1 / 11

Proof techniques

Proof techniques. CHAPTER TWO. Formal Logic vs. Real-world Arguments. Real-world arguments, unlike the formal proofs of Chapter 1, are normally dependent on context, not the structure of the argument.

dewei
Télécharger la présentation

Proof techniques

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Proof techniques CHAPTER TWO

  2. Formal Logic vs. Real-world Arguments • Real-world arguments, unlike the formal proofs of Chapter 1, are normally dependent on context, not the structure of the argument. • In other words, a real-world argument may not be universally valid, though it be valid in some important context. • Terminology: What is a conjecture? CSC 333

  3. Argument Context • In real-world situations, we often are interested only in the truth of an argument in a particular context. • Example:“If Mary Beth (or some other student) makes an A in CSC 333, then she must be a bright, hard worker.” • Call this the Mary Beth Theorem. CSC 333

  4. Examination of the Mary Beth Theorem • Can we state the M.B. Theorem formally? • Yes. Let “Mary Beth makes an A in CSC 333” be proposition P, and “Mary Beth is a bright, hard worker” be Q. • We can state the M.B. Theorem as P -> Q. • (Or perhaps more properly, let R be “Mary Beth is bright”, and let S be “Mary Beth is a hard worker”; thus, Q can be decomposed as R ^ S • So, we can state the M.B. Theorem as P-> (R ^ S). • We can easily establish a truth table for this. • As stated in the text, if we can’t translate a real-world argument into a formal proof, we should look askance at the argument. CSC 333

  5. Attacking the M.B. Theorem • Disproving by counterexample: • Assume that Mary Beth can be shown to be an imbecile, although she has an A in CSC 333. • This would be a case where R (Mary Beth is bright) is false, making (R ^ S) false. • So, in at least one case P does NOT imply Q. CSC 333

  6. Which is easier? • Proving a conjecture using a formal proof, i.e., showing that for all truth values of the propositions, the theorem holds. OR • Disproving a conjecture by showing one instance in which the theorem “folds” (does not hold), i.e., a counterexample. • Note that showing one example in which it holds is insufficient as a proof. • Aside: How does this apply to software testing? CSC 333

  7. Exhaustive Proofs • If we have a finite population to which we are applying the M.B. Theorem, • say, the students in CSC 333 in spring of 2010, • And we can show the truth of the M.B. Theorem for all those students, • then we have proved the M.B. Theorem by exhaustion. • Aside: How does this apply to software testing? CSC 333

  8. Direct Proof • If we want to prove the M.B. Theorem for all students who ever enroll in CSC 333, we might attempt a direct proof: • Assume P is true. • Show that the conjecture is universally true because R ^ S inevitably follows from P. • For the M.B. Theorem, we can’t show this. • See text for an example of such a proof (p. 92). CSC 333

  9. Contraposition • Proving by contraposition: • We already know that P -> Q is logically equivalent to ~Q -> ~P. • So, if we prove, for example, that if Mary Beth is NOT both bright [R] and a hard worker [S], then Mary Beth will NOT get an A in CSC 333, we have proved the original conjecture. • Note that the if the contrapositive can be shown to be false in at least one case, this disproves the original conjecture. • What if Mary Beth is lazy but cheats cleverly? CSC 333

  10. Contradiction • Proving by contradiction: • Show that P is true and Q is false is a contradiction, i.e., it is always false. • In other words, show that it cannot ever be true that P is true and Q is false. • See example 10, p. 95. • Proving that something is not true is usually more difficult than assuming it is true and then showing a contradiction. CSC 333

  11. Terms • Conjecture • Inductive reasoning • Deductive reasoning • Counterexample • Direct proof • Contraposition • Contradiction CSC 333

More Related