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(LMCS, p. 317) V.1. First Order Logic. This is the most powerful, most expressive logic that we will examine.
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1 (LMCS, p. 317) V.1 First Order Logic This is the most powerful, most expressive logic that we will examine. Our version of first-order logic will use the following symbols: variables connectives (,,,, ) function symbols relation symbols constant symbols equality ( ) quantifiers (, )
2 (LMCS, p. 318) V.2 Formulas for a first-order language L are defined inductively as follows: There are two kinds of atomic formulas: (s t), where s and t are terms, and (rt 1 t n ), where r is an n ary relation symbol and t 1,, t n are terms. If F is a formula, then so is ( F). If F and G are formulas, then so are (F G), (F G), (F G), (F G). If F is a formula and x is a variable, then ( x F) and ( x F) are formulas.
3 (LMCS, p. 318) V.3 Notational Conventions Drop outer parentheses Adopt the previous precedence conventions for the propositional connectives. Quantifiers bind more strongly than any of the connectives. Thus y (rxy) y (rxy) means ( y (rxy)) ( y (rxy)).
4 (LMCS, p. 318) V.4 The subformulas of a formula F : The only subformula of an atomic formula F is F itself. The subformulas of F are F itself and all the subformulas of F. The subformulas of F G are F G itself and all the subformulas of F and all the subformulas of G. ( is any of,,, r ). The subformulas of x F are x F itself and all the subformulas of F. The subformulas of x F are x F itself and all the subformulas of F.
5 (LMCS, p ) V.5 An occurrence of a variable x in a formula F is: bound if the occurrence is in a subformula of the form x G or of the form x G (such a subformula is called the scope of the quantifier that begins the subformula). Otherwise the occurrence of the variable is said to be free. A formula with no free occurrences of variables is called a sentence.
6 (LMCS, p ) V.6 Given a bound occurrence of x in F, we say that x is bound by an occurrence of a quantifier Q if (i) the occurrence of Q quantifies the variable x, and (ii) subject to this constraint the scope of this occurrence of Q is the smallest in which the given occurrence of x occurs.
7 (LMCS, p. 319) V.7 It is easier to explain scope, and quantifiers that bind variables, with a diagram: An occurrence of the variable x bound by A x ( ( r x y ) ( y (( x ( r z x) ) ( r x z))) A E Scopes of quantifiers are underlined
8 (LMCS, p. 319) V.8 The following figure indicates all the bound and free variables in the previous formula: free occurrences A x ( ( r x y ) ( y (( x ( r z x) ) ( r x z))) A E bound occurrences
9 (LMCS, p ) V.9 Examples b < 3 is also an atomic sentence, which says four is less than three. False in N. c. x y (x < y) says that for every number there is a larger number. True in N. d. y x (x < y) says that there is a number that is larger than every other number. False in N.
10 (LMCS, p ) V.10 f. x ( (0 < x) y (y y x) ) says that every positive number is a square. False in N. h. x y ((x < y) z ( (x < z) (z < y) )) says that if one number is less than another, then there is a number properly between the two. False in N.
11 (LMCS, p ) V.11 We will use the shorthand notation F i 1 i n to mean the same as the notation F 1 F n. Likewise, we will use the notations F i 1 i n and F 1 F n.
12 (LMCS, p ) V.12 English to First-order Given a first order formula F(x) we can find first order sentences to say a. There is at least one number such that F(x) is true in N. x F(x) b. There are at least two numbers such that F(x) is true in N. x y ( (x y) F(x) F(y) )
13 (LMCS, p ) V.13 c. There are at least n numbers (n fixed) such that F(x) is true in N. x 1 x n (( 1 i<j n (x i x j ) ) ( 1 i n F(x i ) )) d. There are infinitely many numbers that make F(x) true in N. x y ( (x < y) F(y) ) e. There is at most one number such that F(x) is true in N. x y ( (F(x) F(y)) (x y) )
14 (LMCS, p. 325) V.14 Definable Relations To better understand what we can express with first order sentences we need to introduce definable relations. Given a first order formula F(x 1,..., x k ) we say F is true at a k tuple (a 1,..., a k ) of natural numbers if the expression F(a 1,..., a k ) is a true statement about the natural numbers.
15 (LMCS, p. 325) V.15 Example Let F(x, y) be the formula x < y. Then F is true at (a, b) iff a is less than b. Example Let F(x, y) be z (x z y). Then F is true at (a, b) iff a divides b, written a b. [Note: Don t confuse a b with a b. The first is true or false. The second has a value. Check that a 0 for any a, including a = 0.]
16 (LMCS, p. 325) V.16 Definition For F(x 1,..., x k ) a formula let F N be the set of k tuples (a 1,..., a k ) of natural numbers for which F(a 1,..., a k ) is true in N. F N is the relation on N defined by F. Definition A k ary relation r N k is definable in N if there is a formula F(x 1,..., x k ) such that r = F N.
17 (LMCS, p ) V.17 Example (Definable Relations) x is an even number is definable in N by y (x y + y) x divides y is definable in N by z (x z y) x is prime is definable in N by (1 < x) y ((y x) ( (y 1) (y x) )) x y modulo n is definable in N by z ( (x y + n z) (y x + n z) )
18 (LMCS, p. 326) V.18 We will adopt the following abbreviations: x y for the formula z (x z y) prime(x) for the formula (1 < x) y ( (y x) ( (y 1) (y x) )). Note that in the definition of prime(x) we have used the previous abbreviation. To properly write prime(x) as a first order formula we need to replace that abbreviation; doing so gives us (1 < x) y ( z (y z x) ( (y 1) (y x) ))
19 (LMCS, p. 326) V.19 Abbreviations are not a feature of first order logic, but rather they are a tool used by people to discuss first order logic. Why do we use abbreviations? Without them, writing out the first order sentences that we find interesting would fill up lines with tedious, hard to read symbolism.
