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FOL Semantics (6) Consider a world with objects A, B, and C We'll look at a logical languge with constant symbols X, Y, and Z, function symbols f and g, and predicate symbols p, q, and rS T D A U F E E L V W N X M Y Z A B K C U D B E L I E V E F Word search puzzle words to find S E E H E A R R E A D F E E L T A L K H O P E K N O W T H I N K G U E S S D R E A M B E L I E V E U N D E R S T A N D Reprinted with permission by the Dana Alliance for Brain Initiatives "FORMING WORDS" This puzzle gives you five key words toCurves in R2 Three descriptions (1) Graph of a function f R !R (That is y= f(x)) Such curves must pass the vertical line test Example When we talk about the \curve" y= x2, we actually mean to say the graph of the function f(x) = x2That is, we mean the set

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ƒ~ƒfƒBƒAƒ€ ‚Z¶ "¯Œ^ ƒƒ"ƒY-Let z = f(x,y),whichmeans"z is a function of x and y"Inthiscasez is the endogenous (dependent) variable and both x and y are the exogenous (independent) variables To measure the the e ffect of a change in a single independent variable (x or y) on the dependent variable (z) we use what is known as the PARTIAL DERIVATIVE The partial derivativeZ x a f(t)dt This value is f(x) (b) d dx Z b a f(t)dt This value is zero (it's the derivative of a constant) (c) Z b a d dx f(x)dx This value is f(b)−f(a) (d) Z b a f(x)dt This is a little tricky since the integral is in terms of t, but f is an expression in x, it will be constant with respect to t The answer f(x)(b

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 Let f(A, B) = A' B Simplified expression for function f(f(x y, y)z) is (A) x' z (B) xyz xy' z (D) None of these Answer ExplanationOf elements x, y in F there are unique elements x y and xy (often written xy) in F for which the following conditions hold for all elements x, y, z in F (i) x y = y x (commutativity of addition) (ii) (x y) z = x (y z) (associativity of addition) (iii) There is an element 0 ∈ F, called zero, such that x0 = x (existence of anIf (x, y, z) (x, y, z) is a point in space, then the distance from the point to the origin is r = x 2 y 2 z 2 r = x 2 y 2 z 2 Let F r F r denote radial vector field F r = 1 r 2 〈 x r, y r, z r 〉 F r = 1 r 2 〈 x r, y r, z r 〉 The vector at a given position in space points in the direction of unit radial vector 〈 x r, y r, z

This list of all twoletter combinations includes 1352 (2 × 26 2) of the possible 2704 (52 2) combinations of upper and lower case from the modern core Latin alphabetA twoletter combination in bold means that the link links straight to a Wikipedia article (not a disambiguation page) As specified at WikipediaDisambiguation#Combining_terms_on_disambiguation_pages,Answers to Selected Problems on Simplification of Boolean Functions 316 Implement the following functions with threelevel NOR gate circuits 317 Implement the following expressions with threelevel NAND circuits By using different combinations of these, you will be able to implement the function with 2 X 2 X 2 = 8 different twolevel gateDepartment of Computer Science and Engineering University of Nevada, Reno Reno, NV 557 Email Qipingataolcom Website wwwcseunredu/~yanq I came to the US

• f(A) = {Y, Z} • the preimage of Y is b • the preimages of Z are a, c and d • f({c,d}) = { Z } Discrete Mathematics by Section 16 and Its Applications 4/E Kenneth Rosen TP 3 Injections, Surjections and Bijections Let f be a function from A to B Definition f is onetoone (denoted 11) or injective ifDe nition 1 The Cartesian product (or cross product) of A and B, denoted by A B, is the set A B = f(a;b) ja 2A and b 2Bg 1the elements (a;b) of A B are ordered pairsSimilarly, f(b) is substituted for W, which is substituted for Z, which gives f(b)/Z Most General Unifier If s is any unifier of a set of expressions E , and g is the most general unifier of that set of expressions, then for s applied to E there exists another unifier s' such that Es = Egs' (where gs' is the composition of unifiers)

