 Chapter 2 Fundamentals of Logic Dept of Information management National Central University Yen-Liang Chen.

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Chapter 2 Fundamentals of Logic Dept of Information management National Central University Yen-Liang Chen

2.1 Basic connectives and truth table Assertions, called statements or propositions, are declarative sentences that are either true or falseAssertions, called statements or propositions, are declarative sentences that are either true or false New statements can be obtained from existing ones in two ways.New statements can be obtained from existing ones in two ways. –Transform a given statement p into the statement  p –Combine two or more statements into a compound statement

Forming a compound statement

Ex 2.1 s: Phyllis goes out for a walks: Phyllis goes out for a walk t: The moon is outt: The moon is out u: It is snowingu: It is snowing (t  u)  s(t  u)  s t  (  u  s)t  (  u  s)  (s  (u  t))  (s  (u  t))

Ex 2.2 “ If I weigh more than 120 pounds, then I shall enroll in an exercise class ”“ If I weigh more than 120 pounds, then I shall enroll in an exercise class ” p: I weigh more than 120 poundsp: I weigh more than 120 pounds q: I shall enroll in an exercise classq: I shall enroll in an exercise class the four cases of p  qthe four cases of p  q

A word of caution In our everyday language, we often find situations where an implications is used when the intention actually calls for a biconditional.In our everyday language, we often find situations where an implications is used when the intention actually calls for a biconditional. If you do your homework, then you will get to watch the baseball game.If you do your homework, then you will get to watch the baseball game.

p  (q  r)  (p  q)  r

p  (p  q), p  (  p  q)

Key ideas A compound statement is a tautology if it is true for all truth value assignments and a contradiction if it is false for all truth value assignmentsA compound statement is a tautology if it is true for all truth value assignments and a contradiction if it is false for all truth value assignments To show (p 1  p 2  …  p n )  q a valid argument, we need to show this statement is a tautology. If any p i is not true, then no matter what q is the statement is true. Thus, we only need to show that q follows from (p 1  p 2  …  p n ), when all of them are true.To show (p 1  p 2  …  p n )  q a valid argument, we need to show this statement is a tautology. If any p i is not true, then no matter what q is the statement is true. Thus, we only need to show that q follows from (p 1  p 2  …  p n ), when all of them are true. Premises and conclusionPremises and conclusion

2.2 Logic equivalence: the laws of logic Ex 2.7, p  q is equivalent to  p  qEx 2.7, p  q is equivalent to  p  q Definition 2.2. Two statements are said to be logically equivalent, s1  s2, when the statement s1 is true if and only if the statement s2 is trueDefinition 2.2. Two statements are said to be logically equivalent, s1  s2, when the statement s1 is true if and only if the statement s2 is true

(pq)(pq)(qp) (pq)(pq)(qp) (pq)(pq)(qp) (pq)(pq)(qp)

(p  q)  (p  q)  (p  q)

DeMorgan ’ s law  (p  q)  p  q;  (p  q)  p  q  (p  q)  p  q;  (p  q)  p  q

The distributive law p  (q  r)  (p  q)  (p  r)p  (q  r)  (p  q)  (p  r) p  (q  r)  (p  q)  (p  r)p  (q  r)  (p  q)  (p  r)

Observations When s1  s2, then s1  s2 is a tautology; when  s1  s2; then  s1  s2 is a tautologyWhen s1  s2, then s1  s2 is a tautology; when  s1  s2; then  s1  s2 is a tautology When s1  s2 and s2  s3, then s1  s3When s1  s2 and s2  s3, then s1  s3

The laws of logic  p  p  p  p  (p  q)  p  q  (p  q)  p  q  (p  q)  p  q  (p  q)  p  q p  q  q  p;p  q  q  p; p  q  q  pp  q  q  p p  (q  r)  (p  q)  rp  (q  r)  (p  q)  r p  (q  r)  (p  q)  rp  (q  r)  (p  q)  r p  (q  r)  (p  q)  (p  r)p  (q  r)  (p  q)  (p  r) p  (q  r)  (p  q)  (p  r)p  (q  r)  (p  q)  (p  r) p  p  p p  p  p p  F  p p  T  p p  p  T p  p  F p  T  T p  F  F p  (p  q)  p p  (p  q)  p

Observation Definition 2.3, s d, the dual of s, is obtained by replacing  with ,  with , T with F and F with T.Definition 2.3, s d, the dual of s, is obtained by replacing  with ,  with , T with F and F with T. Theorem 2.1. The principle of duality. Let s and t be statements that contain no logical connectives other than  and . If s  t, then s d  t d.Theorem 2.1. The principle of duality. Let s and t be statements that contain no logical connectives other than  and . If s  t, then s d  t d.