20 (LMCS, p. 326) V.20 Given: x y abbreviates z (x z y. Then (u + 1) (u u + 1) is an abbreviation for z ((u + 1) z u u + 1) Now let us write out z 2 to obtain z (z z 1 + 1) Unfortunately, this last formula does not define the set of elements in N that divide 2. It is a first order sentence that is simply false in N the square root of 2 is not a natural number.
21 (LMCS, p. 326) V.21 We have stumbled onto one of the subtler points of first order logic, namely, we must be careful with substitution. The remedy for defining z divides 2 is to use another formula, like w (x w y), for x divides y.
22 (LMCS, p. 326) V.22 We obtain such a formula by simply renaming the bound variable z in the formula for x y. With this formula we can correctly express z divides 2 by w (z w 2). The danger in using abbreviations in first order logic is that we forget the names of the bound variables in the abbreviation.
23 (LMCS, p. 327) V.23 Our solution: add a to the abbreviation to alert the reader to the necessity for renaming the bound variables that overlap with the variables in the term to be substituted into the abbreviation. For example, prime (y + z) alerts the reader to the need to change the formula for prime(x), say to ( (1 < x) v (v x) ( (v 1) (v x) )), so that when we substitute y + z for x in the formula, no new occurrence of y or z becomes bound.
24 (LMCS, p. 327) V.24 Thus we could express prime(y + z) by (1 < y + z) v ( (v (y + z)) ( (v 1) (v y + z) )). Examples [Expressing statements in first-order logic.] a. The relation divides is transitive. x y z ( ((x y) (y z) ) (x z) ) b. There are an infinite number of primes. x y ( (x < y) prime (y) )
25 (LMCS, p. 327) V.25 d. There are an infinite number of pairs of primes that differ by the number 2. (Twin Prime Conjecture) x y ( (x < y) prime (y) prime (y + 2) ) e. All even numbers greater than two are the sum of two primes. (Goldbach s Conjecture) ( ((2 x) ) x (2 < x) y z ( prime (y) prime(z) (x y + z) ))
26 (LMCS, p. 329) V.26 Other Number Systems Our first order language L = {+,, <, 0, 1} can just as easily be used to study other number systems, in particular, the integers Z = (Z, +,, <, 0, 1), the rationals Q = (Q, +,, <, 0, 1), and the reals R = (R, +,, <, 0, 1). However, first order sentences that are true in one can be false in another.
27 (LMCS, p ) V.27 Example Consider the following first order sentences: a. x y (x < y) b. y x (x < y) c. x y ((x < y) z ( (x < z) (z < y) )) d. x y ( (0 < x) (x y y) ) e. x y (x < y).
28 (LMCS, p ) V.28 Then we have N Z Q R a. true true true true b. false true true true c. false false true true d. false false false true e. false false false false We arrive at the simple conclusion that the notion of truth depends on the structure being considered!
29 (LMCS, p. 332) V.29 First Order Syntax for (Directed) graphs The first order language of (directed) graphs is L = {r}, where r is a binary relation symbol. The only terms are the variables x. Atomic formulas look like (x y) or (rxy). Example The subformulas of are x ((rxy) y (ryx)) x ((rxy) y (ryx)) (rxy) y (ryx) rxy y (ryx) ryx.
30 (LMCS, p ) V.30 First Order to English b a d Graph A b c a e Graph B c d b. x (rxx) says the (directed) graph is irreflexive. False for Graph A, true for Graph B. c. x y ( (rxy) (ryx) ) says the (directed) graph is symmetric. False for Graph A, true for Graph B.
31 (LMCS, p ) V.31 b a d Graph A b c a e Graph B c d e. x y (rxy) says all possible edges are present. False for both examples. f. x y (rxy) says for every vertex x there is an edge going from x to some vertex y. True for both examples.
32 (LMCS, p ) V.32 English to First-Order a. The (directed) graph has at least two vertices. x y ( (x y)). b. Every vertex has an edge attached to it. x y ( (rxy) (ryx) ). c. Every vertex has at most two edges directed from it to other vertices. x y z w ( ((rxy) (rxz) (rxw) ) ( (y z) (y w) (w z) ) ).
33 (LMCS, p ) V.33 Statements About Graphs The degree of a vertex is the number of (undirected) edges attached to it. A path of length n from vertex x to vertex y is a sequence of vertices a 1,..., a n+1 with each (a i, a i+1 ) being an edge, and with x = a 1, y = a n+1. Two vertices are adjacent if there is an edge connecting them.
34 (LMCS) V.34 Some Definable Relations a. The degree of x is at least one. y (rxy) b. The degree of x is at least two. y z ( (y z) (rxy) (rxz)) Statements About Graphs a. Some vertex has degree at least two. x y z ( (y z) (rxy) (rxz) ) b. Every vertex has degree at least two. x y z ( (y z) (rxy) (rxz) )
35 (LMCS, p. 339) V.35 Semantics for First Order Logic Given a first order L structure S = (S, I), the interpretation I gives meaning to the symbols of the language L. We associate with each term t(x 1,, x n ) the n ary term function t S. We associate with each formula F(x 1,, x n ) an n ary relation F S S n : [The superscript S is omitted in the following for ease of reading.]
