Axioms — Fill in the Blanks! Anything Wrong? false (3.1) Axiom, Associativity of ≡ : = � (3.15) ¬ p ≡ p ≡ false � (3.2) Axiom, Symmetry of ≡ : ¬ p ≡ p (3.3) Axiom, Identity of ≡ : = � (3.11) ¬ p ≡ q ≡ p ≡ ¬ q � (3.8) Axiom, Definition of false : ¬ p ≡ ¬ p ≡ q ≡ ¬ q (3.9) Axiom, Distributivity of ¬ over ≡ : = � (3.14) ( p �≡ q ) ≡ ¬ p ≡ q , twice � (3.10) Axiom, Definition of �≡ : p �≡ p �≡ q ≡ ¬ q (3.24) Axiom, Symmetry of ∨ : = � (3.14) ( p �≡ q ) ≡ ¬ p ≡ q , twice � (3.25) Axiom, Associativity of ∨ : p ≡ p ≡ ¬ q ≡ ¬ q (3.26) Axiom, Idempotency of ∨ : = � (3.2) Symmetry of ≡� (3.27) Axiom, Distributivity of ∨ over ≡ : p ≡ p (3.28) Axiom, Excluded Middle : = � (3.3) Identity of ≡� true Plan for Today Textbook Chapter 3: Propositional Calculus Mathematics for Computing Equational Logic as Calculus of Boolean Expressions COMP SCI 1FC3 Equivalence, true McMaster University, Winter 2013 Negation, Inequivalence, false Disjunction Wolfram Kahl Conjunction kahl@cas.mcmaster.ca Calculating puzzle solutions: Knights and Knaves 30 January 2013 Theorems Equivalence Axioms, and Theorem (3.4) A theorem is (3.1) Axiom, Associativity of ≡ : (( p ≡ q ) ≡ r ) ≡ ( p ≡ ( q ≡ r )) either an axiom or the conclusion of an inference rule where the premises (3.2) Axiom, Symmetry of ≡ : p ≡ q ≡ q ≡ p are theorems (3.3) Axiom, Identity of ≡ : true ≡ q ≡ q or a Boolean expression proved (using the inference rules) equal to an axiom or a previously proved theorem . The least interesting theorem , shorter proof: Such proofs will be presented in the calculational style . Proving (3.4) true : Note: true “ theorem ” is a syntactic concept − → proofs “ validity ” is a semantic concept − → truth tables = � (3.2) Symmetry of ≡ , with p : = true � For the propositional logic E , theoremhood and validity true ≡ q ≡ q — This is (3.3) Identity of ≡ coincide: All theorems in E are valid: E is sound All valid Boolean expressions are theorems in E : E is complete
Equivalence Axioms and Theorems Proving Equivalences Using Equality Chains (3.1) Axiom, Associativity of ≡ : (( p ≡ q ) ≡ r ) ≡ ( p ≡ ( q ≡ r )) (3.6) Proof Method : To prove that P ≡ Q is a theorem, trans- form P to Q or Q to P using Leibniz. (3.2) Axiom, Symmetry of ≡ : p ≡ q ≡ q ≡ p Justification: Each transformation of P to Q using Leibniz can be turned into a proof of P ≡ Q : (3.3) Axiom, Identity of ≡ : true ≡ q ≡ q true — This is (3.4) Theorems and Metatheorems: = � (3.3), true = q = q � (3.4) true P ≡ P P (3.5) Reflexivity of ≡ : p ≡ p = � Hint 0 � = � Hint 0 � (3.6) Proof Method : To prove that P ≡ Q is a theorem, trans- R P ≡ R form P to Q or Q to P using Leibniz. = � Hint 1 � = � Hint 1 � (3.7) Metatheorem : Any two theorems are equivalent. . . . . . . = � Hint 2 � = � Hint 2 � Q P ≡ Q \ ThisIs { ... } — also Before First Step (3.6) Proof Method : To prove that P ≡ Q is a theorem, trans- form P to Q or Q to P using Leibniz. true — This is (3.4) true — This is (3.4) = � (3.3), true = q = q � = � (3.3), true = q = q � P ≡ P P ≡ P P = � Hint 0 � P ≡ R = � Hint 0 � = � Hint 0 � R P ≡ R = � Hint 1 � = � Hint 1 � = � Hint 1 � . . . . . . . . . = � Hint 2 � P ≡ Q = � Hint 2 � = � Hint 2 � P ≡ Q Q \begin{calc} (3.22) Principle: Structure proofs to avoid repeating the same \true \ThisIs{(3.4)} \CalcStep{=}{(3.3), $\true = q = q$} subexpression on many lines. P \equiv P \CalcStep{=}{Hint 0} A More Subtle Puzzle A More Subtle Puzzle — Solution 1 We now have only two brothers (identical twins). One of A T : Arthur is truthful them is named Arthur and the other has a different name. � ( Arthur answers X ) ≡ X � Axiom: A T ≡ One of the two always lies and the other always tells the A T ≡ Arthur answers A T ≡ A T truth, but we are not told whether Arthur is the liar or the truth-teller. One day you meet the two brothers together, = � (3.2) symmetry: p ≡ q ≡ q ≡ p , twice � and you wish to find out which one is Arthur. Arthur answers A T You are allowed to ask just one of them a question answer- able by yes or no, and again the question may not contain A T ≡ � Henry answers X ≡ ¬ X � Axiom: more than three words. What question would you ask? A T ≡ Henry answers ¬ A T ≡ ¬¬ A T = � (3.12) Double negation ¬¬ p = p � A T ≡ Henry answers ¬ A T ≡ A T Raymond Smullyan: To Mock a Mockingbird = � (3.2) symmetry: p ≡ q ≡ q ≡ p , twice � Henry answers ¬ A T
