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truth_conditions

Course: PHIL 220, Winter 2010
School: UBC
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Truth 6. and Possible Worlds We have defined logical entailment, consistency, and the connectives , , , all in terms of belief. In view of the close connection between belief and truth, described in the first section, we should see what each of these has to do with truth. Suppose A is true, and A entails B. Does B then have to be true as well, or could it be false? If it is possible for B to be false here, then...

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Truth 6. and Possible Worlds We have defined logical entailment, consistency, and the connectives , , , all in terms of belief. In view of the close connection between belief and truth, described in the first section, we should see what each of these has to do with truth. Suppose A is true, and A entails B. Does B then have to be true as well, or could it be false? If it is possible for B to be false here, then thats bad news. For then even a perfect thinker, starting with truth, could end up in error. Fortunately we shall see that this is not possible. Logical entailment cannot lead from truth to something false. Entailment, we say, is truth preserving. 6.1 Possible Worlds There are many other links between the logical terms and truth. In order to investigate all of these it will be useful to introduce the notion of a possible world. This will allow to define the structure of propositions more precisely than before, and see just what it means for a proposition to be true. Recall the idea of an expansion o f an epistemic state knowledge is added to the state, without removing any. This notion of expansion naturally leads to the idea of a learning path o f epistemic states. A learning path is a sequence of epistemic states where each state is an expansion of the previous one. A learning path might be seen as the intellectual biography of a perfect thinker, as she acquires more and more knowledge. One question about learning paths is whether they can ever come to an end. Is there any epistemic state thats so big it cant be expanded further? Theoretically, at least, a perfect thinker might get to the stage where she has a firm, definite opinion on every issue. She has unshakable beliefs about the doings of every single beetle in the Amazon basin, knows everything there is to know about every person on the planet, even the number of hairs on their head. Her epistemic state is every bit as detailed and complicated as the real world itself! No human could have such an epistemic state, of course; its just a theoretical entity. Such an epistemic state can be called maximal, since it cannot be expanded. It is not contained in any epistemic state (apart from itself of course). We will call such a maximal epistemic state a possible world. Definition A possible world is a maximal epistemic state, i.e. one which cannot be expanded. It is not contained in any other epistemic state. There are many possible worlds, in this sense, as there are many possible combinations of beliefs. A possible world doesnt have to match the real world, it only has to be a possible state of belief for a perfect thinker. It has to be coherent, make sense, and so on, but it doesnt have to be true. 1 The opposite of a maximal epistemic state (possible world) is a minimal epistemic state. This is a state with no information at all, so that it is contained in every epistemic state, even itself. It is clear that there can only be one minimal state. (Why?) Assuming it exists, we will call this state KO, and refer to it as the empty epistemic state. Definition KO, the empty epistemic state, is the unique state that is contained in every epistemic state. Even though KO is the state of maximum ignorance, it is not entirely empty of all belief. It does contain some beliefs that every rational thinker must share, such as 1+1=2, all green things are green, if it happens, it happens, and other trivial claims. Such beliefs, known as logical truths, are present in every possible epistemic state, and are thus present in KO as well. Using the empty epistemic state, we can define baseless propositions. In general, a proposition makes background assumptions, without which it makes no sense. For example, The Loch Ness Monster weighs at least 40 tons assumes that there is a monster in Loch Ness, among other things. Now consider a proposition that makes no assumptions at all, so that its meaningful even in KO, the empty state. We will call such a proposition baseless. Definition A baseless proposition is one whose base is KO, the empty epistemic state. If you like geometry, you can think of a baseless proposition as like a position vector, which starts from the origin O of the space. KO can be thought of as the origin of logical space. All