36 (LMCS, p. 339) V.36 F( x) is atomic: F is t 1 ( x ) t 2 ( x ) : F( a) holds iff t 1 ( a) = t 2 ( a) F is rt 1 ( x) t n ( x) F( a) holds iff r(t 1 ( a),..., t n ( a)) holds. F is G Then F( a ) holds iff G( a) does not hold.
37 (LMCS, p. 339) V.37 F is G H Then F( a ) holds iff G( a ) or H( a ) holds. F is G H Then F( a ) holds iff G( a ) and H( a ) holds. F is G H Then F( a ) holds iff G( a ) does not hold, or H( a ) holds.
38 (LMCS, p. 339) V.38 F is G H Then F( a ) holds iff both or neither of G( a ) and H( a ) holds. F( x) is y G(y, x) Then F( a ) holds iff G(b, a ) holds for every b S. F( x) is y G(y, x) Then F( a ) holds iff G(b, a ) holds for some b S.
39 (LMCS, p. 340) V.39 Example S = {a, b} r a b a 0 1 b 1 0 a b f b a F(x) = y z ((rfxfy) (rfyfz)). F S =? Let H 1 (x, y, z) = rfxfy H 2 (x, y, z) = rfyfz H(x, y, z) = H 1 (x, y, z) H 2 (x, y, z) G(x, y) = z H(x, y, z) F(x) = y G(x, y).
40 (LMCS, p. 340) V.40 Then x y z fx fy fz H 1 (x, y, z) H 2 (x, y, z) H(x, y, z) a a a b b b a a b b b a a b a b a b a b b b a a b a a a b b b a b a b a b b a a a b b b b a a a x y G(x, y) a a 0 a b 1 b a 1 b b 0 x F(x) a 0 b 0
41 (LMCS, p. 341) V.41 S = {a, b} r a b a 0 1 b 1 0 a b f b a F(x, y) = z ((rxfz) (fy z)) (fy fx). F S =? Let G 1 (x, y, z) = rxfz G 2 (x, y, z) = fy z G(x, y, z) = G 1 (x, y, z) G 2 (x, y, z) F 1 (x, y) = z G(x, y, z) F 2 (x, y) = fy fx F(x, y) = F 1 (x, y) F 2 (x, y).
42 (LMCS, p. 341) V.42 Then x y z fx fy fz G 1 (x, y, z) G 2 (x, y, z) G(x, y, z) a a a b b b a a b b b a a b a b a b a b b b a a b a a a b b b a b a b a b b a a a b b b b a a a so x y F 1 (x, y) F 2 (x, y) F(x, y) a a a b b a b b 0 1 1
43 (LMCS, p. 342) V.43 Let F be a sentence and S a structure. Then F is true in S provided one of the following holds: F is rt 1 t n and r S (t S 1,..., ts n ) holds; F is t 1 t 2 and t S 1 = ts 2 ; F is G and G is not true in S; F is G H and at least one of G, H is true in S;
44 (LMCS, p. 342) V.44 F is G H and both of G, H are true in S; F is G H and G is not true in S or H is true in S; F is G H and both or neither of G, H is true in S; F is x G(x) and G S (a) is true for every a S; F is x G(x) and G S (a) is true for some a S. If F is not true in S, then we say F is false in S.
45 (LMCS, p ) V.45 Given a first order language L, let F be a sentence, S a set of sentences, and S a structure for this language. S = F means F is true in S. F is valid if it is true in all L structures. S = S means every sentence F in S is true in S. Sat(S) means S is satisfiable. S = F means every model of S is a model of F. We say F is a consequence of S.
46 (LMCS, p. 343) V.46 The Propositional Skeleton (A simple test for having a one element model.) The propositional skeleton, Skel(F), of a formula is defined as follows: Delete all quantifiers and terms. Replace with 1. Replace the relation symbols r with propositional variables R.
47 (LMCS, p. 343) V.47 Theorem F has a one element model iff Skel(F) is satisfiable. If Skel(F) is satisfiable, then choose an evaluation e that makes it true. Let r S (a,..., a) hold iff e(r) = 1 for r R. Let f S (a,..., a) = a for f F.
48 (LMCS, p. 343) V.48 To illustrate this, we consider the first order sentence F given by x y ( (x < y) z ((x < z) (fz y))). Then Skel(F) is the propositional formula P P 1 This is satisfiable by evaluating P as 0, so F has a one element model on the set {a}, namely < a a 0, and fa = a
49 (LMCS, p. 344) V.49 Equivalent Sentences The sentences F and G are equivalent, written F G, if they are true of the same L structures S, that is, for all structures S we have S = F iff S = G. x ( (x 0) y (x y 1)) and x y ( (x 0) (x y 1)) are equivalent. Proposition The sentences F and G are equivalent iff F G is a valid sentence.
50 (LMCS, p. 344) V.50 Equivalent Formulas Two formulas F(x 1,..., x n ) and G(x 1,..., x n ) are equivalent, written F(x 1,..., x n ) G(x 1,..., x n ), iff F and G define the same relation on any L structure S, that is, F S = G S. (x 0) y (x y 1) and y ( (x 0) (x y 1)) are equivalent.
51 (LMCS, p. 345) V.51 Proposition The formulas F( x) and G( x) are equivalent iff x(f( x) G( x)) is a valid sentence. Proposition The relation is an equivalence relation on sentences as well as on formulas. This is immediate from the definition of and the fact that ordinary equality (=) is an equivalence relation.