Two substitution rules Suppose that the compound statement P is a tautology. If p is a primitive statement that appears in P and we replace each occurrence of p by the same statement q, then the resulting compound statement P1 is also a tautology.Suppose that the compound statement P is a tautology. If p is a primitive statement that appears in P and we replace each occurrence of p by the same statement q, then the resulting compound statement P1 is also a tautology. Let P be a compound statement where p is an arbitrary statement that appears in P, and let q be such a statement such that p  q. Suppose that in P we replace one or more occurrences of p by q. Then this replacement yields the compound statement P1. Under these circumstances P  P1.Let P be a compound statement where p is an arbitrary statement that appears in P, and let q be such a statement such that p  q. Suppose that in P we replace one or more occurrences of p by q. Then this replacement yields the compound statement P1. Under these circumstances P  P1.

Ex 2.10 P:  (p  q)  (  p  q) is a tautologyP:  (p  q)  (  p  q) is a tautology P1:  ((r  s)  q)  (  (r  s)  q)P1:  ((r  s)  q)  (  (r  s)  q) P2:  ((r  s)  (t  u))  (  (r  s)  (t  u))P2:  ((r  s)  (t  u))  (  (r  s)  (t  u))

Ex 2.11 Let P: (p  q)  r be a compound statement.Let P: (p  q)  r be a compound statement. –Because (p  q)  p  q, if P1: (  p  q)  r, then P1  P. Let P: p  (p  q) be a compound statement.Let P: p  (p  q) be a compound statement. –Because  p  p, if P1:  p  (  p  q), then P1  P.

Ex 2.12, Ex 2.13  [(p  q)  r]   [(p  q)  r]   [  (p  q)  r]   [  (p  q)  r]   (p  q)  r   (p  q)  r  (p  q)  r(p  q)  r  (p  q)   (  p  q)   p  q  p  q

Definitions Implication p  qImplication p  q contrapositive,  q  pcontrapositive,  q  p converse, q  pconverse, q  p inverse  p  qinverse  p  q

Ex 2.16, Ex 2.17 (p  q)  (  p  q) (p  q)  (  p  q)  (p  q)  (  p  q) (p  q)  (  p  q)  (p  q)  (p  q) (p  q)  (p  q)  p  (q  q) p  (q  q)  p  F  pp  F  p  [  [(p  q)  r]  q]   [(p  q)  r]  q  [(p  q)  r]  q  (p  q)  (q  r)  [(p  q)  q]  r  q  r

Simplifying the switch network (p  q  r)  (p  t  q)  (p  t  r)  p  [r  (t  q)] (p  q  r)  (p  t  q)  (p  t  r)  p  [r  (t  q)]

2.3 Logic implication: rules of inference (p 1  p 2  …  p n )  q is a valid argument, if the premises are true, then the conclusion is also true.(p 1  p 2  …  p n )  q is a valid argument, if the premises are true, then the conclusion is also true. If any one of p 1, p 2, …, p n is false, the implication is automatically true.If any one of p 1, p 2, …, p n is false, the implication is automatically true. To establish the validity of a given argument is to show that the statement (p 1  p 2  …  p n )  q is a tautology.To establish the validity of a given argument is to show that the statement (p 1  p 2  …  p n )  q is a tautology. The conclusion is deduced or inferred from the truth of premises.The conclusion is deduced or inferred from the truth of premises.

Ex 2.19 [(p  r)  (  q  p)  r]  q [(p  r)  (  q  p)  r]  q

Ex 2.20 [p  ((p  r)  s)]  (r  s) [p  ((p  r)  s)]  (r  s)

Key concepts Definition 2.4. If p and q are arbitrary statements such that p  q is a tautology, then we say that p is logically implies q and we write p  q to denote this situation.Definition 2.4. If p and q are arbitrary statements such that p  q is a tautology, then we say that p is logically implies q and we write p  q to denote this situation. When p  q, we refer to p  q as a logical implication.When p  q, we refer to p  q as a logical implication. If p  q, then p  q is a tautology, and we have p  q and q  p. Conversely, suppose that p  q and q  p, then we have p  q.If p  q, then p  q is a tautology, and we have p  q and q  p. Conversely, suppose that p  q and q  p, then we have p  q.