learning paths begin at KO. Consider a baseless proposition that takes you from KO to a possible world KW. This proposition (call it W) contains all the information that is in KW, so we shall call it a possible world as well. If A is a baseless proposition, and W is a possible world, then A and W are either consistent or inconsistent. If A and W are consistent, then there must be some proposition that entails them both. But, since W is maximal, no proposition entails W except W itself. Therefore, if A is consistent with W, then W A. If A is inconsistent with W, then W A, of course.1 Thus, for any world W and proposition A, W either entails A or it entails A. 6.2 Propositions Defined Properly Lets start with some baseless proposition A. We have seen that, for every possible world W, either W entails A or it entails A. Thus A partitions the class of possible worlds into two separate groups; one group of worlds that entail A, and another that entail A. Let us call the 1We found out earlier that, if two propositions are inconsistent, then each entails the negation of the other. 2 first group the A-worlds. You might think of these as the worlds according to which A is true. According to each of the other worlds, A is false. Perhaps the fundamental fact about propositions is this: The (baseless) proposition A is the disjunction of all the A-worlds2 The proposition A, in other words, is the information that is in common between all the Aworlds; it is the point of strongest agreement. This fact enables us to investigate the relation between truth and belief. We said before that, when someone utters a sentence, they are trying to cause a change in the listeners epistemic state. We can now see more clearly what that change is. Suppose Fred tells Betty: Your cat has shredded my logic assignment. Let the proposition expressed by this sentence be A. Now consider the minimal state where A is fully believed, which we will call KA. This is actually the strongest epistemic state that is weaker than all the A-worlds. Since Fred wants Betty to believe A, Fred wants Bettys epistemic state to be stronger than KA. Fred doesnt want Betty to lose any of her current knowledge, so her new epistemic state should be an expansion of her old one, Kold. So Bettys new state, Knew, should be an expansion of KA, and also an expansion of Kold. Moreover, since Fred isnt saying anything apart from A, Knew should be the weakest such state. This state we will call (Kold KA), since it is the weakest state that is stronger than both Kold and KA. Thus: The proposition A is associated with the expansion from K to (K KA). In other words, coming to believe the proposition A means that your epistemic state moves up to the closest state in which A is believed. 6.3 Truth The possible worlds are all inconsistent with each other. This follows from the definition of consistency. Since a possible world is only entailed by itself, there cannot be any proposition that entails two distinct possible worlds. Therefore, no two possible worlds can be consistent. (Any two completely opinionated people will disagree, unless they completely agree.) Note however that two worlds can agree on some things; they just cant agree on everything. We shall assume that precisely one of the possible worlds is true. (By true, I mean that it exactly matches the real world, the world that actually exists. One of the possible worlds is like a perfectly detailed map that corresponds exactly to the actual territory.) This assumption that 2In a more rigorous treatment of the subject, possible worlds are defined first, and then propositions are defined in terms of them. Such a treatment would be harder to understand, however. 3 there is one true world can be split into two parts: (i) at least one possible world is true, and (ii) no more than one possible world is true. The first assumption says that thought is capable, at least, of capturing reality. It would be rather depressing if every possible world were false. The second assumption is grounded in the fact that the possible worlds are mutually inconsistent, so that only one can be believed. If two or more of the worlds could be true, then belief would always be seriously different from truth, even under the best possible circumstances. This would also be depressing. Thus, when I assume that precisely one of the possible worlds is true, I am hoping for the best. Belief seems to have no purpose other than to represent truth, so Im hoping that it can, in principle, succeed at this. Im hoping that thought is not essentially dysfunctional. (This hope is bolstered by the thought that to say otherwise seems incoherent.) This assumption that there is one true possible world is not granted by all. It is part of a philosophical view called realism, so those who deny it are usually called anti-realists. I dont require everyone here to be a realist, but