52 (LMCS, p. 346) V.52 Fundamental Equivalences x F x ( F) x F x ( F) ( x F) G x (F G) if x is not free in G ( x F) G x (F G) if x is not free in G ( x F) G x (F G) if x is not free in G ( x F) G x (F G) if x is not free in G ( x F) G x (F G) if x is not free in G ( x F) G x (F G) if x is not free in G F ( x G) x (F G) if x is not free in F F ( x G) x (F G) if x is not free in F x (F G) ( x F) ( x G) x (F G) ( x F) ( x G)
53 (LMCS, p. 346) V.53 Notes on the Fundamental Equivalences The first two, combined with the fundamental laws from propositional logic, are very useful in practice, e.g., the statement f is continuous at x: ε((ε > 0) δ((δ > 0) y( x y < δ fx fy < ε))), negates to f is not continuous at x: ε((ε > 0) δ((δ > 0) y(( x y < δ) ( fx fy < ε)))
54 (LMCS, p. 346) V.54 Comment If x occurs free in G then we cannot conclude ( x F) G x (F G), for example, ( x(x < 0)) (0 < x) and x((x < 0) (0 < x)) are not equivalent, for in the natural numbers the first is true of positive numbers x, whereas the second is false in the natural numbers.
55 (LMCS, p. 346) V.55 For implication we have: ( x F) G ( x F) G x ( F) G x ( F G) x (F G) To see that x (F G) ( x F) ( x G) need not be true consider the following example: x ((0 x) (0 < x)) and ( x (0 x)) ( x (0 < x)) And to see that x (F G) ( x F) ( x G) need not be true consider the example: x ((0 x) (0 < x)) and ( x (0 x)) ( x (0 < x))
56 (LMCS, p. 347) V.56 Replacement and Substitution Equivalent propositional formulas lead to equivalent first order formulas as follows. If F(P 1,..., P n ) and G(P 1,..., P n ) are equivalent propositional formulas, then for any sequence H 1,..., H n of first order formulas we have F(H 1,..., H n ) G(H 1,..., H n ) Example (P Q) P Q holds, so we have, in first order logic, ( ( x(x x 1)) ( x y(x y y x)) ) ( x(x x 1) ) ( x y(x y y x) )
57 (LMCS, p. 347) V.57 Equivalence is compatible with our connectives and quantifiers in the obvious sense. Lemma Suppose F 1 G 1 and F 2 G 2. Then F 1 G 1 F 1 F 2 G 1 G 2 F 1 F 2 G 1 G 2 F 1 F 2 G 1 G 2 F 1 F 2 G 1 G 2 x F 1 x G 1 x F 1 x G 1.
58 (LMCS, p. 347) V.58 This leads us to the basic replacement theorem for first order logic. Replacement Theorem If F G then H( F ) H( G ). Example Since ( ( x(x x 1)) ( x y(x y y x)) ) ( x(x x 1) ) ( x y(x y y x) ) we have ( x y(x < y)) ( ( x(x x 1)) ( x y(x y y x)) ) ( x y(x < y)) ( x(x x 1) ) ( x y(x y y x) )
59 (LMCS, p. 348) V.59 Substitution in first-order logic often requires the need to rename variables. We need to be careful with renaming variables to avoid binding any newly introduced occurrences of variables. Given a first order formula F, define a conjugate of F to be any formula F obtained by renaming the occurrences of bound variables of F so that no free occurrences of variables in F become bound. Keep bound occurrences of distinct variables distinct.
60 (LMCS, p ) V.60 Conjugates are equivalent: If F is a conjugate of F, then F F. Thus y(x y 1) w(x w 1). Substitution Theorem If F(x 1,..., x n ) G(x 1,..., x n ) and t 1,..., t n are terms, then Thus from follows F (t 1,..., t n ) G (t 1,..., t n ). y(x y 1) y( (x y 1)) u((y + w) u 1) u( ((y + w) u 1))
61 (LMCS, p. 349) V.61 Prenex Form A formula F is in prenex form if it looks like Q 1 x 1 Q n x n G where the Q i are quantifiers and G has no occurrences of quantifiers. A formula with no occurrences of quantifiers is called an open formula. The formula x ( (rxy) u (ruy) ) is not in prenex form, but it is equivalent to the prenex form formula x u ( (rxy) (ruy) ).
62 (LMCS, p ) V.62 Theorem Every formula is equivalent to a formula in prenex form. The following steps put F in prenex form: Rename the quantified variables so that distinct occurrences of quantifiers bind distinct variables, and no free variable is equal to a bound variable
63 (LMCS, p ) V.63 Thus change z((rzy) y((rxy) y(ryx))) to z((rzy) u((rxu) w(rwx))) Eliminate all occurrences of and using G H ( G H) ( H G) G H G H The above example becomes z( (rzy) u((rxu) w(rwx)))
64 (LMCS, p ) V.64 Pull the quantifiers to the front by repeated use of the following (F G) ( F G) (F G) ( F G) G ( x H) x (G H) G ( x H) x (G H) G ( x H) x (G H) G ( x H) x (G H) ( x G) H x (G H) ( x G) H x (G H) ( x G) H x (G H) ( x G) H x (G H) x G x G x G x G
65 (LMCS, p ) V.65 Applying these steps to z( (rzy) u((rxu) w(rwx))) gives z( (rzy) u((rxu) w(rwx))) z( (rzy) u( ((rxu) w(rwx)))) z u( (rzy) ((rxu) w(rwx))) z u( (rzy) ( (rxu) ( w(rwx)))) z u( (rzy) ( (rxu) w( (rwx)))) z u( (rzy) w( (rxu) ( (rwx)))) z u w( (rzy) ( (rxu) ( (rwx))))
66 (LMCS, p. 352) V.66 Valid Arguments Now we are working with sentences in a fixed first order language L. An argument F 1,, F n F is valid (or correct) in first order logic provided every structure S that makes F 1,..., F n true also makes F true, that is, for all S S = {F 1,..., F n } implies S = F.
