Reasoning for Humans: Clear Thinking in an Uncertain World PHIL 171 - PowerPoint PPT Presentation

Reasoning for Humans: Clear Thinking in an Uncertain World PHIL 171 Eric Pacuit Department of Philosophy University of Maryland pacuit.org

Recap: Truth Tables ( ϕ ∧ ψ ) ( ϕ ∨ ψ ) ϕ ψ ϕ ψ T T T T T T T F F T F T F T F F T T F F F F F F ϕ ψ ( ϕ → ψ ) ϕ ψ ( ϕ ↔ ψ ) T T T T T T T F F T F F F T T F T F F F T F F T ϕ ¬ ϕ T F F T 1

Construct a truth table with columns for ϕ 1 , ϕ 2 , . . . , ϕ n , and ψ . Is there a row in which ϕ 1 , ϕ 2 , . . . , ϕ n are all true and ψ is false ? yes no The argument is invalid (there The argument is is a counterexample) valid 2

Weird Cases: Which arguments are valid? B ⇒ ( A ∨ ¬ A ) A ∨ ¬ A ⇒ B A ∧ ¬ A ⇒ B 3

Weird Cases • Suppose that an argument contains a tautology as a premise. Is the argument valid? 4

Weird Cases • Suppose that an argument contains a tautology as a premise. Is the argument valid? Maybe. 4

Weird Cases • Suppose that an argument contains a tautology as a premise. Is the argument valid? Maybe. • Suppose that the argument contains a contradiction as a premise. Is the argument valid? 4

Weird Cases • Suppose that an argument contains a tautology as a premise. Is the argument valid? Maybe. • Suppose that the argument contains a contradiction as a premise. Is the argument valid? Yes. 4

Weird Cases • Suppose that an argument contains a tautology as a premise. Is the argument valid? Maybe. • Suppose that the argument contains a contradiction as a premise. Is the argument valid? Yes. • Suppose that the argument contains a tautology as a conclusion. Is the argument valid? 4

Weird Cases • Suppose that an argument contains a tautology as a premise. Is the argument valid? Maybe. • Suppose that the argument contains a contradiction as a premise. Is the argument valid? Yes. • Suppose that the argument contains a tautology as a conclusion. Is the argument valid? Yes. 4

Weird Cases • Suppose that an argument contains a tautology as a premise. Is the argument valid? Maybe. • Suppose that the argument contains a contradiction as a premise. Is the argument valid? Yes. • Suppose that the argument contains a tautology as a conclusion. Is the argument valid? Yes. • Suppose that the argument contains a contradiction as a conclusion. Is the argument valid? 4

Weird Cases • Suppose that an argument contains a tautology as a premise. Is the argument valid? Maybe. • Suppose that the argument contains a contradiction as a premise. Is the argument valid? Yes. • Suppose that the argument contains a tautology as a conclusion. Is the argument valid? Yes. • Suppose that the argument contains a contradiction as a conclusion. Is the argument valid? Maybe. 4

ϕ 1 , ϕ 2 , . . . , ϕ n ⇒ ψ denotes an argument with premises ϕ 1 , . . . , ϕ n and conclusion ψ . ϕ 1 , ϕ 2 , . . . , ϕ n | = ψ means that the argument is valid. ϕ 1 , ϕ 2 , . . . , ϕ n �| = ψ means that the argument is invalid. 5

P , P → Q ⇒ Q A , A → B ⇒ B Q , Q → P ⇒ P ¬ A , ¬ A → B ⇒ B ( Q ∨ R ) , ( Q ∨ R ) → P ⇒ P ( P → Q ) , ( P → Q ) → ( R ∨ ( Q → S )) ⇒ ( R ∨ ( Q → S )) 6

P , P → Q | = Q A , A → B | = B Q , Q → P | = P ¬ A , ¬ A → B | = B ( Q ∨ R ) , ( Q ∨ R ) → P | = P ( P → Q ) , ( P → Q ) → ( R ∨ ( Q → S )) | = ( R ∨ ( Q → S )) 6

ϕ, ϕ → ψ | = ψ Every way of replacing ϕ with a formula and ψ with a formula results in a valid argument. 7

Name Valid inference rule Modus Ponens ϕ, ϕ → ψ | = ψ Modus Tollens ϕ → ψ, ¬ ψ | = ¬ ϕ Disjunctive Syllogism ϕ ∨ ψ, ¬ ϕ | = ψ Transitivity ϕ → ψ, ψ → χ | = ϕ → χ 8

Q , P → Q ⇒ P B , A → B ⇒ A P , Q → P ⇒ Q B , ¬ A → B ⇒ ¬ A P , ( Q ∨ R ) → P ⇒ ( Q ∨ R ) ( R ∨ ( Q → S )) , ( P → Q ) → ( R ∨ ( Q → S )) ⇒ ( P → Q ) 9

Q , P → Q �| = P B , A → B �| = A P , Q → P �| = Q B , ¬ A → B �| = ¬ A P , ( Q ∨ R ) → P �| = ( Q ∨ R ) ( R ∨ ( Q → S )) , ( P → Q ) → ( R ∨ ( Q → S )) �| = ( P → Q ) 9

ψ, ϕ → ψ �| = ϕ Some way of replacing ϕ with a formula and ψ with a formula results in an invalid argument. 9

Name Invalid inference rule Denying the Antecedent ¬ ϕ, ϕ → ψ �| = ¬ ψ Affirming the Consequent ψ, ϕ → ψ �| = ϕ Affirming a Disjunct ϕ ∨ ψ, ϕ �| = ¬ ψ 10

( Q ∧ P ) , ( P ∧ Q ) → S ⇒ S ( Q ∧ P ) ≈ ( P ∧ Q ) Q , P → ¬ Q ⇒ ¬ P Q ≈ ¬¬ Q P , ¬ P ∨ Q ⇒ Q ¬ P ∨ Q ≈ P → Q 11

( Q ∧ P ) , ( P ∧ Q ) → S | = S ( Q ∧ P ) ≈ ( P ∧ Q ) Q , P → ¬ Q | = ¬ P Q ≈ ¬¬ Q P , ¬ P ∨ Q | = Q ¬ P ∨ Q ≈ P → Q 11

Observation . If ϕ and ψ are tautologically equivalent, denoted ϕ ≈ ψ , then for all formulas χ , ϕ | = χ if and only if ψ | = χ . χ | = ϕ if and only if χ | = ψ . 12

P Q ¬¬ Q P Q P ∧ Q Q ∧ P T T T T T T T T F F T F F F F T T F T F F F F F F F F F P Q ¬ P ∨ Q P → Q T T T T T F F F F T T T F F T T 13

P Q P ∧ Q P ∧ ¬ Q T T T F T F F T F T F F F F F F P ≈ ( P ∧ Q ) ∨ ( P ∧ ¬ Q ) 14

¬ P ∧ ¬ Q P ∧ ¬ Q P ∧ Q ¬ P ∧ Q Q Q Q Q P P P P P Q T T T F F T F F 15

P Q P P ∧ ¬ Q P ∧ Q Q Q P P 16

P Q R T T T T T F T F T Q P T F F F T T F T F R F F T F F F 17