The rule of inferences The rule of Modus PonensThe rule of Modus Ponens – (method of affirming), the rule of detachment – [p  ( p  q)]  q – [(r  s)  [(r  s)  (  t  u)]  (  t  u) The rule of syllogismThe rule of syllogism –[( p  q)  ( q  r)]  ( p  r)

Ex 2.24 [ (p)  (p  q)  (  q  r) ]   r[ (p)  (p  q)  (  q  r) ]   r

the rule of Modus Tollens (method of denying), [  q  ( p  q)]   p(method of denying), [  q  ( p  q)]   p Ex 2.25 Ex 2.25 [(p  r)  (r  s)  (t  s)  (  t  u)  (  u)]   p[(p  r)  (r  s)  (t  s)  (  t  u)  (  u)]   p

Some notes Some arguments look similar in appearance but are indeed invalid.Some arguments look similar in appearance but are indeed invalid. –[q  ( p  q)]  p –[  p  ( p  q)]  q the rule of conjunction, [(p)  (q)]  (p  q) the rule of conjunction, [(p)  (q)]  (p  q) the rule of disjunctive syllogism, [( p  q)  (  p)]  q the rule of disjunctive syllogism, [( p  q)  (  p)]  q

the rule of contradiction the rule of contradiction [(  p)  (F)]  (p) [(  p)  (F)]  (p)

The rule of contradiction When we want to establish the validity of the argument (p 1  p 2  …  p n )  q, we can establish the validity of the logically equivalent argument (p 1  p 2  …  p n  q)  FWhen we want to establish the validity of the argument (p 1  p 2  …  p n )  q, we can establish the validity of the logically equivalent argument (p 1  p 2  …  p n  q)  F

Ex 2.30

Ex 2.31

Ex 2.32 [(  p  q)  (q  r)  r  p]  F[(  p  q)  (q  r)  r  p]  F

Another inference rule [( p)  (q  r)]  [( p  q)  r][( p)  (q  r)]  [( p  q)  r] [((p 1  p 2  …  p n )  (q  r))  [(p 1  p 2  …  p n  q)  r][((p 1  p 2  …  p n )  (q  r))  [(p 1  p 2  …  p n  q)  r] This result tells us that if we want to establish the validity of the first argument, we may be able to do so by establishing the validity of the corresponding argument.This result tells us that if we want to establish the validity of the first argument, we may be able to do so by establishing the validity of the corresponding argument.

Ex 2.33

2.4 The use of Quantifiers Definition 2.5. A declarative sentence is an open statement ifDefinition 2.5. A declarative sentence is an open statement if –(1) it contains one or more variables, and –(2) it is not a statement, but –(3) it becomes a statement when the variables in it are replaced by certain allowable choices. These allowable choices constitute what is called the universe or universe of discourse. The universe comprises the choices we wish to consider or allow for the variables in the open statement.These allowable choices constitute what is called the universe or universe of discourse. The universe comprises the choices we wish to consider or allow for the variables in the open statement.

definitions Existential quantifier (  ) and universal quantifier (  ) are used to quantify the open statements.Existential quantifier (  ) and universal quantifier (  ) are used to quantify the open statements. In an open statement p(x) the variable x is called a free variable. In the statement  x p(x) the variable x is called a bound variable — it is bound by the existential quantifier . Similarly, in the statement  x p(x) the variable x is bound by the universal quantifier .In an open statement p(x) the variable x is called a free variable. In the statement  x p(x) the variable x is called a bound variable — it is bound by the existential quantifier . Similarly, in the statement  x p(x) the variable x is bound by the universal quantifier .