you should understand what the view says and see what its based on. Standard, classical logic assumes realism, so it makes sense to teach you about realism in this class. There are other logics that dont assume realism, such as intuitionistic logic and paraconsistent logic, which you can learn about in other philosophy classes. So, for this class, lets assume that there is one true possible world. This proposition contains all the truth there is: the whole history of the universe, in perfect detail, all of science and math, the whole lot. Its like a Gods eye view of reality. Lets call this world WT, for the Truth. What is it for an ordinary proposition, one that is not a possible world, to be true? It simply has to be part of the truth. In other words, A is true just in case WT A, so that WT is an A-world. Definition A proposition A is true just in case WT A. We can now state the two most important theorems, that connect entailment and consistency to truth. Theorem Proof: If A is and true, A B, then B is true as well. (Entailment preserves truth.) Suppose A is true, i.e. WT A. Then, since A B, and entailment is transitive, it follows that WT B. In other words, B is true as well. Theorem If A is inconsistent with B, then A and B are not both true. (True propositions are mutually consistent.) Suppose that A and B are both true, i.e. WT A and WT B. Then, since some proposition entails both A and B, it follows that they are consistent. Thus, given that A is inconsistent with B, if follows that they are not both true. Proof: 4 6.4 Set Theory and Venn Diagrams We have seen that every proposition A is associated with a class of worlds, known as the Aworlds these are the possible worlds that each entail A. This fact enables us to represent the logical relations of entailment and consistency on a diagram, known as a Venn diagram. Youve probably used Venn diagrams before, to represent relations between sets, or classes, and operations on classes. A set (or class) is a collection of objects, in no particular order. The objects in a set are known as its members. A set is sometimes written down by listing its members, separated by commas, between braces. For example, {2, 4, 5} is the set whose members are 2, 4 and 5. There is one set with no members, written {} or . If two sets have the same members, then they are actually the same set. We assume that all objects are members of the universal set U. Here are the basic relations: 1. X is a subset o f Y, or X Y. This means that every member of X is also a member of Y. This relation is shown on the following diagram. The universal set U is the whole rectangle. 2. X and Y are disjoint. This means that they have no members in common. You can see, in the Venn diagram, that the loops dont overlap, so that no object can be inside both. 5 Here are the basic operations on sets. These are ways to form a new set out of one or two old sets. 1. The intersection o f X and Y. This is the set of all the objects that are in both X and Y. In the diagram, the shaded area is the intersection of X and Y, written X Y. 2. The union of X and Y. This is the set of all the objects that are either in X, or in Y, or both. The shaded area on the next diagram is the union of X and Y, written X Y. 3. The complement of X. This is the set of all the objects that are in the universal set, but not in X. It is the shaded area in the diagram below. The complement of X can be written Xc. ________________________________________________________ 6 Exercise 6.1 1. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, X = {2, 4, 6, 8, 10}, Y = {6, 7, 8, 9}, Z = {2, 4}. a. What is the relation between X and Z? b. What is the relation between Y and Z? c. What is the relation between X and Y? d. What is X Y? (Write down the elements, inside braces.) e. What is X Z? f. What is Y Z? g. What is Y Z? h. What is X Y? i. What is Xc? j. What is Yc? 2. On a Venn diagram, draw X and Y as intersecting sets, where neither is a subset of the other. Shade the set X Yc. 3. On a similar diagram, shade the set (X Y)c. __________________________________________________________ 6.5 Entailment Now we can get back to logic. You can draw a proposition on a Venn diagram, because each proposition is associated with a set o f possible worlds. To show a proposition A, we can draw a loop around the set of A-worlds, i.e. the worlds that entail A. Let the set of all the A-worlds be written as t(A), the truth set o f A. Logical relations can then be shown as relations between sets. The rectangle now represents the set of all possible worlds. 