Negation Axioms and Theorems A More Subtle Puzzle — Solution 2 false ≡ ¬ true (3.8) Axiom, Definition of false : A L : Arthur lies � � A L ≡ Arthur answers X ≡ ¬ X Axiom: (3.9) Axiom, Distributivity of ¬ over ≡ : ¬ ( p ≡ q ) ≡ ¬ p ≡ q Can be used as: A L ≡ Arthur answers ¬ A L ≡ ¬¬ A L ¬ ( p ≡ q ) = ( ¬ p ≡ q ) = � (3.12) ¬¬ p = p � ( ¬ ( p ≡ q ) ≡ ¬ p ) = q A L ≡ Arthur answers ¬ A L ≡ A L (3.10) Axiom, Definition of �≡ : ( p �≡ q ) ≡ ¬ ( p ≡ q ) = � (3.2) symmetry: p ≡ q ≡ q ≡ p , twice � Theorems: Arthur answers ¬ A L (3.11) ¬ p ≡ q ≡ p ≡ ¬ q ( ¬ p ≡ ¬ q ) ≡ ( p ≡ q ) � � A L ≡ Henry answers X ≡ X Axiom: (3.12) Double negation : ¬¬ p ≡ p (3.13) Negation of false : ¬ false ≡ true A L ≡ Henry answers A L ≡ A L (3.14) ( p �≡ q ) ≡ ¬ p ≡ q = � (3.2) symmetry: p ≡ q ≡ q ≡ p , twice � (3.15) ¬ p ≡ p ≡ false Henry answers A L Raymond Smullyan posed many puzzles about an island that You encounter two people A and B . What are A and B if has two kinds of inhabitants: A says “We are of the same type.”? 4 knights , who always tell the truth, and A V : A is a knave knaves , who always lie. Axiom: A V ≡ A says X ≡ ¬ X You encounter two people A and B . What are A and B if ≡ A says ( A V ≡ B V ) ≡ ¬ ( A V ≡ B V ) A V A says “We are both knaves.”? 1 = � (3.2) symmetry of ≡� A says “At least one of us is a knave.”? A says ( A V ≡ B V ) ≡ ≡ ¬ ( A V ≡ B V ) A V 2 = � (3.9) ¬ ( p ≡ q ) ≡ ¬ p ≡ q � A says “If I am a knight, then so is B .”? 3 A says ( A V ≡ B V ) ≡ ≡ ≡ ¬ B V A V A V A says “We are of the same type.”? 4 = � (3.2) symmetry of ≡ : p ≡ q ≡ q ≡ p � A says “ B is a knight” and A says ( A V ≡ B V ) ≡ ¬ B V 5 B says “The two of us are opposite types.”? Avoid Repetition in Proofs! You encounter two people A and B . What are A and B if A says “We are of the same type.”? 4 (3.22) Principle: Structure proofs to avoid repeating the same A V : A is a knave subexpression on many lines. Axiom: A V ≡ A says X ≡ ¬ X A says X ≡ A V ≡ ¬ X Textbook, p. 48 A says ( A V ≡ B V ) = � Def. “says” � ≡ ¬ ( A V ≡ B V ) A V = � (3.9) ¬ ( p ≡ q ) ≡ ¬ p ≡ q � A V ≡ A V ≡ ¬ B V = � (3.2) symmetry of ≡ : p ≡ q ≡ q ≡ p � ¬ B V
Inequivalence Theorems (3.23) Heuristic of Definition Elimination (3.16) Symmetry of �≡ : ( p �≡ q ) ≡ ( q �≡ p ) To prove a theorem concerning an operator ◦ that is defined in terms of another, say • , expand the definition (3.17) Associativity of �≡ : (( p �≡ q ) �≡ r ) ≡ ( p �≡ ( q �≡ r )) of ◦ to arrive at a formula that contains • ; exploit prop- (( p �≡ q ) ≡ r ) ≡ ( p �≡ ( q ≡ r )) (3.18) Mutual associativity : erties of • to manipulate the formula, and then (possibly) p �≡ q ≡ r ≡ p ≡ q �≡ r (3.19) Mutual interchangeability : reintroduce ◦ using its definition. Textbook, p. 48 “Unfold-Fold strategy” A says “ B is a knight” and 5 Inequivalence Theorems: Symmetry B says “The two of us are opposite types.”? ≡ P H ≡ X P says X (3.16) Symmetry of �≡ : ( p �≡ q ) ≡ ( q �≡ p ) B says ( A H �≡ B H ) Proving (3.16) Symmetry of �≡ : A says B H = � Def. “says” � p �≡ q = � Def. “says” � B H ≡ ( A H �≡ B H ) — Unfold A H ≡ B H = � (3.10) Definition of �≡ � = � (3.14) ( p �≡ q ) ≡ ¬ p ≡ q � ¬ ( p ≡ q ) B H ≡ ¬ A H ≡ B H = � (3.2) Symmetry of ≡� = � symmetry of ≡ : p ≡ q ≡ q ≡ p � ¬ ( q ≡ p ) ¬ A H — Fold = � (3.10) Definition of �≡ � = � idempotency of conjunction � q �≡ p ¬ A H ∧ ¬ A H = � Above: “ A says B H ” � ¬ A H ∧ ¬ B H (3.21) Heuristic Identify applicable theorems by matching the structure of expressions or subexpressions. The operators that appear in a boolean expression and the shape of its subexpres- sions can focus the choice of theorems to be used in ma- nipulating it. Obviously, the more theorems you know by heart and the more practice you have in pattern matching, the easier it will be to develop proofs. Textbook, p. 47
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