Table of standard equiences 30/57 372 TABLES FOR PART I Propositional Logic Lecture 2 (Chapter 7) September 9, 2016 Equiences for connectives Commutativity: Associativity: P Q == Q P, (P Q) R == P (Q R), P Q == Q P, (P Q) R == P (Q R), P Q == Q P (P Q) R == P (Q R) Idempotence: Double Negation: P P == P, P == P P P == P Inversion: True/False-elimination: True == False, P True == P, False == True P False == False, P True == True, P False == P Negation: Contradiction: P == P False P P == False Excluded Middle: P P == True Distributivity: De Morgan: P (Q R) == (P Q) (P R), (P Q) == P Q, P (Q R) == (P Q) (P R) (P Q) == P Q Implication: Contraposition: P Q == P Q P Q == Q P Bi-implication: Self-equience: P Q == (P Q) (Q P ) P P == True For the collection of all standard equiences, see page 372 of the book! You will have to know them by heart (including their names!). Start memorising them today! Calculation 31/57 == is a decent equience 32/57 Recall the following calculation: P Q P Q == { Double Negation } P Q Can we conclude P Q == P Q? 1. What about applying two standard equiences in a row? Does it preserve equience? Lemma 6.1.1 1. (Reflexivity:) P == P 2. (Symmetry:) If P == Q, then Q == P 3. (Transitivity:) If P == Q and Q == R, then P == R 2. First step: not a literal application of Implication. Can we do substitutions? 3. Second step: literal application of Double Negation. Is it safe to apply standard equiences in a larger context?
Substitution 35/57 Substitution 36/57 Substitution is the replacement of all occurrences of a letter by a formula. Substitution is the replacement of all occurrences of a letter by a formula. Examples: Examples: 1. If we substitute Q P for P in the id equience 2. If we substitute R for Q in the id equience P Q == P Q, (Q P ) Q == (Q P ) Q, then we get the id equience then we get the id equience (Q P ) Q == (Q P ) Q. ( R P ) R == ( R P ) R. Substitution 37/57 Substitution Rule 38/57 Substitution is the replacement of all occurrences of a letter by a formula. SUBSTITUTION PRESERVES EQUIVALENCE Examples: 3. If we (simultaneously) substitute Q P for P and R for Q in the id equience P Q == P Q, then we get the id equience Important remarks: 1. Substitution operates on entire equiences 2. If you substitute for some letter P in an equience, then you have to replace all occurrences of P in that equience! (Q P ) R == (Q P ) R.
Leibniz s rule 40/57 Leibniz s rule 41/57 Leibniz s rule is about the replacement of a subformula by an equient subformula. Leibniz s rule is about the replacement of a subformula by an equient subformula. Example: From the id equience Schematically: P Q == P Q we can make new id equiences by replacing P Q in some complex formula by P Q, for instance: P P == Q == Q ( P (P Q)) R == ( P ( P Q)) R Proving tautologies method 1 42/57 Proving tautologies method 1 (example) 43/57 Prove with a calculation that (P P ) is a tautology. We have the following calculation: To prove with a calculation that P is a tautology: (P P ) Give calculation that shows P == True. P P == { Double Negation } P P == { Excluded Middle } True So (P P ) is a tautology.
Proving tautologies method 2 44/57 Proving tautologies method 2 (example) 45/57 Prove with a calculation that (Q R) ( R Q) is a tautology. Lemma 6.1.3 If P == Q, then P Q is a tautology, and vice versa. To prove with a calculation that P is a tautology: Give a calculation that shows P == Q. First, we establish, with a calculation, that (Q R) == ( R Q): (Q R) ( Q R) Q R == { Double negation } R Q Explanation: Substituting Q for P and R for Q in P Q == P Q (Implication) we get, by the substitution rule: Q R == Q R. (The application of this equience in the calculation involves an application of Leibniz.) Proving tautologies method 2 (example) 45/57 Proving tautologies method 2 (example) 45/57 Prove with a calculation that (Q R) ( R Q) is a tautology. Prove with a calculation that (Q R) ( R Q) is a tautology. First, we establish, with a calculation, that (Q R) == ( R Q): (Q R) ( Q R) Q R == { Double negation } R Q Explanation: Substituting Q for P and R for Q in (P Q) == P Q (De Morgan) we get, by the substitution rule: ( Q R) == Q R. First, we establish, with a calculation, that (Q R) == ( R Q): (Q R) ( Q R) Q R == { Double negation } R Q Explanation: Substituting Q for P in P == P (Double negation) we get, by the substitution rule: Q == Q. (The application of this equience in the calculation involves an application of Leibniz, and is followed by an application of Commutativity.)