1. A entails B. This holds just in case t(A) t(B). You might have expected the exact opposite here! (Many students do.) Since A is a stronger proposition than B, and contains more information, you might expect t(A) to be a bigger set than t(B). In fact, as you see below, its a smaller set. You have to remember that each point on the diagram is a possible world, which is a very strong proposition. A is the disjunction of all the members of t(A), i.e. the common content between all of them. Now, the more worlds there are in t(A), the less there is in common between them all. So, the stronger the proposition A is, the smaller the set t(A). 7 Proof: 1. Suppose A B. For any A-world W, W A. Then, since A B, W B as well. So any Aworld is also a B-world, i.e. t(A) t(B). 2. Suppose t(A) t(B). It follows that every A-world entails B. Now, A is the disjunction of the A-worlds, so A is the strongest proposition that it entailed by every A-world. So A is stronger than B, i.e. A B. ___________________________________________________ Exercise 6.2 Consider the three propositions: A = Janet is 5' 8" tall B = Janet is between 5' 5" and 5' 9" tall C = Janet is between 5' and 6' tall 1. Which proposition entails the other two? (i.e. which is the strongest?) 2. Which proposition is the weakest? 3. Which proposition has the widest range of possible heights for Janet? 4. Which proposition has the smallest range of possible heights for Janet? _______________________________________________________ 6.6 Consistency Propositions A and B are consistent just in case there is some proposition that entails them both. Suppose that proposition is C. Then each C-world will entail both A and B, obviously, and so each C-world is both an A-world and a B-world. Thus the set of A-worlds intersects the set of Bworlds. If there is no such intersection, then it is easy to see that the propositions are inconsistent. We can show this on Venn diagrams. First, this is what inconsistent propositions look like: 8 The sets of worlds, as you can see, are disjoint. Consistent propositions, on the other hand, have a non-empty intersection, like this: 6.7 Truth and the Connectives Suppose A is true, and B is also true. Doesnt it follow that the conjunction AB is true as well? Yes it does, and we can now show this. In the work that follows it will be helpful to say A is true in W when W A, and A is false in W when W A. These terms are a little misleading, as truth is really a matter of correspondence to the real world. What is really meant by A is true in W is that W claims that A is true. Theorem AB is true in all and only the worlds where A and B are both true. I.e. t(AB) = t(A) t(B). This is most easily seen using a Venn diagram. AB is the weakest proposition that is entailed by A, and entailed by B. Thus t(AB) must be the largest set of worlds that is a subset of t(A), and also a subset of t(B). This set is, of course, the intersection of t(A) with t(B). Proof: 9 The shaded area on this diagram is the class of (AB)-worlds. We can see that this is the intersection of the A-worlds with the B-worlds. We see that there are basically four kinds of world in this diagram. There are worlds where (i) A and B are both true, (ii) A is true and B is false, (iii) A is false and B is true, and (iv) A is false and B is false. The proposition A&B is true only in worlds of the first kind. In view of this fact, the information in this diagram can also be presented in a table, known as a truth table, as follows: A T T F F B T F T F AB T F F F The symbol T means true (in those worlds), and F means false (in those worlds). If you look at the two columns on the left, under A and B, you will see that there is one row for each of the four kinds of world. The column on the right, under AB, says whether AB is true or false in each of these kinds of world. Note that the rows of a truth table are always placed in this order, from top to bottom: TT, TF, FT, FF. This is an arbitrary choice, but if we all do it the same way then your work will be much easier to mark! Truth tables are much more commonly used than Venn diagrams. Partly this is because they can handle more than three propositions at once, unlike Venn diagrams. They are also neat and compact. If a sentential connective has a complete truth table, then its called truth functional. This means that the truth value of the output proposition is determined by the truth values of the input propositions. (In this case, the inputs are A and B, and the output is A&B.) 10 We can look at disjunction () in the same way. It turns out that the class of (AB)-worlds, i.e. t(AB), is the union of t(A) with t(B). This may be shown on a Venn diagram: The shaded area shows the class of (AB)-worlds, i.e. t(AB). We may also use a truth table: A T T F F B T F T F AB T T T F Note that the two columns on the left of this table are the same as for &. In the right-hand column, under A B, note that there are more Ts than before. AB is true if either A is true, B is true, or both. Its only false when A and B are both false. I havent shown you an argument that this table is correct, but its fairly obvious. Since A B is the strongest proposition entailed by both A and B, t(AB) must be the smallest set of which t(A) and t(B) are both subsets. This set is just t(A) t(B). Note that , like &, is a truth-functional connective. The negation of A, you will recall, is the weakest proposition that is inconsistent with A. Therefore, t(A) is the largest set that is disjoint with A. This set is just the complement of t(A). Thus t(A) is the shaded area on the diagram below. 