67 (LMCS, p. 352) V.67 Proposition An argument F 1,, F n F in first order logic is valid iff F 1 F n F is a valid sentence; and this holds iff {F 1, F n, F} is not satisfiable.
68 (LMCS, p. 352) V.68 Example The work in Chapter 3 concerned equations. In first-order logic we treat equations as universally quantified sentences: x(s( x) t( x)) The argument x y u v (x y u v) x y z ((x y) z x (y z)) is valid. This is evident by considering that if a structure S satisfies the premiss then all multiplications give the same value. But then the multiplication must be associative.
69 (LMCS, p. 352) V.69 The argument y x (rxy) x y (rxy) is valid. To see this suppose S is a structure satisfying the premiss. Then choose an element a from S such that x (rxa) holds. From this we can conclude that x y (rxy) also holds since we can let y be a. We have demonstrated the validity of the above arguments by appealing to our reasoning skills in mathematics. It is fair to ask if one can now sit down and write out all the axioms and rules that one would ever need to justify valid arguments in first-order logic.
70 (LMCS, p. 352) V.70 This question was posed by Hilbert and Ackerman in 1928 in their slender book, the elementary and famous Principles of Logic. They proposed a collection of first-order axioms and rules of inference, and noted that they were correct rules of reasoning. But were they enough, that is, was their system complete? They said in their book that their proof system was sufficient for doing all the things they tried with first-order logic, and indeed asked if it was complete.
71 (LMCS) V.71 The brilliant young student Kurt Gödel in Vienna answered their question within a year, in the affirmitive, and used it for his PhD Thesis. This is the famous Gödel Completeness Theorem, discussed in Chapter 6 of the text. It says S = ϕ iff S ϕ. Another question that was prominent at the time was whether or not the powerful logical system of Principia Mathematica, developed by Whitehead and Russell in 3 volumes ( ), was powerful enough to do all mathematics, forever.
72 (LMCS) V.72 This system of logic was higher-order, meaning that one could quantify over things other than just the variables for the elements of the universe. One could also quantify over relation symbols (and function symbols). Thus they could write r(rx ry) as a definition of x y. It was known, from looking at the examples in Principia, that all known mathematical proofs could be written up in this system (if one so desired). So perhaps this is all one would ever need. Within a year of his PhD Gödel had answered this (much more difficult) question as well.
73 (LMCS) V.73 To everyones astonishment he put together a (lowly) first-order sentence F in the language of the natural numbers N = {N, +,, <, 0, 1} that he could show was True of N But Principia Mathematica was not strong enough to prove it. (Assuming that the proof system of Principia did not lead to a contradiction G G, for then one could derive every formula!) This is the famous Gödel Incompleteness Theorem.
74 (LMCS, p. 353) V.74 Counterexamples To show that an argument F 1,, F n F is not valid it suffices to find a single structure S such that each of the premisses F 1,, F n is true in S, but the conclusion F is false in S. Such a structure S is a counterexample to the argument. The argument x y (rxy) y x (rxy) is not valid; a two element graph gives a counterexample: a b
75 (LMCS, p ) V.75 Skolemization Skolem, following the work of Löwenheim (1915), gave us a technique to convert a first-order sentence F into a sentence F in prenex form, with only universal quantifiers, such that F is satisfiable iff F is satisfiable. Universally quantified sentences are apparently much easier to understand. This has provided one of the powerful techniques in automated theorem proving.
76 (LMCS, p ) V.76 One takes a given assertion in mathematics and expresses it as a first-order sentence F. This means we want the assertion to be true iff F is valid (i.e., true in all structures). This sentence F is valid iff F is not satisfiable. One replaces F by a universal sentence F obtained by Skolem s process. Then F is valid iff F is not satisfiable. One puts the matrix of F in conjunctive form.
77 (LMCS, p ) V.77 Then it is easy to replace F by a finite set of clauses C such that F is not satisfiable iff C is not satisfiable. Namely the clauses come from the conjuncts of the matrix of F. Now one has F is valid iff C is not satisfiable. To show C is not satisfiable one can apply resolution theorem proving. Now we look at the details of Skolem s process, called skolemization.
78 (LMCS, p ) V.78 Lemma Given the sentence y G(y), augment the language with a new constant c and form the sentence G(c). Then Sat ( y G(y) ) iff Sat ( G(c) ). Given the sentence x 1 x n y G( x, y), augment the language with a new n ary function symbol f and form the sentence x 1 x n G ( x, f( x)). Then Sat ( x y G( x, y) ) iff Sat ( x G ( x, f( x)) ).
79 (LMCS, p. 354) V.79 Proof In each step we can expand a model of the sentence on the left hand side to one on the right hand side by adding the appropriately designated element for the constant, or by adding an appropriate function f. Conversely, we can take a model of the sentence on the right hand side and remove the constant symbol, respectively function symbol, and get a model for the left hand side.
