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1 Section 1.5 Rules of Inference. 2 Definitions Theorem: a statement that can be shown to be true Proof: demonstration of truth of theorem –consists of.

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Presentation on theme: "1 Section 1.5 Rules of Inference. 2 Definitions Theorem: a statement that can be shown to be true Proof: demonstration of truth of theorem –consists of."— Presentation transcript:

1 1 Section 1.5 Rules of Inference

2 2 Definitions Theorem: a statement that can be shown to be true Proof: demonstration of truth of theorem –consists of series of arguments called axioms or postulates –these are statements of underlying assumptions about mathematical structures, hypotheses of theorem to be proved, and previously proved theorems

3 3 More Definitions Lemma: simple theorem used in proof of other theorems Corollary: proposition that can be established directly from a proved theorem Conjecture: statement whose truth value is unknown

4 4 Rules of Inference Means to draw conclusions from other assertions Rules of inference provide justification of steps used to show that a conclusion follows from a set of hypotheses The next several slides illustrate specific rules of inference

5 5 Addition A true hypothesis implies that the disjunction of that hypothesis and another are true p  p  q or p  (p  q)

6 6 Simplification If the conjunction of 2 propositions is true, then each proposition is true p  q  p or (p  q)  p

7 7 Conjunction If p is true and q is true, then p  q is true p q  p  q or ((p)  (q))  p  q

8 8 Modus Ponens If a hypothesis and implication are both true, then the conclusion is true p p  q  q or (p  (p  q))  q

9 9 Modus Tollens If a conclusion is false and its implication is true, then the hypothesis must be false  q p  q  p or [  q  (p  q)]   p

10 10 Hypothetical Syllogism If an implication is true, and the implication formed using its conclusion as the hypothesis is also true, then the implication formed using the original hypothesis and the new conclusion is also true p  q q  r  p  r or [(p  q)  (q  r)]  (p  r)

11 11 Disjunctive Syllogism If a proposition is false, and the disjunction of it and another proposition is true, the second proposition is true p  q  p  q or, [(p  q)   p]  q

12 12 Using rules of inference We can use the rules of inference to form the basis for arguments A valid argument is an implication in which, when all hypotheses are true, the conclusion is true: (p 1  p 2  …  p n )  q When several premises are involved, several rules of inference my be needed to show that an argument is valid

13 13 Example Let p = “It is Monday” and p  q = “If it is Monday, I have Discrete Math today” Since these statements are both true, then by Modus Ponens: (p  (p  q))  q we can conclude “I have Discrete Math today” (q)

14 14 Another Example Let  q = “I don’t have Discrete Math today” and p  q = “If it is Monday, I have Discrete Math today” If both of the above are true, then by Modus Tollens: [  q  (p  q)]   p we can conclude “It is not Monday” (  p)

15 15 Yet Another Example Let p  q = “If I am taking this class, I passed the test” and q  r = “If I passed the test, I’m a happy camper” If both of the above are true, then by hypothetical syllogism: [(p  q)  (q  r)]  (p  r) we can conclude “If I am taking this class, I’m a happy camper” (p  r)

16 16 A More Complicated Example Randy works hard If Randy works hard, then he is a dull boy If Randy is a dull boy, then he will not get the job Construct an argument using rules for inference to show that the hypotheses: Imply the conclusion: Randy will not get the job

17 17 Example Continued Let p = Randy works hard, q = He is a dull boy, r = He will get the job So we want to prove: (p  (p  q)  (q   r))   r 1.Applying modus ponens to the first part: p p  q  q 2. We now have: (q  (q   r))   r 3. Applying modus ponens again, substituting q for p and  r for q: q q   r  r 4. We have a valid argument!

18 18 Fallacies A fallacy is an argument based on contingencies rather than tautologies; some examples: –Fallacy of affirming the conclusion: [(p  q)  q]  p This is not a tautology because it’s false when p is false and q is true –Fallacy of denying the hypothesis: [(p  q)  p]   q Like the previous fallacy, this is not a tautology because it is false when p is false and q is true

19 19 Rules of Inference for Quantified Statements Universal instantiation:  xP(x)  P(c) if c  U Universal generalization: P(c) for arbitrary c  U   xP(x) Note: c must be arbitrary

20 20 Rules of Inference for Quantified Statements Existential instantiation:  xP(x)  P(c) for some c  U Note that value of c is not known; we only know it exists Existential generalization: P(c) for some c  U   xP(x)

21 21 Example Let P(x) = “A man is mortal”; then  xP(x) = “All men are mortal” Assuming p = “Socrates is a man” is true, show that q = “Socrates is mortal” is implied This is an example of universal instantiation: P(Socrates) = “Socrates is mortal”; Since  xP(x)  P(c) Also, by modus ponens: (p  (p  q))  q

22 22 Section 1.5 Rules of Inference - ends -


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