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### 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.
- 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?

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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.

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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.

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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.

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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?

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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?

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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).

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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?

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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.

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Terms

- Conjecture
- Inductive reasoning
- Deductive reasoning
- Counterexample
- Direct proof
- Contraposition
- Contradiction

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