Ex 2.36 p(x): x  0p(x): x  0 r(x): x 2 -3x-4=0r(x): x 2 -3x-4=0 q(x): x 2  0q(x): x 2  0 s(x): x 2 -3>0s(x): x 2 -3>0  x [p(x)  r(x)]  x [p(x)  q(x)]  x [q(x)  s(x)]  x [r(x)  s(x)]  x [r(x)  p(x)]

Ex 2.37 p(x): x is a rational number, q(x): x is a real numberp(x): x is a rational number, q(x): x is a real number –  x [p(x)  q(x)] e(t): triangle t is equilateral, a(t): triangle t has three angles of 60 e(t): triangle t is equilateral, a(t): triangle t has three angles of 60  –  t [e(t)  a(t)]  x [sin 2 x+cos 2 x=1]  x [sin 2 x+cos 2 x=1]  m  n [41=m 2 +n 2 ]  m  n [41=m 2 +n 2 ]

Ex 2.39 For n:=1 to 20 do A[n]:=n  n-nFor n:=1 to 20 do A[n]:=n  n-n  n (A[n]  0)  n (A[n]  0)  n (A[n+1]=2A[n])  n (A[n+1]=2A[n])  n [(1  n  19)  (A[n]<A[n+1])  n [(1  n  19)  (A[n]<A[n+1])  m  n [(m  n)  (A[m]  A[n])]  m  n [(m  n)  (A[m]  A[n])]

Definitions p(x) and q(x) are called logically equivalent, written as  x [p(x)  q(x)], when p(a)  q(a) is true for each replacement a from the universe. We say that p(x) logically implies q(x), written as  x [p(x)  q(x)], when p(a)  q(a) is true for each replacement a from the universe.p(x) and q(x) are called logically equivalent, written as  x [p(x)  q(x)], when p(a)  q(a) is true for each replacement a from the universe. We say that p(x) logically implies q(x), written as  x [p(x)  q(x)], when p(a)  q(a) is true for each replacement a from the universe.  x [p(x)  q(x)] if and only if  x [p(x)  q(x)] and  x [q(x)  p(x)]  x [p(x)  q(x)] if and only if  x [p(x)  q(x)] and  x [q(x)  p(x)]  x p  q; contrapositive,  x [  q  p]; converse,  x [q  p]; inverse  x [  p  q];  x p  q; contrapositive,  x [  q  p]; converse,  x [q  p]; inverse  x [  p  q];

Examples Ex 2.40.Ex 2.40. –s(x): x is a square; e(x): x is a equilateral; –  x [s(x)  e(x)]; contrapositive, converse, inverse Ex 2.41.Ex 2.41. –p(x):  x  >3; q(x) x>3; –  x [p(x)  q(x)]; contrapositive, converse, inverse Ex 2.42. r(x): 2x+1=5; s(x): x 2 =9Ex 2.42. r(x): 2x+1=5; s(x): x 2 =9 –  x [r(x)  s(x)]  x [r(x)]   x [s(x)]; –but we have  x [r(x)  s(x)]   x [r(x)]   x [s(x)]

Table 2.22  x [r(x)  s(x)]   x [r(x)]   x [s(x)]  x [r(x)  s(x)]   x [r(x)]   x [s(x)]  x [r(x)  s(x)]   x [r(x)]  x [s(x)]  x [r(x)  s(x)]   x [r(x)]  x [s(x)]  x [r(x)  s(x)]   x [r(x)]   x [s(x)]  x [r(x)  s(x)]   x [r(x)]   x [s(x)]  x [r(x)  s(x)]  x [r(x)]   x [s(x)]  x [r(x)  s(x)]  x [r(x)]   x [s(x)]

Ex 2.43  x [p(x)  (q(x)  r(x))]   x [(p(x)  q(x))  r(x)]  x [p(x)  (q(x)  r(x))]   x [(p(x)  q(x))  r(x)]  x [p(x)  q(x)]  x(  p(x)  q(x))  x [p(x)  q(x)]  x(  p(x)  q(x))  x  p(x)  x p(x)  x  p(x)  x p(x)  x  [p(x)  q(x)]  x [  p(x)  q(x)]  x  [p(x)  q(x)]  x [  p(x)  q(x)]  x  [p(x)  q(x)]  x [  p(x)  q(x)]  x  [p(x)  q(x)]  x [  p(x)  q(x)]

Rules for negation  [  x p(x)]   x  p(x)  [  x p(x)]   x  p(x)  [  x p(x)]   x  p(x)  [  x p(x)]   x  p(x)  [  x  p(x)]   x  p(x)   x p(x)  [  x  p(x)]   x  p(x)   x p(x)  [  x  p(x)]   x  p(x)   x p(x)  [  x  p(x)]   x  p(x)   x p(x)

