1. Discrete Structures (DS)
Prof. Dr. Aman Ullah Khan
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Text Book
Discrete Mathematics
and Its Applications
Kenneth H. Rosen
Fourth Edition
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3. Chapter 3 Mathematical Reasoning
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Methods of Proof
A theorem is a statement that can be shown to be true.
The proof of a theorem is “the sequence of statements that form an argument” to demonstrate that the theorem is true.
Axioms or postulates are the underlying assumptions about mathematical structures, the hypothesis of the theorem to be proved and previously proved theorems.
The Rules of Inference are the means to draw conclusions from other assertions and tie together the steps of a proof.
Fallacies are “the incorrect reasoning”.
A Lemma is simple theorem used in the proof of other theorems.
A Corollary is a proposition that can directly established from a theorem that has been proved.
A Conjecture is a statement whose truth value is unknown.
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Rules of Inference
Rules of Inference for Propos ional Logic provides justifications of the steps used to show that a conclusion follows from a set of hypothesis.
A rule has two parts “hypotheses” and a “conclusion”. The conclusion is established (true) if all the hypotheses are true.
The hypotheses are written in a column and the conclusion below a bar. The symbol denotes “therefore”. The Inference rule shown below states that if the both the implication and its hypothesis are known to be true, then conclusion of this implications is true. p
p→q ∴q
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6. Inference Rules Table 3.1.1 Rules of Inference Rule of Inference
Tautology
Name
p→(p∨q)
Addition
(p∧q) →p
Simplification
((p)∧(q)) →(p∧q)
Conjunction
[p∧(p→q)] → q
Modus ponens
[¬q∧(p→q)] → ¬p
Modus tolens
[(p→q)∧(q→r)] → (p→r)
Hypothesis syllogism
[(p∨q) ∧(¬p)] → q
Disjunctive syllogism p
∴ p∨q
p∧q ∴p p q
∴p∧q p
p→q ∴q ¬q
p→q
∴ ¬p
p→q
q→r
∴ p→r
p∨ q ¬p
∴ q
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Examples
Example 1: Suppose that the implication “If it snows today, then we will go skiing” and its hypothesis “it is snowing today”. Then by modus ponens it follows that the conclusion of the implication, “we will go skiing” is true.
Example 2: The implication “if n is divisible by 3, then n2 is divisible by 9”. Consequently, if n is divisible by 3, then by modus ponens, it follows that n2 is divisible by 9.
Example 3: Show which Inference rule is used by the following argument: “It is below freezing now”. Therefore, it is either below freezing or raining now.
Solution: Let p be the proposition “It is below freezing now” and q be the proposition “It is raining now.” Then, this argument is of the form of addition rule namely: Hence, it uses the addition rule. p
∴ p∨q
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Examples ..
Example 4: Show which Rule of Inference is the basis of the following argument: “It is below freezing and raining now. Therefore, it is below freezing now”.
Solution: Let p be the proposition “It is below freezing now” and q be the proposition “It is raining now.” This argument is of the form: Hence, it uses the Simplification rule.
Example 5: State which rule of Inference is used in the argument: “If it rains today, then we will not have a barbecue toady. If it do not have barbecue today, then we will have barbecue tomorrow.”
Solution: Let p be the proposition “It is raining today”, q be the proposition “we will not have a barbecue toady” and r be the proposition “we will have a barbecue tomorrow.” Then this argument is of the form: Hence, this argument is hypothetical syllogism.
p∧q ∴p
p→q
q→r
∴ p→r
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Examples …
Example 6: Show that the hypothesis “It is not sunny this afternoon and it is colder than yesterday,” “We will go swimming only if it is sunny,” “If we do not go to swimming, then we will take a canoe trip,” and “if we take a canoe trip, then we will be home by sunset” lead to the conclusion “We will be home by sunset.”
