# 1 Introduction to Abstract Mathematics Valid AND Invalid Arguments 2.3 Instructor: Hayk Melikya

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1 Introduction to Abstract Mathematics Valid AND Invalid Arguments 2.3 Instructor: Hayk Melikya melikyan@nccu.edu

2 Introduction to Abstract Mathematics Argument An argument is a sequence of propositions (statements ), and propositional forms. All statements but the final one are called assumptions or hypothesis. The final statement is called the conclusion. An argument is valid if: whenever all the assumptions are true, then the conclusion is true. The conclusion is said to be inferred, entails or deduced from the truth of the premises If today is Wednesday, then yesterday is Tuesday. Today is Wednesday. Yesterday is Tuesday.

3 Introduction to Abstract Mathematics Entailment v A collection of statements P 1,…,P n (premises) entails statement Q (conclusion) if and only if: –Whenever all premises hold the conclusion holds –For every interpretation I that makes all P j hold (true), I also makes Q hold (true) v Notations for valid arguments: P 1,…,P n Q or P 1,…,P n ├ Q v Example Premises: P 1 = “If Socrates is human then Socrates is mortal” P 2 = “Socrates is human” Conclusion: Q = “Socrates is mortal”

4 Introduction to Abstract Mathematics Valid Argument Forms To sat that an argument form is valid if no matter what particular propositions are substituted for the propositional variables in its promises if the resulting promises are true then the conclusion is also true pqp → qq TTTT TFFF FTTT F FTF p and p  q then q Modus Ponens Modus ponens is Latin meaning “method of affirming”. If you are a fish, then you drink water. You drink water. You are a fish.

5 Introduction to Abstract Mathematics Modus Tollens P  Q, ~ Q ├ ~ P If p then q. ~q ~p pqp → q~q~p Modus tollens is Latin meaning "method of denying”. If you are a fish, then you drink water. You are not a fish. You do not drink water.

6 Introduction to Abstract Mathematics Equivalence A student is trying to prove that propositions P, Q, and R are all true. She proceeds as follows. First, she proves three facts: P implies Q Q implies R R implies P. Then she concludes, ``Thus P, Q, and R are all true.'' Proposed argument: Is it valid?

7 Introduction to Abstract Mathematics Valid Argument? Conclusion true whenever all assumptions are true. assumptions conclusion To prove an argument is not valid, we just need to find a counterexample.

8 Introduction to Abstract Mathematics Exercises

9 Introduction to Abstract Mathematics More Exercises

10 Introduction to Abstract Mathematics Contradiction If you can show that the assumption that the statement p is false leads logically to a contradiction, then you can conclude that p is true. You are working as a clerk. If you have won Mark 6, then you would not work as a clerk. You have not won Mark 6.

11 Introduction to Abstract Mathematics Propositional Logic v Method #1: –Go through all possible interpretations and check the definition of valid argument v Method #2: –Use derivation rules to get from the premises to the conclusion in a logically sound way u “derive the conclusions from premises”

12 Introduction to Abstract Mathematics Method #1 v Section 1.3 in the text proves many arguments/inference rules using truth tables v Suppose the argument is: P 1,…,P N therefore Q Create a truth table for formula F=(P 1 & … & P N  Q) Check if F is a tautology

13 Introduction to Abstract Mathematics But Why? v Formula A entails formula B iff (A  B) is a tautology v In general: –premises P 1,…,P N entail Q Iff formula F=(P 1 & … & P N  Q) is a tautology P 1 P 2. P N Q

14 Introduction to Abstract Mathematics Examples P v Q v R ~R P v Q P v Q v R ~R Q valid/invalid? (example 1.3.1 in the book, p. 30)

15 Introduction to Abstract Mathematics Examples v P  Q v Q v entails v P v valid/invalid?

16 Introduction to Abstract Mathematics Example P v Q ~P  ~Q P  Q v valid/invalid? v Any argument with a contradiction in its premises is valid by default…

17 Introduction to Abstract Mathematics Method #2 : Derivations v To prove that an argument is valid: –Begin with the premises –Use valid/sound inference rules –Arrive at the conclusion

18 Introduction to Abstract Mathematics Inference Rules (Logic Rules) v But what are these “inference rules”? They are simply… - valid arguments! v Example: –X  Y –X  Y  Z  W –therefore –Z  W by modus ponens

19 Introduction to Abstract Mathematics Example v (X  Y  Z  W)  K v X  Y v therefore v Z  W v How? v (X  Y  Z  W)  K v X  Y  Z  Wby conjunctive simplification v X  Y v Z  W by modus ponens

20 Introduction to Abstract Mathematics Derivations v The chain of inference rules that starts with the premises and ends with the conclusion is called a derivation: –The conclusion is derived from the premises Such a derivation makes a proof of argument’s validity

21 Introduction to Abstract Mathematics Example v (X  Y  Z  W)  K v X  Y v therefore v Z  W v How? v (X  Y  Z  W)  K v X  Y  Z  Wby conjunctive simplification v X  Y v Z  W by modus ponens derivation

22 Introduction to Abstract Mathematics Definition Let H 1, H 2, … H k and Q be propositional expressions. A propositional expression Q is sad to be logical consequence of H 1, H 2, … H k if Q is true whenever all H 1, H 2, … H k are evaluated to be true. This relationship is expressed symbolically by writing H 1, H 2, … H k Q. Or H 1, H 2, … H k ├ Q This is, what is called rule of inference or valid argument. It is normally read as H 1, H 2, …, and H k yield Q H 1, H 2, …, and H k entails Q.