Proving tautologies method 2 (example) 45/57 Logical Consequence 46/57 Prove with a calculation that (Q R) ( R Q) is a tautology. First, we establish, with a calculation, that (Q R) == ( R Q): Recall: P == Q means { (a) whenever P is 1, then also Q is 1 (b) whenever Q is 1, then also P is 1 (Q R) ( Q R) Q R == { Double negation } R Q From (Q R) == R Q it follows (by Lemma 6.1.3) that (Q R) ( R Q) is a tautology. Define: P == Q means P Q 1 1 0 1/0 { (a) whenever P is 1, then also Q is 1 Pronounce P == Q as P is stronger than Q. P == Q: 1s are carried over from P to Q. Logical Consequence (example 1) 47/57 Logical Consequence (example 2) 48/57 { } ==? P ==? P Q { } ==? P Q ==? P Q P Q P P Q 0 0 1 1 0 1 1 1 1 0 0 0 1 1 0 1 extra true P Q P Q P Q 0 0 1 0 0 1 1 1 1 0 0 1 1 1 1 1 So P is stronger than P Q (i.e., P == P Q). So P Q and P Q are incomparable.
Standard weakenings 49/57 Basic properties of == 50/57 Also: - -weakening: P Q == P P == P Q P Q == Q and Q == P Q. P Q P Q P P Q 0 0 0 0 0 0 1 0 0 1 1 0 0 1 1 1 1 1 1 1 Lemma 7.3.1 (1a) P == P. (2) If P == Q, then Q == P, and vice versa. (3) If P == Q and Q == R, then P == R. Extremes: False == P P == True False is strongest of all True is weakest of all Basic properties of == 51/57 Basic properties of == 52/57 Lemma 7.3.2 P == Q if, and only if, P == Q and P == Q. So, if you need to prove P == Q or P == Q by a calculation, then it is enough to prove P == Q. But P == Q (or P == Q) alone is not enough to conclude P == Q! Lemma 7.3.4 P == Q if, and only if, P Q is a tautology.
Substitution Rule for == 53/57 Leibniz for ==? 54/57 The Substitution Rule also works for == and == : Recall Leibniz s Rule for making new equiences: SUBSTITUTION PRESERVES WEAKENING/STRENGHTENING Example We have the following id weakening: P Q == P R and hence, according to the Substitution Rule, if we substitute (Q R) for P and (P Q) for Q, we get another id weakening: (Q R) (P Q) == (Q R) R. P P == Q == Q Can we replace == by == in this rule? Examples Note that, by - -weakening, P Q == P Q. Now consider: 1. (P Q) = (P Q); 2. R (P Q) == R (P Q); and 3. (P Q) R = (P Q) R. Conclusion: replacing == by == does not yield a id rule! Monotonicity 55/57 Example 56/57 We do have the following weaker variant of Leibniz s Rule: Prove with a calculation that (P Q) ( Q (P R)) is a tautology. Monotonicity: (1) If P == Q, then P R == Q R (2) If P == Q, then P R == Q R Example: Since P == P Q by - -weakening, we have: P R == { - -weakening + Monotonicity } (P Q) R. First, we establish that (P Q) == Q (P R): (P Q) ( P Q) P Q == { Double Negation } P Q == { - -weakening + Monotonicity } Q (P R) So, by Lemma 7.3.4, the formula (P Q) ( Q (P R)) is a tautology.
Formal System for Calculation 57/57 We now have a precisely defined formal system for calculating with abstract propositions: standard equiences and standard weakenings; inference rules (viz. reflexivity, symmetry, transitivity, substitution, Leibniz for equality, Monotonicity for weakening) It gives a method to prove in a structured manner that two abstract propositions are equient, or one is stronger/weaker than the other; an abstract proposition is a tautology.