Solution: Let p :: “It is sunny this afternoon,” q :: “It is colder than yesterday,” r :: “We will go swimming,” s :: “We will take a canoe trip,” and t :: “We will be home by sunset.” Then the hypothesis is:
¬p∧q :: “It is not sunny this afternoon and it is colder than yesterday,”
r→p :: “We will go swimming only if it is sunny,”
¬r→s :: “If we do not go to swimming, then we will take a canoe trip,” and
s→t :: “If we take a canoe trip, then we will be home by sunset”
Construction of Conclusion:
Step #
Step
Reason 1
¬p∧q
Hypothesis 2 ¬p
Simplification using step 1 3
r → p 4 ¬r
Modus tollens using steps 2 and 3 5
¬r → s 6 S
Modus ponens using steps 4 and 5 7
S → t 8 t
Modus ponens using steps 6 and 7
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Examples …
Example 7: Show that the hypothesis “If you send me an message, then I will finish writing the program,” “If you do not send me an message, then I will go to sleep early,” and “If I go to sleep early, then I will wake up feeling refreshed.” lead to the conclusion “If I do not finish writing the program, Then I will wake up feeling refreshed.”
Solution: Let p :: “you send me an message,” q :: “I will finish writing the program,” r :: “I will go to sleep early,” s :: “I will wake up feeling refreshed,” and t :: “We will be home by sunset.” Then the hypotheses are:
p→q :: “If you send me an message, then I will finish writing the program,”
¬p→r :: “If you do not send me an message, then I will go to sleep early,”
r→s :: “If I go to sleep early, then I will wake up feeling refreshed.” and the conclusion
¬q→s :: “If I do not finish writing the program, Then I will wake up feeling refreshed.”
Construction of Conclusion:
Step #
Step
Reason 1
p → q
Hypothesis 2
¬q → ¬p
Contrapositive of step 1 3
¬p → r 4
¬q → r
Hypothetical syllogism using steps 2 and 3 5
r → s 6
Hypothetical syllogism using steps 4 and 5
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Fallacies
Several common fallacies arise in incorrect arguments. These fallacies resembles rules of inference but are based on contingencies rather than tautologies.
The proposition [(p→q)∧q] →p is not a tautology, since it is false when p is false and q is true. However, there are many incorrect arguments that treat this as a tautology. This type of incorrect reasoning is called the fallacy of affirming the conclusion.
The proposition [(p→q)∧¬p] →¬q is not a tautology, since it is false when p is false and ¬q is true. However, there are many incorrect arguments that treat this as a tautology. This type of incorrect reasoning is called the fallacy of denying the hypothesis.
Many incorrect arguments are based on a fallacy called begging the question or circular reasoning. This fallacy occurs when one or more steps of a proof are based on the truth of the statement being proved. In other words, this fallacy arises when a statement is proved using itself, or a statement equivalent to it.
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12. Examples: Fallacy of Affirming the Conclusion
Example 8: Is the following argument valid?
If you do every problem in this book, then you will learn discrete mathematics. You learned discrete mathematics.
Therefore, you did every problem in this book.
Solution: Let p :: “you did every problem in this book,” q :: “You learned discrete mathematics.” Then, this argument is of the form p→q and q, then p. This is an example of incorrect argument using the fallacy of affirming the conclusion. Indeed, it is possible to learn discrete mathematics in some other way than by doing every problem in this book.
Example 9. Let p :: “n≡1 (mod 3)” and let q :: “n2≡1 (mod 3).” The implication p→q :: “if n≡1 (mod 3)”, then n2≡1 (mod 3).” is true. If q is true so that n2≡1 (mod 3), does it follow that p is true namely n≡1 (mod 3)?
Solution: It would be incorrect to conclude that p is true, since it is possible that n≡2 (mod 3). If the incorrect conclusion that p is true is made, this would be an example of the fallacy of affirming the conclusion.
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13. Examples: Fallacy of Denying the Hypothesis
Example 10: Let p and q be the propositions of example 8. If the implication p→q is true and ¬p is true, is it correct to conclude that ¬q is true? In other words, is it correct to conclude that you did not learn discrete mathematics if you did not do every problem in the book, assuming that if you do every problem in this book, then you will learn discrete mathematics?
Solution: It is possible that you learned discrete mathematics even if you did not do every problem in this book. This argument is of the form p→q and ¬p imply ¬q, which is an example of Fallacy of Denying the Hypothesis.
