, we have ®1; : : : ; ®n¡1; An ` A and ®1; : : : ; ®n¡1; »An ` A So, by the Deduction Theorem, we have ®1; : : : ; ®n¡1 ` An ⊃ A and ®1; : : : ; ®n¡1 `» An ⊃ A And, from the lemma and MP, we have ®1; : : : ; ®n¡1 ` A Next, we eliminate ®n¡1, and after n steps, we get ` A So, if A is a tautology, i.e., if j= A, then ` A. So, we only have to prove our lemma Lemma 5.3.1 Let A1;:::; An be the atomic letters in a w® A, and let v be a truth value assignment to A1;:::; An. If v(Ai) = T then ®i = Ai; if v(Ai) = F then ®i =»Ai Then, ®1; : : : ; ®n ` ® 5.3 Completeness 67 Proof by induction on the number of occurrences of logical connectives in A. Case 1 n=0. Then the lemma reduces to showing either A ` A or » A `» A and this is trivial. Case 2 A =»B, where B has fewer than n connectives. Case 2a v(B) = T . So, v(A) = F . By the induction hypothesis, ®1; : : : ; ®n ` B And by lemma x and MP, we have ®1; : : : ; ®n `»» B Now, since v(A) = F , ® =»A =»» B. QED Case 2b v(B) = F . So, v(A) = T . By the induction hypothesis, ®1; : : : ; ®n `» B But since v(A) = T , ® = A =»B.
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