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William Yeh March 10, 2003 Project 1 xxP ( P( x( x) )QxQ ( x())x) Part A: ?<=>? A-1. They are equivalent. <=> A-2. xxP ( P( x( x) )QxQ ( x())x) We will prove => xxP ( P( x( x) )QxQ ( x())x) (1) P( x) Q( x) => (3) P(x) => Q (x ) => first, using a direct proof. / (2) (for an arbitrary xU) (1) Universal Instantiation (2) Simplification (4) (2) Simplification x( P( x)) x(Q( x)) => (5) (since x was arbitrary) (3) Universal Generalization => (since x was arbitrary) (4) Universal Generalization (6) x( P( x)) x(Q( x)) => (7) (5) + (6) Conjunction xxP ( P( x( x) )QxQ ( x())x) We will now prove the other direction: => with a direct proof. xxP ( P( x( x) )QxQ ( x())x) (1) / x( P( x)) => (2) (1) Simplification x(Q( x)) => (3) (1) Simplification P(x) => Q (x ) => (4) (for an arbitrary xU) (2) Universal Instantiation (5) (for an arbitrary xU) (3) Universal Instantiation P( x) Q( x) => x( P( x) Q( x)) => (6) (4) + (5) Conjunction (7) (since x was arbitrary for both P(x) and Q(x)) (6) Universal Generalization Since we have proven both implication directions of the statement, we can conclude that the statements are indeed equivalent. A-3. xxP ( P( x( x) )QxQ ( x())x) Why does imply ? x( P( x) Q( x)) The statement , being true, has the meaning, that for every such object in some universe, each of those objects satisfies two conditions. For example, let our universe be all students at Rutgers. Let P(x) mean “x is a student” and Q(x) mean “x attends Rutgers University.” For all x in our universe, that is, students at Rutgers, P(x) and Q(x) is true, thus is true. xP( x) xQ( x) With this being said, the statement is implied by the first. For the first statement to be true, every object in some universe must satisfy two conditions. What this second statement is saying is that for every object x in a universe, some condition is satisfied, and for every object x in the same universe, some other condition is also satisfied. x(Q P( x)) Returning to our example, we know that every student at Rutgers is both a student and attends Rutgers University. So we ask ourselves, is every student at Rutgers a student (that is, )? Yes, of course. Then, is every student at Rutgers attending Rutgers University (that is, )? Yes again. In summary, if we know that every object in an universe satisfies two conditions, we know that every object in that universe satisfies one of the conditions, as well as, every object in that universe satisfies the second condition. xxP ( P( x( x) )QxQ ( x())x) Why does imply ? xP( x) xQ( x) The statement has the meaning that for every x in a universe, some condition is true. Also, for every x in the same universe, some other condition is true. Knowing that those two conditions are true for all x in an universe, we can conclude that for all x in a universe, those two conditions are true. Using the same example mentioned previously, all students at Rutgers, are both, students, and attend Rutgers. Knowing that all students at Rutgers satisfies those two conditions, we can say that all students at Rutgers are students and attend Rutgers. xxP ( P( x( x) )QxQ ( x())x) Part B: ?<=>? B-1. The second implies the first. <= B-2. xxP ( P( x( x) )QxQ ( x())x) We will prove => using a proof by cases. xxP (Q P( x( x) )) xQ( x) We start off with knowing that is true. With this in mind, we know that one of or MUST be true. And these will be our cases. x(Q P( x)) Case 1: is true. We do not care whether is true or not. xP( x) xQ( x) x( P( x)) ) Q( x)) P(x) => (1) (2) (assumed premise) (3) (for an arbitrary xU) / (2) Universal Instantiation P( x) Q( x) => (4) (x still arbitrary) x( P( x) Q( x)) => (3) Addition (5) (since x was arbitrarily chosen from our universe) (4) Universal Generalization x(Q P( x)) Case 2: is true. We do not care whether is true or not. xP( x) xQ( x) (1) x(Q P( x))) Q( x)) Q (x ) => (2) (assumed premise) (3) (for an arbitrary xU) / (2) Universal Instantiation Q( x) P( x) => (4) (x still arbitrary) (3) Addition P( x) Q( x) <=> (5) (4) Commutative Laws x( P( x) Q( x)) => (6) (since x was arbitrarily chosen from our universe) (5) Universal Generalization xxP ( P( x( x) )QxQ ( x())x) We have shown that in