80 (LMCS, p. 354) V.80 Definition A first order formula F is universal if it is in prenex form and all quantifiers are universal, that is, F is of the form x G, where G is quantifier free. G is called the matrix of F. Example x y z ((x y) (y z) (x z)) } {{ } matrix
81 (LMCS, p. 354) V.81 Theorem Given a first order sentence F, there is an effective procedure for finding a universal sentence F (usually in an extended language) such that Sat(F) iff Sat(F ). Furthermore, we can choose F such that every model of F can be expanded to a model of F, and every model of F can be reducted to a model of F. Proof First, we put F in prenex form. Then we just apply the Lemma repeatedly until there are no existential quantifiers.
82 (LMCS, p. 354) V.82 The process of converting a sentence to such a universal sentence is called skolemizing. The new constants and functions are called skolem constants and skolem functions. Example To skolemize the sentence F = x y ((x < y) z ((x < z) (z < y))) we first put it in prenex form F = x y z ((x < y) (x < z) (z < y))
83 (LMCS, p. 354) V.83 Applying the Lemma, we introduce a new binary function symbol, say f, and arrive at the universal sentence F = x y ((x < y) (x < f(x, y)) (f(x, y) < y)) The structure Q = (Q, <), consisting of the real numbers with the usual <, satisfies F. If we choose f(a, b) = a+b 2 for a, b Q, we see that the expansion (Q, <, f) of Q satisfies F. Now we extend these ideas to sets of sentences.
84 (LMCS, p. 356) V.84 Theorem Given a set of first order sentences S, there is a set S of universal sentences (usually in an extended language) such that Sat(S) iff Sat(S ). Furthermore, every model of S can be expanded to a model of S, and every model of S can be reducted to a model of S. Proof We skolemize each sentence in S as before, making sure that distinct sentences do not have any common skolem constants or functions.
85 (LMCS, p. 356) V.85 This theorem is at the heart of the translation from first order logic to clause logic, as we shall see in the next section. Example We skolemize the set of sentences { x y z (x < y + z), x y z ( (x < y + z))} and obtain a set of universal sentences { y (a < y + fy), x ( (b < y + gy))}.
86 (LMCS, p. 356) V.86 The Reduction of First-Order Logic to Clause Logic Now we show that first order formulas can be reduced to the predicate clause logic, and thus to the propositional clause logic. First we want to look at some basic translations between clauses and first order formulas. A clause C = {L 1 ( x),..., L k ( x)} is considered as the universal sentence F C = x (L 1 ( x) L k ( x)) in first-order logic.
87 (LMCS, p. 356) V.87 For indeed S = C iff S = F C. There is also a translation from universal first order sentences to clauses. Lemma One can effectively construct, for any given universal sentence F, a finite set C F of clauses such that for any structure S S = F iff S = C F.
88 (LMCS, p. 356) V.88 Proof Let F be x G( x) with G( x) quantifier free. Put G( x) in conjunctive form G( x) G 1 ( x) G k ( x), where each G i ( x) is a disjunction of literals. Then x G( x) x (G 1 ( x) G k ( x)) F 1 F k where F i = x G i ( x). Thus S = F iff S = {F 1,, F k }. Let C i be the clause C Fi associated with the matrix of F i. Then S = F iff S = C F
89 (LMCS, p. 356) V.89 Example Let F = x y z (rx ry rz). First, we put the matrix of F in conjunctive form: rx ry rz rx (ry rz) ( rx ry) ( rx rz). Now we can read off the clauses from the conjuncts, namely, C F = { { rx, ry}, { rx, rz} }.
90 (LMCS, p. 358) V.90 Definition For S a set of universal sentences let C S = C F, F S the union of the collection of all sets C F of clauses obtained from the universal sentences F in S. Lemma Given a set S of universal sentences and S a structure, S = S iff S = C S. Thus Sat(S) iff Sat(C S ).
91 (LMCS, p. 358) V.91 Now we can formulate a connection between satisfiability of a set of first order sentences (not necessarily universal) and a set of clauses. Lemma Given a set S of first order sentences, let S be a skolemization of S. Then Sat(S) iff Sat(C S ).
92 (LMCS, p. 358) V.92 Example For S = { x y (rxy), y x (rxy)} we have S = { y (ray), x (rxb)} C S = {{ray}, { rxb}}. We can easily derive the empty clause from C S (in one resolution step), so C S is not satisfiable. Thus S is also not satisfiable. Consequently, the sentence F = ( x y (rxy)) ( y x (rxy)), obtained by conjuncting the members of S, is not satisfiable. So F is a valid sentence. Clearly F ( x y (rxy)) ( y x (rxy))
93 (LMCS, p. 359) V.93 Now we can state the fundamental transformation from first order logic to clause logic. Theorem Given S, a set of sentences, and F, a sentence, S = F iff Sat( C S( F) ), where S( F) is the set S { F}, and the prime symbol refers to taking the skolemization, as before. Thus the argument S F is valid iff one cannot satisfy the set C S( F) of clauses.
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The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,More information
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AN INTRODUCTION TO SET THEORY Professor William A. R. Weiss October 2, 2008 2 Contents 0 Introduction 7 1 LOST 11 2 FOUND 19 3 The Axioms of Set Theory 23 4 The Natural Numbers 31 5 The Ordinal NumbersMore information


text:
Ever since the dreaded attacks which took place on 9/11, Muslims primarily from the gulf region are being viewed with great suspicion in the United States of America along with every other non Muslim country. The western media is to be thanked for profiling each and every Muslim as a possible terror threat, although the majority of Muslims from around the world condemn such extremist actions by a minute fraction of misguided so called "Muslims".