Ex 2.44 p(x): x is odd, q(x): x 2 -1 is evenp(x): x is odd, q(x): x 2 -1 is even  x (p(x)  q(x)). If x is odd, x 2 -1 is even.  x (p(x)  q(x)). If x is odd, x 2 -1 is even.  [  x (p(x)  q(x))]  [  x (p(x)  q(x))]  x  (p(x)  q(x))  x  (p(x)  q(x))  x  (  p(x)  q(x))  x  (  p(x)  q(x))  x  p(x)  q(x))  x  p(x)  q(x))  x [p(x)  q(x)]  x [p(x)  q(x)] There exists an integer x such that x is odd and x 2 -1 is odd.There exists an integer x such that x is odd and x 2 -1 is odd.

examples Ex 2.45Ex 2.45 –  x  y [p(x, y)]   y  x [p(x, y)] Ex 2.46Ex 2.46 –  x  y  z [p(x, y, z)] can be written as  x, y, z [p(x, y, z)] Ex 2.47Ex 2.47 –  x  y [p(x, y)]   y  x [p(x, y)]

Ex 2.48 when a statement involves both existential and universal quantifiers, we must be careful about the order in which the quantifiers are written.when a statement involves both existential and universal quantifiers, we must be careful about the order in which the quantifiers are written. p(x, y): x+y=17p(x, y): x+y=17  x  y p(x, y) is different from  y  x p(x, y)  x  y p(x, y) is different from  y  x p(x, y)

Ex 2.49 What is the negation of  x  y[(p(x,y)  q(x,y))  r(x,y)]What is the negation of  x  y[(p(x,y)  q(x,y))  r(x,y)]

Ex 2.50

2.5 Quantifiers, definitions and the proofs of theorems Ex 2.52.Ex 2.52. –For all n in 2, 4, 6, …, 26, we can write n as the sum of at most three perfect squares. –Table 2.4 shows this by the method of exhaustion. –The method is reasonable when we dealing with a fairly small universe. –When the universe is very large, it is impossible to use the method of exhaustion.

The rule of universal specification. If p(x) is an open statement for a given universe, and if  x p(x) is true, then p(a) is true for each a in the universe.If p(x) is an open statement for a given universe, and if  x p(x) is true, then p(a) is true for each a in the universe. Note that this a is a specific but arbitrarily chosen member from the prescribed universe.Note that this a is a specific but arbitrarily chosen member from the prescribed universe.

Ex 2.53 (b)(c)

The rule of universal generalization. If an open statement p(x) is proved to be true when x is replaced by a specific but arbitrarily chosen element c from our universe, then the universally quantified statement  x p(x) is true. If an open statement p(x) is proved to be true when x is replaced by a specific but arbitrarily chosen element c from our universe, then the universally quantified statement  x p(x) is true. Furthermore, the rule extends beyond a single variable. That is, the same holds for  x  y [p(x, y)],  x  y  z [p(x, y, z)] or more variables.Furthermore, the rule extends beyond a single variable. That is, the same holds for  x  y [p(x, y)],  x  y  z [p(x, y, z)] or more variables.

Ex 2.54

Ex 2.56

Theorems Proving The rule of universal specification and the rule of universal generalization can be applied to prove theorems.The rule of universal specification and the rule of universal generalization can be applied to prove theorems. Theorem 2.2. If k and l are both odd, then k+l is even.Theorem 2.2. If k and l are both odd, then k+l is even. Theorem 2.3. If k and l are both odd, then k  l is also odd.Theorem 2.3. If k and l are both odd, then k  l is also odd.

Theorem 2.4 If m is an even integer, the m+7 is odd.If m is an even integer, the m+7 is odd. Theorem 2.4 uses three different ways to prove the theorem.Theorem 2.4 uses three different ways to prove the theorem. (1) p  q, if m is even then m+7 is even(1) p  q, if m is even then m+7 is even (2)  q  p, if m+7 is even then m is odd(2)  q  p, if m+7 is even then m is odd (3)p  q  F, if m and m+7 are both even, then it is a contradiction.(3)p  q  F, if m and m+7 are both even, then it is a contradiction.

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