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Propositional Logic: Consistency and
completeness
Reading: Metalogic Part II, 24, 15, 28-
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 |= 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):
A B B ⊃ A (A ⊃ (B ⊃ A)) T T T T T F T T F T F T F F T T
Case 2 (PS2)
62 5 Propositional Logic: Consistency and completeness
X Y Z A B C B ⊃ C A ⊃ (B ⊃ C) A ⊃ B A ⊃ C Y ⊃ Z X ⊃ (Y ⊃ Z) T T T T T T T T T T T F F F T F F T T F T T T F T T T T F F T T F F T T F T T T T T T T T F T F F T T T T T F F T T T T T T T F F F T T T T T T
Case 3 (PS3)
A B ∼ B ∼ A ∼ B ⊃∼ A A ⊃ B (∼ B ⊃∼ A) ⊃ (A ⊃ B) T T F F T T T T F T F F F T F T F T T T T F F T T T T T
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 defining it.
Definition 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 A′, if there is no sentence or propositional form A such that A and A′.
Lemma 5.2.1 The Propositional Calculus with respect to the transformation of A into ∼ A.
Proof. By the definition 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.
Definition 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.
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 T T T Xi T F F Xi F T F Xi F F F
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 A B A ∨ B Xi T T T Xi T F T Xi F T T Xi F F F
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 T T T Xi T F F Xi F T F Xi F F T
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 T F T
So, we have the following rules of inference:
Xi ⊃ A ` Xi ⊃∼∼ A
Next, we put in the following axiom schema:
(X 1 ∧ X 2 ∧ · · · ∧ 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 simplified 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 A 1 ,... , An be the atomic letters in a wff A, and let v be a truth value assignment to A 1 ,... , An. If v(Ai) = T then αi = Ai; if v(Ai) = F then αi =∼ Ai Then, we show
α 1 ,... , αn ` α
For example, consider the wff A ⊃ B. We have the following truth table A B A ⊃ B T T T T F F F T T F F T
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,
α 1 ,... , αn ` B
And by lemma x and MP, we have
α 1 ,... , αn `∼∼ B
Now, since v(A) = F , α =∼ A =∼∼ B. QED Case 2b v(B) = F. So, v(A) = T. By the induction hypothesis,
α 1 ,... , αn `∼ B
But since v(A) = T , α = A =∼ B. QED. Case 3 A = B ⊃ C, where B and C each have fewer than n connectives. Case 3a v(B) = F. So, v(A) = v(B ⊃ C) = T. By the induction hypothesis,
α 1 ,... , αn `∼ B
But, by lemma x and MP, we have
α 1 ,... , αn ` B ⊃ C
QED Case 3b v(C) = T. So, v(A) = v(B ⊃ C) = T. By the induction hypothesis,
α 1 ,... , αn ` C
And by lemma x and MP, since C ` B ⊃ C, we have
α 1 ,... , αn ` B ⊃ C
QED Case 3c v(B) = T and v(C) = F. So, v(A) = v(B ⊃ C) = F. Then we have
α 1 ,... , αn ` B
and α 1 ,... , αn `∼ C
So, by lemma x and MP, we have
α 1 ,... , αn `∼ (B ⊃ C)
But α =∼ (B ⊃ C). QED.
68 5 Propositional Logic: Consistency and completeness
5.4 Homework Exercises
5.4.1 Questions
5.4.2 Answers