This has caused yet another arrest in the USA where a man of Saudi origin, who arrived at the Detroit Metro Airport, was arrested because he brought along a pressure cooker in his luggage. The man is currently behind bars while his lawyer in the United States, James Howarth has stated that having such a pressure cooker does not make him a criminal.The man, whose name is Hussain Al Khawahir, has been accused by the American authorities of using an altered Saudi Arabian passport (certain pages were missing) and also for making an alleged false statement to the U.S Customs officer regarding the pressure cooker. Hussain appeared in the U.S district court and waived his rights to a detention hearing.
It is said that an American lady informed the FBI that she had spotted a young Saudi man who lived in her neighborhood walking around with a huge pressure cooker in his possession. The woman, who has not been named, added that the grey color of the pressure cooker resembled the one which was used in the Boston Marathon bombing which killed 3 people and injured 264.Al Khawahir's lawyer, Howarth, still maintains that his client knew nothing about the use of a pressure cooker in the terrorist attack on the Marathon.
The United States Attorney, Barbara McQuade stated in talks with the media that the unique combination of allegations and facts which include false statements and altered passports has raised the concerns of the authorities. She added that even though they do not want to jump to conclusions, they have to conduct a full length investigation for the safety of the people.According to the formal complaint which has been filed in the U.S. District Court, Al Khawahir flew in to the United States from Saudi Arabia via Amsterdam and informed the Customs officer that he had purchased the pressure cooker for his nephew, who is a student in the U.S., because they aren't sold in America.
The Customs officer also noticed that pages 33 and 34 had been removed from the passport. On this Al Khawahir stated that he does not know how it happened as only his wife and 3 little children have access to it.It should also be noted that the Saudi man does not speak English and it is possible that a misunderstanding took place with the officials, the lawyer stated. He added that people of Saudi Arabia often travel with the pressure cooker due to how they prepare their traditional food.Al Khawahir's nephew, Nasser Al Marzooq said that it was all a misunderstanding and he needed the device to cook lamb.
Cultural Issues Abroad
- How Brazilians treat Saudis and Arabs?
- Saudis are treated as "Cash Machines" abroad
- Why some Saudis cannot adjust Abroad?
- Tips on Conversing with an American Citizen
- How can Muslims avoid pork in the USA / UK?

subset: commoncrawl


text:
Besides being a great athlete and one of the nicest Jets players, Curtis Martin is something else: a smart man. Why? He takes care of his linemen.
Martin is continuing a tradition he started when he played for the New England Patriots, financially rewarding his linemen for opening up those large holes he rumbles through.
Martin gives each lineman $100 after every game in which he rushes for at least 100 yards. He has had seven 100-yard games, which amounts to $3,500 divided five ways. After yesterday's 17-10 victory over the Bills, in which Martin gained 38 yards on 20 carries, there was one more regular-season game and the playoffs still to go, and he may also give the linemen some sort of other gift at the end of the season.
''Our offensive line is the key,'' Martin said earlier this season. ''Those guys do a great job.''
Martin has had a solid season. Entering the Bills game, he was fourth in the American Football Conference in both rushing yardage with 1,146 yards and in rushing touchdowns with 8. He came to the Jets with a reputation as a player who often got hurt, and though he missed some time this year with leg injuries, he spent most of the year on the field and not in the trainer's room.
What Martin has provided the Jets is an explosive running threat. If opponents try to stop the Jets' passing game, he breaks lose. If they use eight-man fronts to try and stop him, the wide receivers Keyshawn Johnson and Wayne Chrebet have big days.Continue reading the main story
Martin's provision of a running game -- as well as pass blocking -- has helped quarterback Vinny Testaverde have a great year. Testaverde had a gaudy quarterback rating of 128.3 and had thrown 133 consecutive passes coming into the Bills game without getting sacked. He was the only quarterback in the National Football League who had not thrown an interception on third down.
Although the linemen have done a nice job of pass protecting and run blocking, none of them will be starters in the Pro Bowl.
The linemen -- center Kevin Mawae, guards Todd Burger and Matt O'Dwyer and tackles Jumbo Elliott and Jason Fabini -- enjoy blocking for Martin because he is sincere in his appreciation of them. But Martin remains a mystery in the locker room. Few players know him well because he rarely opens up about himself to teammates. He is not aloof, but is extremely quiet and private.
Unlike some running backs, however, Martin understands that a handshake, a thank you and a little cash can go a long way.
Rebellion on and Off the Field
Carolina Panthers linebacker Kevin Greene shoved his assistant coach on the sideline. Pittsburgh Steelers quarterback Kordell Stewart screamed at Coach Bill Cowher. Ugly, ugly, ugly -- and these incidents were just from last Sunday.
Some fans may not realize these kinds of incidents are common, but most of them are not caught on tape.
A more subtle form of player disagreement also occurs: ignoring a play that is called by the coaches. That happened during the Detroit Lions' loss to the San Francisco 49ers last Monday night.
In the fourth quarter, the Lions' coaches called a play that was supposed to go to someone other than Herman Moore, the Pro Bowl wide receiver, who has been frustrated this season by what he perceives as a reduced role.
Moore did not like the call. While in the huddle, according to several Lions, Moore told quarterback Frank Reich, who was in for the injured Charlie Batch, to ignore the called play and simply lob the ball to him in the end zone. That was what Reich did, and the Lions scored on a 24-yard pass.
Defying a called play would normally anger coaches. But when Moore went to the sideline, he was congratulated for the touchdown.
