Valid Logical Reasoning 2/41 If Alfred is a duck, then Alfred can - PowerPoint PPT Presentation

Valid Logical Reasoning 2/41 If Alfred is a duck, then Alfred can swim Alfred is a duck Alfred can swim Predicate Logic Lectures 2–3 (Chapters 8-10) a = ‘Alfred is a duck’ b = ‘Alfred can swim’ val a ⇒ b ∧ a | = = b / department of mathematics and computer science / department of mathematics and computer science Invalid Logical Reasoning Valid Logical Reasoning? 3/41 4/41 Every animal with feathers is a bird If Alfred is a duck, then Alfred can swim Alfred is an animal with feathers Alfred can swim Alfred is a bird Alfred is a duck a = ‘Every animal with feathers is a bird’ a = ‘Alfred is a duck’ b = ‘Alfred is an animal with feathers’ b = ‘Alfred can swim’ c = ‘Alfred is a bird’ val val a ⇒ b ∧ b | � = = a a ∧ b | � = = c / department of mathematics and computer science / department of mathematics and computer science

Predicate Logic Unary predicate (example) 5/41 6/41 Propositional logic only describes how to reason about Consider the statement 4 m < 5 . complete statements about things; it does not describe how Whether the statement is true or false depends on the value of m : to reason about things themselves. 5 4 on Z Predicate logic will extend propositional logical with: − 4 − 3 − 2 − 1 0 1 2 3 4 on R ◮ names for individuals; ◮ predicates to express properties of individuals; 4 m < 5 is a unary predicate ◮ quantifiers to quantify over individuals. / department of mathematics and computer science / department of mathematics and computer science Equivalence of predicates Universal quantification 7/41 8/41 Predicates P ( x ) and Q ( x ) are equivalent if for all x we have that P ( x ) evaluates to true if, and only if, We express the statement Q ( x ) evaluates to true. “for all m ∈ Z it holds that 4 m < 5 ” as ∀ m [ m ∈ Z : 4 m < 5] Example = m < 5 Universal quantifier ∀ m turns the predicate 4 m < 5 into a proposition. val 4 m < 5 = (on R and Z ) 4 (This proposition is false; take, e.g., m = 2 , 3 , . . . .) val = = m ≤ 1 (on Z , but not on R ) . / department of mathematics and computer science / department of mathematics and computer science

Universal quantification Existential quantification 9/41 10/41 We express the statement In general, if P ( x ) and Q ( x ) are predicates, then we write “there exists m ∈ Z such that 4 m < 5 ” ∀ x [ P ( x ) : Q ( x ) ] as � �� � � �� � domain predicate ∃ m [ m ∈ Z : 4 m < 5] for “all x satisfying P satisfy Q .” Existential quantifier ∃ m turns the predicate 4 m < 5 into a proposition. (This proposition is true; take m = 1 , 0 , − 1 , . . . .) / department of mathematics and computer science / department of mathematics and computer science Existential quantification Domain of Quantification 11/41 12/41 What are the truth values associated with the following propositions? ◮ ∃ x [ x ∈ R : x 2 = 1] val In general, if P ( x ) and Q ( x ) are predicates, then we write = = True ◮ ∃ x [ x ∈ R ∧ x > 0 : x 2 = 1] val ∃ x [ P ( x ) : Q ( x ) ] = = True � �� � � �� � domain predicate ◮ ∃ x [ x ∈ R ∧ x > 2 : x 2 = 1] val = = False for ◮ ∀ x [ x ∈ R : x 2 > 1] val = = False “there exists x satisfying P that also satisfies Q .” ◮ ∀ x [ x ∈ R ∧ x > 0 : x 2 > 1] val = = False ◮ ∀ x [ x ∈ R ∧ x > 2 : x 2 > 1] val = = True / department of mathematics and computer science / department of mathematics and computer science