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Transitive Let x;y;z 2Z and suppose that x ˘y and y ˘z Therefore 2j(x y) and 2j(y z) So there exist k;' 2Z such that 2k = x y and 2' = y z Add these equations to get 2k 2' = x 2y z Subtract 2y from both sides to get 2(k ' y) = x z Note that k' y 2Z, because k;';y 2Z and Z is closed under addition and subtractionB F(A,B,C,D) = D (A' C') 6 a Since the universal gates {AND, OR, NOT can be constructed from the NAND gate, it is universalThe curl of F is Here are two simple but useful facts about divergence and curl Theorem 1651 ∇ ⋅ (∇ × F) = 0 In words, this says that the divergence of the curl is zero Theorem 1652 ∇ × (∇f) = 0 That is, the curl of a gradient is the zero vector Recalling that gradients are conservative vector fields, this says that the

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All A\B, we obtain a well de ned function h X!Y where h(x) = f(x) for x2nd h(x) = g(x) for x2B 6 Homeomorphisms De nition 19 Let X;Yb topological spaces A bijection f X!Yis called a homeomorphism if both fand f 1 are continuous De nition Suppose f X !Y be both injective and continuous, with X;Y being topological spaces Let Z = fLet x and y be any two elements in the domain (Z), such that f(x) = f(y) f(x) = f(y) x 3 = y 3 x = y So, f is an injection Surjection condition Let y be any element in the codomain (Z), such that f(x) = y for some element x in Z (domain) f(x) = y x 3 = y x = ∛y which may not be in Z For example, if yZ b a fX;Y (xjy)dx Conditioning on Y = y is conditioning on an event with probability zero This is not de ned, so we make sense of the left side above by a limiting procedure P(a X bjY = y) = lim !0 P(a X bjjY yj < ) We then de ne the conditional expectation of X given Y = y to be

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2π = πa4b2 2 Find the flux of → F = x,y2,z upward through the firstoctant part S of the cylindrical surface x2 z2 = a2 for 0 6 y 6 b Solution The surface S is one of five surfaces that form the boundary of the solid region D consisting of the part of the cylinder x2z2 = a2 for 0 6 y 6 b that lies within the firstoctant The otherExample 5 X and Y are jointly continuous with joint pdf f(x,y) = (e−(xy) if 0 ≤ x, 0 ≤ y 0, otherwise Let Z = X/Y Find the pdf of Z The first thing we do is draw a picture of the support set (which in this case is the firstChapter 8 Integrable Functions 81 Definition of the Integral If f is a monotonic function from an interval a,b to R≥0, then we have shown that for every sequence {Pn} of partitions on a,b such that {µ(Pn)} → 0, and every sequence {Sn} such that for all n ∈ Z Sn is a sample for Pn, we have {X (f,Pn,Sn)} → Abaf 81 Definition (Integral) Let f be a bounded function from an interval

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3For any two numbers aand bwith a1 (x) y g 2 (x) then ZZ D f(x;y)dA= Z b a Z g 2(x) g 1(x) f(x;y)dydx (44) Remark 333 Notice that the function F(x;y) does not appear in the formula It was simply a device used to derive this formula When computing an integral, it does not come into play, as the next example will show Remark 334 When writing ZZ D f(x;y)ds an iteratedCutaneous lesions on hands of casepatient 3 (A, B) and casepatient 5 are shown Negative staining electron microscopy of samples from casepatient 3 (D) and casepatient 5 (E, F) show ovoid particles ( ≈250 nm long, 150 nm

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 y (x, y, z) = x (x y) (X Y Z ) , is an example of POS expression , since all the sum terms in the expression doesn't have all the literals X , Y and Z , so we have to express it in such a way that it will have all the three literals in each termSuppose that x,y∈X and ε,δ>0If z∈B(x,δ)∩B(y,ε),then B(z,α) ⊂B(x,δ)∩B(y,ε) (102) where α=min{δ−d(x,z),ε−d(y,z)},see Figure 102 This is a formal consequence of the triangle inequality For example let us show that B(z,α) ⊂ B(x,δ)By the definition of α,we have that α≤δ−d(x,z) or that d(x,zZ b a f(x)dx= F(b) F(a), where Fis any antiderivative of f 2 Give the de nition of the de nite integral Z b a f(x)dx= lim n!1 i=1 f(x i) x iOr you may be more speci c and use right endpoints Z b a dtso F0(x) = 2x2 1x4, assuming x>0 (b) y= Z 3x p x et t dt y0= 3e 3x 3x 1 2 p x e p x p x = e x e p x 2x 10 The idea here was to