11 The truth table for negation is then as follows. A T F A F T You can see that A and A have opposite truth values, in every world. _________________________________________________________ Exercise 6.3 1. For each pair of propositions below, use a truth table to show that the propositions are identical, i.e. they are true in exactly the same class of possible worlds. a. AB, BA. b. A, A. c. A, AA. d. G, GG. e. (AB), A B. f. (AB), A B. g. A (BC), (AB) (AC). h. A (BC), (AB) (AC). i. (AB) C, (AB) C 2. Show the following logical relations on Venn diagrams. (For each part, draw a separate diagram in which the claim(s) hold(s). Note that there should be no shading.) Make your diagram as general as possible; i.e. it should not show any logical relation that has not been asked for. a. B A b. B A, C A. c. (BC) A 12 d. B C, C A e. (B C) A f. A (BC) g. A B, A C. 13
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The cat (Felis catus), also known as the domestic cat or housecat[5] to distinguish it from other felines and felids, is a small furry domesticated carnivorous mammal that is valued by humans for its companionship and for its ability to hunt vermin and ho
Cal Poly Pomona - KIN - 370
Exam #1 Chapter 1: The Nature of Stress According to the definitions of stress, identify the components of which stress is comprised. Describe the two types of stress and the two types of distress. Describe the Yerkes-Dodson Principle. Describe the types
Drexel - MEM - 345
PROBLEM 1.1 KNOWN: Thermal conductivity, thickness and temperature difference across a sheet of rigid extruded insulation. FIND: (a) The heat flux through a 2 m 2 m sheet of the insulation, and (b) The heat rate through the sheet. SCHEMATIC:A = 4 m2k =
Drexel - MEM - 345
PROBLEM 1.2KNOWN: Inner surface temperature and thermal conductivity of a concrete wall. FIND: Heat loss by conduction through the wall as a function of outer surface temperatures ranging from -15 to 38C. SCHEMATIC:ASSUMPTIONS: (1) One-dimensional condu
Drexel - MEM - 345
PROBLEM 1.3KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency of gas furnace and cost of natural gas. FIND: Daily cost of heat loss. SCHEMATIC:ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3
Drexel - MEM - 345
PROBLEM 1.4 KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed thickness. FIND: Thermal conductivity, k, of the wood. SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions,
Drexel - MEM - 345
PROBLEM 1.5 KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions. FIND: Heat loss through window. SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant proper
Drexel - MEM - 345
PROBLEM 1.6 KNOWN: Width, height, thickness and thermal conductivity of a single pane window and the air space of a double pane window. Representative winter surface temperatures of single pane and air space. FIND: Heat loss through single and double pane
Drexel - MEM - 345
PROBLEM 1.7 KNOWN: Dimensions of freezer compartment. Inner and outer surface temperatures. FIND: Thickness of styrofoam insulation needed to maintain heat load below prescribed value. SCHEMATIC:ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimens
Drexel - MEM - 345
PROBLEM 1.8 KNOWN: Dimensions and thermal conductivity of food/beverage container. Inner and outer surface temperatures. FIND: Heat flux through container wall and total heat load. SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat t
Drexel - MEM - 345
PROBLEM 1.9 KNOWN: Masonry wall of known thermal conductivity has a heat rate which is 80% of that through a composite wall of prescribed thermal conductivity and thickness. FIND: Thickness of masonry wall. SCHEMATIC:ASSUMPTIONS: (1) Both walls subjected
Drexel - MEM - 345
PROBLEM 1.10 KNOWN: Thickness, diameter and inner surface temperature of bottom of pan used to boil water. Rate of heat transfer to the pan. FIND: Outer surface temperature of pan for an aluminum and a copper bottom. SCHEMATIC:ASSUMPTIONS: (1) One-dimens
Drexel - MEM - 345
PROBLEM 1.11 KNOWN: Dimensions and thermal conductivity of a chip. Power dissipated on one surface. FIND: Temperature drop across the chip. SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform heat dissipation, (4) Neg
Drexel - MEM - 345
PROBLEM 1.12 KNOWN: Heat flux gage with thin-film thermocouples on upper and lower surfaces; output voltage, calibration constant, thickness and thermal conductivity of gage. FIND: (a) Heat flux, (b) Precaution when sandwiching gage between two materials.
