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

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

Dec 19, 2023 16 likes •317 views

Reasoning for Humans: Clear Thinking in an Uncertain World PHIL 171 Eric Pacuit Department of Philosophy University of Maryland pacuit.org Recap: Truth Tables ( ) ( ) T T T T T T T F F T F T F

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Reasoning for Humans: Clear Thinking in an Uncertain World

PHIL 171

Eric Pacuit

Department of Philosophy University of Maryland pacuit.org

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Recap: Truth Tables

ϕ ψ (ϕ ∧ ψ) T T T T F F F T F F F F ϕ ψ (ϕ ∨ ψ) T T T T F T F T T F F F ϕ ψ (ϕ → ψ) T T T T F F F T T F F T ϕ ψ (ϕ ↔ ψ) T T T T F F F T F F F T ϕ ¬ϕ T F F T

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Construct a truth table with columns for ϕ1, ϕ2, . . ., ϕn, and ψ. Is there a row in which ϕ1, ϕ2, . . ., ϕn are all true and ψ is false? The argument is valid The argument is invalid (there is a counterexample) no yes

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Weird Cases: Which arguments are valid?

B ⇒ (A ∨ ¬A) A ∨ ¬A ⇒ B A ∧ ¬A ⇒ B

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Weird Cases

  • Suppose that an argument contains a tautology as a premise. Is the

argument valid?

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Weird Cases

  • Suppose that an argument contains a tautology as a premise. Is the

argument valid? Maybe.

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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?

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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.

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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?

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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.

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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?

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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.

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ϕ1, ϕ2, . . . , ϕn ⇒ ψ denotes an argument with premises ϕ1, . . . , ϕn and conclusion ψ. ϕ1, ϕ2, . . . , ϕn | = ψ means that the argument is valid. ϕ1, ϕ2, . . . , ϕn | = ψ means that the argument is invalid.

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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))

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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))

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

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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

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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)

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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)

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

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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

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

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

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  • 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.

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

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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)

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P Q ¬P ∧ ¬Q P Q P ∧ ¬Q P Q P ∧ Q P Q ¬P ∧ Q

P Q T T T F F T F F 15

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P Q P P Q P ∧ ¬Q P Q P ∧ Q

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

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