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 Canonical Form – In Boolean algebra,Boolean function can be expressed as Canonical Disjunctive Normal Form known as minterm and some are expressed as Canonical Conjunctive Normal Form known as maxterm In Minterm, we look for the functions where the output results in "1" while in Maxterm we look for function where the output results in "0"The CDC AZ Index is a navigational and informational tool that makes the CDCgov website easier to use It helps you quickly find and retrieve specific information Find links to key CDC topic areas in this alphabetical indexY IS FOR YMCA & YOUTH HOSTELS YMCA organizations are located throughout Illinois Illinois has several YOUTH HOSTELS There are also Elderhostel programs Z IS FOR ZIPCODES, ZONE, AND ZOOS The use of postal ZIPCODES began in 1963 Zipcodes in Illinois start with the number 6 The mailing code is IL

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Theorem 36 Let F be any partition of the set S Define a relation on S by x R y iff there is a set in F which contains both x and y Then R is an equivalence relation and the equivalence classes of R are the sets of F Pf Since F is a partition, for each x in S there is one (and only one) set of FA" is a onetoone function on the set Z=nZ (a) Prove that if a and b are units modulo n, then their product is also a unit Conclude that \multiplication by a" is bijective map from the set of units in Z=nZ to the set of units in Z=nZ Hint Given that a has an inverse a 1 and b has an inverse b 1 modulo n, can you write downPage 5 of 7 Problem 215 Given the Boolean function F=xy'zx'y'zw'xywx'ywxy (a)Obtain the truth table of the function (b)Draw the logical diagram using the original Boolean expression

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1 y 1g which is just the square centered at the origin, of side two 2 R2 itself can be identi ed (and we usually do!) with the Cartesian product R R 3 let Cˆ R2 be convex and let S= R C Then Sis called a right cylinder and is just f(z;x) 2 R3 z>0;x2 Cg If, in particular C= f(u;vY = f(x) for a function of one variable, and that the graph of y = f(x) is a curve For functions of two variables the notation simply becomes z = f(x;y) where the two independent variables are x and y, while z is the dependent variable The graph of something like z = f(x;y) is a surface in threedimensional space SuchPage 2 c) d) wx x′y BD B′D′ A′B 312) Simplify the following Boolean functions in products of sums a) F(w, x, y, z) = ∑ (0, 2, 5, 6, 7, 8, 10)

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 Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange David Altizio Homemade Problem Collection Page 3 9(NIMO April 17) For rational numbers aand bwith a>b, define the fractional average of aand bto be the unique rational number cwith the following property when cis written in lowest terms, there exists an integer Nsuch that adding Nto both the numerator and denominator of c gives a, and subtracting Nfrom both theRule02 Groups may overlap each other Rule03 We can only create a group whose number of cells can be represented in the power of 2 In other words, a group can only contain 2 n ie 1, 2, 4, 8, 16 and so on number of cells;

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Let f' be continuous and positive on a,b Prove that $\int_a^bf(x)dx \int_{f(a)}^{f(b)} f^{1}(y)dy=bf(b)af(a)$ I've got the later steps of this covered using integration by parts, and I knoZ b a F0(x)dx ≤ F(b) − F(a) Consequently, Z b a F0(x) − f(x)dx = 0 But F0(x) ≥ f(x) for almost every x ∈ a,b Therefore, F0(x) = f(x) for almost every x in a,b Theorem 23 A function F on a,b is absolutely continuous if and only if F(x) = F(a) Z x a f(t)dt for some integrable function f on a,b Proof The sufficiency4 Definition A literal is any Boolean variable x or its complement x' Truth Tables of Boolean functions Much like the truth tables for logical propositions If f(x,y,z, ) is an nvariable Boolean function, a truth table for f is a table of n1 columns (one

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Let f be the subset of Z × Z defined by f = {(ab, a b) a, b ∈ Z} Is f a function from Z to Z justify your answer Solution Given relation f is defined as f = {(ab, a b) a, b ∈ Z} We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set BH d < i y w t u p z b h d w x m x u p r t w o y v o l y r i u v g p c u b nUnification Solutions Unify, if possible, the following pairs of predicates P( g(h(x)) , f(g(h(b))) , f(x) ) P( y , f(y) , z ) { g(h(b)) / y , b / x , f(b) / z } P

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