Drexel - MEM - 345
PROBLEM 1.13KNOWN: Hand experiencing convection heat transfer with moving air and water. FIND: Determine which condition feels colder. Contrast these results with a heat loss of 30 W/m2 under normal room conditions. SCHEMATIC:ASSUMPTIONS: (1) Temperatur
Drexel - MEM - 345
PROBLEM 1.14KNOWN: Power required to maintain the surface temperature of a long, 25-mm diameter cylinder with an imbedded electrical heater for different air velocities. FIND: (a) Determine the convection coefficient for each of the air velocity conditio
Drexel - MEM - 345
PROBLEM 1.15 KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required to maintain a specified surface temperature for water and air flows. FIND: Convection coefficients for the water and air flow convection processes, hw and ha,
Drexel - MEM - 345
PROBLEM 1.16 KNOWN: Dimensions of a cartridge heater. Heater power. Convection coefficients in air and water at a prescribed temperature. FIND: Heater surface temperatures in water and air. SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) All of t
Drexel - MEM - 345
PROBLEM 1.17 KNOWN: Length, diameter and calibration of a hot wire anemometer. Temperature of air stream. Current, voltage drop and surface temperature of wire for a particular application. FIND: Air velocity SCHEMATIC:ASSUMPTIONS: (1) Steady-state condi
Drexel - MEM - 345
PROBLEM 1.18 KNOWN: Chip width and maximum allowable temperature. Coolant conditions. FIND: Maximum allowable chip power for air and liquid coolants. SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from sides and bottom,
Drexel - MEM - 345
PROBLEM 1.19 KNOWN: Length, diameter and maximum allowable surface temperature of a power transistor. Temperature and convection coefficient for air cooling. FIND: Maximum allowable power dissipation. SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (
Drexel - MEM - 345
PROBLEM 1.20KNOWN: Air jet impingement is an effective means of cooling logic chips. FIND: Procedure for measuring convection coefficients associated with a 10 mm 10 mm chip. SCHEMATIC:ASSUMPTIONS: Steady-state conditions. ANALYSIS: One approach would b
Drexel - MEM - 345
PROBLEM 1.21 KNOWN: Upper temperature set point, Tset, of a bimetallic switch and convection heat transfer coefficient between clothes dryer air and exposed surface of switch. FIND: Electrical power for heater to maintain Tset when air temperature is T =
Drexel - MEM - 345
PROBLEM 1.22KNOWN: Hot vertical plate suspended in cool, still air. Change in plate temperature with time at the instant when the plate temperature is 225C. FIND: Convection heat transfer coefficient for this condition. SCHEMATIC:-0.022 K/sASSUMPTIONS:
Drexel - MEM - 345
PROBLEM 1.23KNOWN: Width, input power and efficiency of a transmission. Temperature and convection coefficient associated with air flow over the casing. FIND: Surface temperature of casing. SCHEMATIC:ASSUMPTIONS: (1) Steady state, (2) Uniform convection
Drexel - MEM - 345
PROBLEM 1.24 KNOWN: Air and wall temperatures of a room. Surface temperature, convection coefficient and emissivity of a person in the room. FIND: Basis for difference in comfort level between summer and winter. SCHEMATIC:ASSUMPTIONS: (1) Person may be a