23 Introduction to Abstract Mathematics Example: PQP  Q P  ( P  Q)(P  (P  Q))  Q TTTTT TFFFT FTTFT FFTFT The following truth table shows P, P  Q ├ Q since it proves that (P  (P  Q))  Q is a tautology

24 Introduction to Abstract Mathematics Not a valid arguments If for some propositional forms H 1, H 2, … H k and Q we have all H 1, H 2, … H k are evaluated to be true and Q false, then we say that Q does not logically follow from H 1, H 2, … H k. We will use the following notation to indicate that by H 1, H 2, … H k ├ ? Q (we call this invalid argument). PQP  Q Q  ( P  Q)(P  (P  Q))  P TTTTT TFFFT FTTTF FFTFT Example: Show that Q, P  Q ├? P ( is not valid argument).

25 Introduction to Abstract Mathematics Theorem (LR): Let P, Q, R, and S be propositional expressions then the followings are valid arguments (rules of logic or inference rules ) 1. P  Q, P├ Q Modus Ponens [MP] 2. P  Q,  Q ├  P Modus Tolens [MT] 3. (P  Q)  (R  S), P  R ├ Q  S Constructive Dilemma [CD] 4. P  Q ├ P Simplification [Simp] 5. P├ P  Q or Q ├ P  Q Addition (Generalization) [Add] 6. P, Q ├ P  Q Conjunction [Conj] 7. P  Q,  P ├ Qor P  Q,  Q ├ P Disjunctive Syllogism [DS](Ellimination) 8. (P  Q)  ( R  S),  Q   S ├  P   R Destructive Dilemma [DD] 9. P  Q, Q  R├ P  RTransitivity [Tr]

26 Introduction to Abstract Mathematics Definition of proof A proof of Q from H 1, H 2, … H k is finite sequence of propositional forms Q 1, Q 2, … Q n such that Q n is same as Q and every Q j is either one of Hi, (i = 1, 2, …, k) or it follows from the proceedings by the logic rules. Note: In these proofs we will follow the following formats: We begin with by listing all the hypotheses (marked as Hyp), then the sequence of propositional forms followed by the reason (short description of rules) that allowed that proposition to be included in proof in the same line and end with the conclusion. To make referencing easier we will number the lines and use abbreviated names of logic rules specified in the Theorem (RL1).

27 Introduction to Abstract Mathematics Example: Prove (P  Q)  (Q  R), P ├ Q Proof: 1. (P  Q)  (Q  R) Hyp 2. PHyp 3. P  Q 3 Add 4. Q  R 3, 1 MP 5. Q 4 S. END

28 Introduction to Abstract Mathematics Example: (P  Q), P  (R  S),  Q ├ (R  S) Proof: 1. (P  Q) Hyp 2. P  (R  S) Hyp 3.  Q Hyp 4.  P 1,3 MT 5. (R  S) 2,4 DS END

29 Introduction to Abstract Mathematics Second Type of Logic rules (Rules of Replacement) v Recall that if two propositional expressions P and Q are logically equivalent P  Q if they have same truth tables. v Now suppose P  Q and P appears in a propositional expression R. If in R some of the appearances of P is replaced by Q then the new resulting propositional expression R’ is logically equivalent to R. v To make proof writing more flexible we will extend rules of logic by adding some simple rules of replacement listed in the following theorem.

30 Introduction to Abstract Mathematics Theorem RR ( See theorem 1.1.1) ( replacement rules) 1. Commutative Law [Com] 2. Associative Law [Assoc] P  Q  Q  P, (P  Q )  R  P  (Q  R) P  Q  Q  P (P  Q )  R  P  (Q  R) 3. Distributive Law [Dist] 4. Contrapositive Law [Contr] P  (Q  R)  (P  Q)  (P  R) (P  Q)  (~ Q  P) P  (Q  R)  (P  Q)  (P  R) 5. DeMorgan Law [DeM] 6. Double Negation [DN] ~ ( P  Q)  (~ P  ~ Q ) ~ ~ (P)  P ~ ( P  Q)  (~ P  ~ Q ) 7. Implication Law [Impl] 8. Equivalence Law [Equiv] (P  Q)  (~ P  Q) P  Q  ( P  Q)  (Q  P) P  Q  (P  Q)  (~ Q  ~ P) 9. Exportation [Exp] 10. Tautology (Identity) [ Taut] (P  Q)  R  P  (Q  R) P  P  P or P  P  P 11. P  t  P and P  c  c 12 P  t  t and P  c  P t-tautolagy and c-contradiction

31 Introduction to Abstract Mathematics Eample: Prove (P  Q), (R  Q) ├ (P  R)  Q Proof: 1. P  Q Hyp 2. R  QHyp 3.  P  Q1 Impl 4.  R  Q2 Impl 5. (  P  Q)  (  R  Q)3,4 Conj 6. (Q   P)  (Q   R) 5 Com 7. Q  (  P   R) 6 Dist 8. Q   (P  R) 7 DeM 9.  (P  R)  Q8 Com 10. (P  R)  Q9 Impl END

32 Introduction to Abstract Mathematics Practice problems 1. Study the Sections 2.3 from your textbook. 2. Be sure that you understand all the examples discussed in class and in textbook. 3. Only after you complete the proof of the Theorem LR from the notes 4. Do the following problems from the textbook: Exercise 2.3, # 2, 3, 7, 8, 15, 26, 36, 43, 44, 46, 51.

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