One- or zero-element domains Domain weakening 13/41 14/41 val NB: P ∧ Q | = = P (all x satisfying P ∧ Q also satisfy P ). Empty domain Examples: ∀ x [ False : P ] val = = True “All gnomes are green!” ∃ x [ x ∈ R ∧ x > 0 : x 2 = 1] ∃ x [ False : P ] val = = False = ∃ x [ x ∈ R : x > 0 ∧ x 2 = 1] val = One-element domain ∀ x [ x ∈ Z ∧ x > 1 : x 2 > 1] ∀ x [ x = n : P ] val = = P [ n for x ] = ∀ x [ x ∈ Z : x > 1 ⇒ x 2 > 1] val ∃ x [ x = n : P ] val = = = P [ n for x ] Domain weakening val Example: ∀ m [ m = 3 : 4 · m < 5] = = 4 · 3 < 5 . ∀ x [ P ∧ Q : R ] val = = ∀ x [ P : Q ⇒ R ] ∃ x [ P ∧ Q : R ] val = = ∃ x [ P : Q ∧ R ] / department of mathematics and computer science / department of mathematics and computer science Domain Splitting De Morgan for quantifiers 15/41 16/41 � ∀ is generalised ∧ see book ∀ x [ x ≤ − 2 ∨ x ≥ 1 : x 2 > 1] ∃ is generalised ∨ = ∀ x [ x ≤ − 2 : x 2 > 1] ∧ ∀ x [ x ≥ 1 : x 2 > 1] val = De Morgan ∃ x [ x ≤ − 2 ∨ x ≥ 1 : x 2 = 1] ¬∀ x [ P : Q ] val = = ∃ x [ P : ¬ Q ] not all = there exists one for which not ¬∃ x [ P : Q ] val not one = all not = ∃ x [ x ≤ − 2 : x 2 = 1] ∨ ∃ x [ x ≥ 1 : x 2 = 1] = = ∀ x [ P : ¬ Q ] val = So: We also have: Domain Splitting ∀ x [ P ∨ Q : R ] val ¬∀ = ∃¬ ; and ¬∀¬ = ∃¬¬ = ∃ ; and = = ∀ x [ P : R ] ∧ ∀ x [ Q : R ] ∃ x [ P ∨ Q : R ] val ¬∃ = ∀¬ . ¬∃¬ = ∀¬¬ = ∀ . = = ∃ x [ P : R ] ∨ ∃ x [ Q : R ] / department of mathematics and computer science / department of mathematics and computer science

Substitution Leibniz 17/41 18/41 Leibniz’s Rule is valid for (quantified) predicates. The Substitution Rule is valid for (quantified) predicates. Example: Example We have the following valid equivalence: We have the following valid equivalence: val val ∀ x [ P ∧ Q : R ] = = ∀ x [ P : Q ⇒ R ] ∃ y [ y = 2 : x ≥ y ] = = x ≥ 2 , and hence, by the Substitution Rule, if we substitute Q for P , ¬ P for and hence, by Leibniz’s Rule, we make another valid equivalence by Q , and R ⇒ S for R , then we get another valid equivalence: applying it in a bigger context: val val ∀ x [ Q ∧ ¬ P : R ⇒ S ] = = ∀ x [ Q : ¬ P ⇒ ( R ⇒ S )] . ∀ x [ x ∈ D : ∃ y [ y = 2 : x ≥ y ] ] = = ∀ x [ x ∈ D : x ≥ 2 ] . / department of mathematics and computer science / department of mathematics and computer science Calculation with quantifiers (example) Binary predicates (example) 19/41 21/41 We show with a calculation that The statement 3 m + n > 3 is a binary predicate. val ∀ x [ P ∧ R : S ] ∧ ∀ x [ Q ∧ R : S ] = = ¬∃ x [ P ∨ Q : ¬ ( R ⇒ S )] : 3 m + n = 3 N ¬∃ x [ P ∨ Q : ¬ ( R ⇒ S )] n val = = { De Morgan } ∀ x [ P ∨ Q : ¬¬ ( R ⇒ S )] val = = { Double Negation } ∀ x [ P ∨ Q : R ⇒ S ] val = = { Domain Splitting } ∀ x [ P : R ⇒ S ] ∧ ∀ x [ Q : R ⇒ S ] 1 val = = { Domain Weakening (2 × ) } − 1 0 1 R m ∀ x [ P ∧ R : S ] ∧ ∀ x [ Q ∧ R : S ] . / department of mathematics and computer science / department of mathematics and computer science

Equivalence of binary predicates Binary predicates (example) 22/41 23/41 The statement 3 m + n > 3 is a binary predicate. Predicates P ( x, y ) and Q ( x, y ) are equivalent if for all x, y we have that P ( x, y ) evaluates to true if, and 3 m + n = 3 N only if, Q ( x, y ) evaluates to true. n Example Let x, y ∈ Z . ¬ ( x = 0 ⇒ y > 2) The predicates val = = { Implication } ¬ ( x = 0 ⇒ y > 2) ¬ ( ¬ ( x = 0) ∨ y > 2) val and = = { De Morgan } 1 ¬¬ ( x = 0) ∧ ¬ ( y > 2) x = 0 ∧ y ≤ 2 − 1 val 0 1 m R = = { Double Negation, Mathematics } assignment are equivalent. x = 0 ∧ y ≤ 2 m=0, n=2 (false prop.) / department of mathematics and computer science / department of mathematics and computer science Binary predicates (example) Binary predicates (example) 24/41 25/41 The statement 3 m + n > 3 is a binary predicate. The statement 3 m + n > 3 is a binary predicate. 3 m + n = 3 3 m + n = 3 N N n n 3 m + 2 > 3 val = = 3 m > 1 val = m > 1 = 3 1 1 − 1 − 1 0 1 m R 0 1 R m assignment m=1, n=2 (true prop.) / department of mathematics and computer science / department of mathematics and computer science