Example 11: Let p and q be as in example 9. Is it correct to assume that if ¬p is true then ¬q is true using the fact that p→q is true? In other words, is it correct to conclude that n2≢1 (mod 3) if n≢1 (mod 3), using the implication: if n≡1 (mod 3), then n2≡1 (mod 3)?
Solution: It is incorrect to conclude that n2≢1 (mod 3) if n≢1 (mod 3), since n2≡1 (mod 3) when n≡1 (mod 3). This is another example of Fallacy of Denying the Hypothesis.
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14. Examples: Fallacy of Begging the Question
Example 12: Is the following argument correct? It supposedly shows that n is even integer whenever n2 is an even integer.
Suppose that n2 is even. Then n2 = 2k for some integer k. Let n = 2l for some integer l. This shows that n is even.
Solution: This argument is incorrect. The statement “let n = 2l for some integer l” occurs in the proof. No argument has been given to show that it is true. This is a circular reasoning because this statement is equivalent to the statement being proved, namely, “n is even.” Of course, the result itself is correct; only the method of proof is incorrect.
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15. Rules of Inference for Quantified Elements
Universal Instantiation is the rule of Inference used to conclude that P(c) is true, where c is a particular member of the universe of discourse, given the premise ∀xP(x).
Universal Generalization is the rule of Inference which states that ∀xP(x) is true, given the premises P(c) is true for all elements c in the universe of discourse.
Existential Instantiation is the rule of Inference which allow us to conclude that there is an element c in the universe of discourse for which P(c) is true, if we know that ∃xP(x) is true.
Existential Generalization is the rule of Inference which is used to conclude that ∃xP(x) is true, when a particular element c with premises P(c) is true. That is, if we know one element c in the universe of discorse for which P(c) is true, then we conclude that ∃xP(x) is true.
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16. Rules of Inference for Quantified Elements …
Table Rules of Inference for Quantified Elements
Rule of Inference
Name
∀xP(x)
∴P(c) if c∈ U
Universal Instantiation
P(c) for an arbitray c∈U
∴ ∀xP(x)
Universal Generalization
∃xP(x)
∴P(c) for some element c∈ U
Existential Instantiation
P(c) for some element c∈ U
∴ ∃xP(x)
Existential Generalization
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17. Example: Rules of Inference for Quantified Elements
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Methods of Proof
Direct Proof
The implication p→q can be proved by showing that if p is true, then q must also be true. This shows that the combination p true and q false never occurs. A proof of this kind is called the direct proof. To carry out such a proof, assume that p is true use rules of inference and theorems already proved to show that q must be true.
Indirect Proof
The implication p→q is equivalent to its contrapositive ¬q→¬p, the implication p→q can be proved by showing that its contrapositive ¬q→¬p, is true. This related implication is proved directly, but any proof technique can be used. An argument of this type is called indirect proof.
Vacuous Proof
Suppose that the hypothesis p of an implication p→q is false. Then the implication p→q is true, because the statement has the form F→T or F→F. Consequently, if it can be shown that p is false, then a proof, called trivial proof of implication p→q can be given.
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Methods of Proof …
Trivial Proof
Suppose that the conclusion q of an implication p→q is true. Then the implication p→q is true, because the statement has the form T→T or F→T. Hence, if it can be shown that q is true, then a proof, called trivial proof of implication p→q can be given.
Proof by Contradiction
Suppose that a contradiction q can be found so that ¬p→q is true, that is, ¬p→F is true. Then the proposition ¬p must be true. Consequently, p must be true. This technique can be used when a contradiction, such as q∧¬q , can be found so that it is possible to show that implication ¬p→q∧¬q is true. An argument of this type is called proof by contradiction.
Proof by Cases
To prove an implication of the form (p1∨p2∨…∨pn) →q the tautology
[(p1∨p2∨… ∨pn) →q ]↔[(p1 → q)∧(p2 → q)∧…∧(pn → q)] can be used a rule of inference. This shows that the original implication with a hypothesis made up of disjunction of the propositions p1,p2,…,pn can be proved by proving each of the n implications pi → q, i=1,2,…,n, individually. Such an argument is called a proof by cases.
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