every case, the statement implies . B-3. xxP ( P( x( x) )QxQ ( x())x) Why does imply ? xxP ( P( x( x) )QxQ ( x())x) The statement, , being true, says that at least one of these following assertions are true. Either all objects in the universe satisfy some condition or all objects in the universe satisfy some other condition. The statement says that all objects in the universe satisfy at least one of two conditions. xxP ( P( x( x) )QxQ ( x())x) If the first statement, , is true, we know that all objects in some universe satisfy some condition. Then, we know for sure that all objects in that same universe satisfy that condition or some other condition. Therefore, if is true, must be true as well. For example, let our universe consist of positive numbers. Let P(x) mean “x is a positive number” and Q(x) to mean “x is a negative number.” For all positive numbers, xP( x) xQ( x) of course, P(x) is true, since all positive numbers are positive numbers. Therefore the first statement, is true. Let us consider the second statement. Are all positive numbers either positive or negative? Of course, we already know that all positive numbers are positive. Therefore, the conclusion is true. B-4. xxP ( P( x( x) )QxQ ( x())x) Why is => not true? xxP ( P( x( x) )QxQ ( x())x) Here’s a simple counterexample. Let our universe consist of all people (except hermaphrodites and neuters). Let P(x) stand for “x is male” and Q(x) represent “x is female.” The first statement asks, are all people male or female? Yes, this is true. Does this imply the conclusion? The conclusion asks, are all people male or all people female? This, of course, is not true. Therefore, => is not true, because we have provided a counterexample which makes the premise true, yet the conclusion is false. xP xQ x) Part C: x( P( x( x) )Q ( x()) ?<=>? C-1. The first implies the second. => C-2. xP xQ x) We will prove => with a direct proof. x( P( x( x) )Q ( x()) xP xQ x) x( P( x( x) )Q ( x()) P( s ) Q( s ) => P(s ) => => Q(s ) (4) (3) (1) (2) (for a particular s) / (1) Existential Instantiation (2) Simplification (2) Simplification xP(x) => (5) (3) Existential Generalization xQ(x) => (6) (4) Existential Generalization xP( x) xQ( x) => (7) (5) + (6) Conjunction C-3. xP xQ x) Why is => true? x( P( x( x) )Q ( x()) x( P( x) Q( x)) Given the statement being true, we know that there is some x in our universe that satisfies two conditions. Knowing that, is the conclusion true? The conclusion states that there is some x that satisfies one condition, P(x), and there is some x that satisfies another condition, Q(x). From the hypothesis, we already know that there is some x that satisfies both P(x) and Q(x). Therefore, that x satisfies P(x) and that x satisfies Q(x) and the conclusion is also true. For example, let our universe be people and let P(x) stand for “x is named Stephen Max” and let Q(x) represent “x teaches cs205.” The hypothesis is true, so we know that there is someone named Stephen Max that teaches cs205. Then, the conclusion asks, is there a person named Stephen Max and is there a person that teaches cs205? The answer is yes, there is someone named Stephen Max, and there is a person that teaches cs205, namely, Stephen Max. C-4. xP xQ x) Why is => not true? x( P( x( x) )Q ( x()) Here’s a simple counterexample. Let our universe be people, and let P(x) stand for “x is named Stephen Max and x teaches cs205” and let Q(x) represent “x is named William Yeh and x takes cs205.” There does exist someone named Stephen Max who teaches cs205. There also exists someone named William Yeh who is taking cs205. Then is the conclusion true? Is there someone named Stephen Max and William Yeh and who teaches cs205 and takes cs205? Maybe, but we cannot know for sure from our given hypothesis. Therefore the conclusion can be either false or true given that the hypothesis is true, which in logic, makes the entire implication false. xP xQ x) Part D: x( P( x( x) )Q ( x()) ?