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8/30/17 Valid arguments in propositional logic Consider the - - PowerPoint PPT Presentation
8/30/17 Valid arguments in propositional logic Consider the - - PowerPoint PPT Presentation
8/30/17 Valid arguments in propositional logic Consider the following argument: CS 220: Discrete Structures and their Applications You need a current password to access department machines You have a current password Logical inference
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8/30/17 3 Verifying argument validity using truth tables
Consider the following argument: p → q p ∨ q ∴ q The argument is valid because whenever the hypotheses are true, the conclusion is true as well.
p q p ∨ q p → q T T T T T F T F F T T T F F F T
Verifying argument validity using truth tables
Consider the following argument: ¬p p → q ∴ ¬q Is it valid?
p q ¬p p → q T T F T T F F F F T T T F F T T
Verifying argument validity using truth tables
Consider the following argument: ¬p p → q ∴ ¬q Is it valid?
p q ¬p p → q T T F T T F F F F T T T F F T T
Is there another way of doing this?
Yes!
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8/30/17 4 Example
Let’s prove the validity of the following argument using inference rules: If it is raining or windy, the game will be cancelled. The game will not be cancelled. Therefore, it is not windy.
Example
In propositional logic: w: It is windy r: It is raining c: The game will be cancelled (r ∨ w) → c ¬c ∴ ¬w
Example
In propositional logic: w: It is windy r: It is raining c: The game will be cancelled (r ∨ w) → c ¬c ∴ ¬w Proof: 1. (r ∨ w) → c Hypothesis 2. ¬c Hypothesis 3. ¬(r ∨ w) Modus tollens, 1, 2 4. ¬r ∧ ¬w De Morgan's law, 3 5. ¬w Simplification, 4
Modus tollens ¬q p → q ∴ ¬p
Inference rules
p p → q Modus ponens ∴ q ¬q p → q Modus tollens ∴ ¬p p Addition ∴ p ∨ q p ∧ q Simplification ∴ p p q Conjunction ∴ p ∧ q p → q q → r Hypothetical syllogism ∴ p → r p ∨ q ¬p Disjunctive syllogism ∴ q p ∨ q ¬p ∨ r Resolution ∴ q ∨ r
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8/30/17 5 Validity of inference rules
We can prove the validity of inference rules as well. Consider Modus Tollens for example: 1. p → q Hypothesis 2. ¬p ∨ q Conditional in terms of disjunction 3. ¬q Hypothesis 4. ¬p Disjunctive syllogism, 2, 3
Inference with quantifiers
We’d like to make inferences such as: Every employee who received a large bonus works hard. Linda is an employee at the company. Linda received a large bonus. ∴ Some employee works hard.
Inference rules for quantified statements
Universal instantiation: c is an element in the domain of P ∀x P(x) ∴ P(c) Example: Sam is a student in the class. Every student in the class completed the assignment. Therefore, Sam completed his assignment.
Inference rules for quantified statements
Universal generalization c is an arbitrary element in the domain of P P(c) ∴ ∀x P(x) Example: Let c be an arbitrary positive integer. 1 ≤ c c ≤ c2 Therefore, every integer is less than or equal to its square.
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8/30/17 6 Inference rules for quantified statements
Existential instantiation ∃x P(x) ∴ (c is a particular element in the domain of P) ∧ P(c) Example: There is an integer that is equal to its square. Therefore, c2 = c, for some integer c.
Inference rules for quantified statements
Existential instantiation ∃x P(x) ∴ P(c) for some c in the domain of P Example: There is an integer that is equal to its square. Therefore, c2 = c, for some integer c.
Inference rules for quantified statements
Existential generalization c is an element in the domain of P P(c) ∴ ∃x P(x) Example: Sam is a student in the class. Sam completed the assignment. Therefore, there is a student in the class who completed the assignment.
particular vs arbitrary elements
What is this distinction between particular and arbitrary elements? Let’s go back to universal generalization: c is an arbitrary element in the domain of P P(c) ∴ ∀x P(x) What would happen if we chose a particular element, let’s say joe instead of an arbitrary element?
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8/30/17 7 particular vs arbitrary elements
What is this distinction between particular and arbitrary elements? Revisiting existential instantiation: ∃x P(x) ∴ (c is a particular element in the domain of P) ∧ P(c) This applied to a particular element, otherwise, it would be true for all elements.
More particular elements
What is the problem with the following proof?
- 1. ∃x P(x)
Hypothesis
- 2. (c is a particular element) ∧ P(c)
Existential instantiation, 1
- 3. ∃x Q(x)
Hypothesis
- 4. (c is a particular element) ∧ Q(c)
Existential instantiation, 3
- 5. P(c)
Simplification, 2
- 6. Q(c)
Simplification, 4
- 7. P(c) ∧ Q(c)
Conjunction, 5, 6
- 8. c is a particular element
Simplification, 2
- 9. ∃x (P(x) ∧ Q(x))