<=>? D-1. They are equivalent. <=> D-2. xP xQ x) We will prove that => with a proof by cases. x( P( x( x) )Q ( x()) xP xQ x) x( P( x( x) )Q ( x()) P( s ) Q( s ) => (1) (2) (for a specific s) / (1) Existential Instantiation P( s) Q( s) Here is where the proof by cases comes in. If is true, then either P(s) or Q(s) MUST be true, and these will be our cases. Q P(s ) Case 1: is true. And we do not care if is true. P(s ) xP(x) => (3) (for a specific s) Assumed Premise (4) (3) Existential Generalization xP( x) xQ( x) => (5) (4) Addition Q P(s ) Case 2: is true. And we do not care if is true. Q(s ) xQ(x) => (3) (for a specific s) (4) Assumed Premise (3) Existential Generalization xQ( x) xP( x) => (5) (4) Addition xP( x) xQ( x) <=> (6) (5) Commutative Laws xP xQ x) We have proven that => for all cases. x( P( x( x) )Q ( x()) xP xQ x) We will prove that => also with a proof by cases. x( P( x( x) )Q ( x()) xQ((x (x xP x)) xQ( x) We know that is true, and therefore either or MUST be true. These will be our cases. xQ(x (x)) Case 1: is true. And we do not care if is true or false. xP xP( x) xQ( x) (1) xP x( P(x( x) ) Q( x)) (2) (assumed premise) / P(s ) P( s ) Q( s ) => => (3) (for a specific s) (4) (for same specific s) x( P( x) Q( x)) => (2) Existential Instantiation (3) Addition (5) (4) Existential Generalization xQ(x (x)) Case 2: is true. And we do not care if is true or false. xP xP( x) xQ( x) (1) xQ x( P(x ( x)) Q( x)) (2) Q(s ) => (assumed premise) (3) (for a specific s) / (2) Existential Instantiation Q( s ) P( s ) => (4) (for same specific s) (3) Addition P( s ) Q( s ) <=> (5) (4) Commutative Laws x( P( x) Q( x)) => (6) (5) Existential Generalization xP xQ x) We have proven that => is true for all cases. x( P( x( x) )Q ( x()) xP xQ x) We have proven both directions of the statement <=> . Therefore, x( P( x( x) )Q ( x()) they are equivalent. D-3. xP xQ x) Why is => true? x( P( x( x) )Q ( x()) The hypothesis states that there is some x in our universe that satisfies at least one of two conditions. The conclusion states that either one of the following is true: something satisfies condition one or something satisfies condition two. Given that there exists something that satisfies one of two conditions (the hypothesis), we know for sure that there exists something that satisfies condition one or there exists something that satisfies condition two. For an example, let our universe be people, let P(x) stand for “x plays for the Houston Rockets” and Q(x) represent “x plays for the Los Angeles Lakers.” We know that there exists someone that either plays for the Rockets or the Lakers. Let our x be Yao Ming. Does Yao Ming play for the Rockets or the Lakers? Yes, so the hypothesis is satisfied. We turn to the conclusion. Our x is still Yao Ming, so we ask ourselves, does Yao Ming play for the Rockets, or does Yao Ming play for the Lakers? Yao Ming plays for the Rockets, so the conclusion is also satisfied. Let’s just take another example to further cement this implication in our minds. Using the same example, let’s find another x to satisfy the hypothesis. Shaquille O’Neal plays for the Lakers. So our hypothesis is satisfied. So is there someone who plays for the Rockets? We don’t know. Is there someone who plays for the Lakers? Yes, we know that Shaquille O’Neal plays for the Lakers. Therefore, the conclusion follows from the hypothesis again. xP xQ x) Why is => true? x( P( x( x) )Q ( x()) The hypothesis states that there exists something in our universe that satisfies a condition, or there exists something in our universe that satisfies some other condition. Given that we know that something does indeed satisfy either the first condition or something exists that satisfies the second condition, it follows that there exists something that satisfies either the first or the second condition. For example, let our universe be people, and let P(x) stand for “x is almost done with Project 1” and let Q(x) stand for “x has not even started Project 1.” So, is there someone who is almost done with Project 1? Yes, his name is William Yeh. So the hypothesis is true. Given that we know that William Yeh is almost done with Project 1, does the conclusion hold? The conclusion asks, is there someone who is almost done with Project 1 or has not even started Project 1? From the hypothesis, we know that William Yeh is almost done with Project 1, and therefore the conclusion is true, and thus the conclusion follows from the hypothesis.