Union Saves for Future
What has made Gene Upshaw a respected union leader among N.F.L. players is that he is always prepared and a step ahead. That may be why the union, spurned on by the National Basketball Association lockout, is taking steps to guarantee that its players have a large financial war chest in case there is labor strife when the collective bargaining agreement expires in 2005.
In the N.B.A., there is concern among players and agents that some of the lower-paid players may be feeling a financial squeeze with the lockout nearing its 200th day. The football union wants to make sure at least some money is available for players if they too wind up in a dispute.
According to agents and union officials, most of the money being invested for the players is coming from the union's licensing arm, which earns millions each year from merchandise sales. Some agents believe that by 2005 the union will have emergency funds of at least several hundred million dollars, and maybe much more.
Would that be enough to support each player for, say, one year if there was a strike? Probably not, but it would help ease the pain and give players some security and leverage.
Giants Hear Bronco Talk
Were the Denver Broncos too cocky before they lost to the Giants last weekend? Some Giants believe they were.
The Broncos are normally chatty, and it is not unusual for them to talk trash to opponents during a game. That is what happened at Giants Stadium. Tight end Shannon Sharpe, some Giants have said, was screaming about how he could not be stopped.
The Giants took all the talk personally.
Especially quarterback Kent Graham, who played one of the best games of his career. Late in the final period, Broncos linebacker Bill Romanowski, never afraid to do or say almost anything on the field, told Graham to shut up when he was speaking to several officials about a call. Graham, one of the classiest of players, lost his temper and yelled a few choice words at Romanowski.
The back-and-forth seemed to energize Graham. Soon after the exchange, he launched the game-winning bomb to wide receiver Amani Toomer. Graham got the last word.
Instant Fairness, Instead
One reason the N.F.L.'s competition committee rejected instant replay for the playoffs was because of possible unfairness. Indianapolis Colts General Manager Bill Polian, among others, believed coaching staffs who worked with replay when it was in use from 1986 to 1991 would have had an advantage over the staffs that did not. . . . Now that Denver quarterback John Elway has his Super Bowl ring, who would he like to see win one? Miami quarterback Dan Marino, whom Elway faces tomorrow night. ''If we can't get it this year,'' Elway said, ''then I'd like to see him win one.''Continue reading the main story


text:
The Ugly Duckling
Read a classic fable with your beginning reader! This story page for "The Ugly Duckling" will help build reading skills, and provides a great starter for comprehension exercises such as identifying characters, identifying setting, and identifying the moral of a story.
Click to find similar content by grade, subject, or standard.


text:
Cannot find your local Phoenix Contact site? Use our selector
The illustration shows a version with 8 slots
Sensor/actuator box, Connection method: M12 socket Plastic, Number of slots: 6, Number of positions: 5, Coding: A - standard, Slot assignment: Double, Status indication: No, Universal; Master cable connection: Spring-cage connection 180°, Shielding: No
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text:
Hey all, This is my first blog on the wiki! However, i've been on this sight everyday and have been reading just about every blog for the past two years and thought i should stop being lazy and just join.
Now on to my first blog.
I've been havin a problem with carl in the last volume. He didnt do anything wrong, but i cant help but notice a couple problems...
First: The time span between issue 105 to issue 108 has to be at least a week, two weeks at the most. in that time, carl's hair grew long enough for him to hide his "exposed skull". This is practically impossible unless carl has been hiding a stash of that generic "grow hair quick" shampoo you see in cartoons.
Second: Why hasnt carl been bandaging up his eye? I mean, it's clear how easily it can get infected! all it takes is a windy day with some dust blowing into his eye for him to be in what i would imagine serious pain. I really hope he didnt stop bandaging it up because Negan told him it looked cool. Why would Carl care about what Negan thought? He absolutely hates Negan.
Anyway, sorry if this has been discussed on another blog already. once again, I love the walking dead. This is just one thing that bothers me a bunch. Feel free to tell me what you think! All replies are greatly appreciated!


text:
Iso Tc 54 Items18 textbooks were found
ISO 7355:1985, Oils of sassafras and nutmeg -- Determination of safrole and cis- and trans-isosafrole content -- Gas chromatographic method on packed columns
ISO 3033-2:2005, Oil of spearmint - Part 2: Chinese type (80 % and 60 %) (Mentha viridis L. var. crispa Benth.), redistilled oil
ISO 22972:2004, Essential oils - Analysis by gas chromatography on chiral capillary columns - General method
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Short, fat and good looking. The Sport Drive has plenty of volume to inflate knobbly tyres. Fast. The flexible connector hose lets you choose presta or schrader valves without having to unscrew, modify and reassemble the pump.
Made by: Lezyne
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New York: Ace Indian golfers Jeev Milkha Singh and Jyoti Randhawa, all set for a wild ride at the US Open starting on Thursday are hoping their historic double act will capture the imagination of the golfing world. Jeev, the current Asian Tour's UBS Order of Merit leader, and 2002 Asian number one Randhawa carry the region's charge at Winged Foot Golf Club where they will face the likes of title holder Michael Campbell and world no. one Tiger Woods. This is the first time that two Indians are featuring together in a US Open, with Jeev making his second appearance and Randhawa his debut. They earned their places in the year's second major through qualifying in England earlier this month. "It should be a big day for Indian golf," said Jeev. "Whenever you qualify for a major, you want to play well for personal reasons. But coming from an Asian background, both of us being here is fantastic for the Asian Tour. The more Asians we have in big tournaments and majors, the more recognition we'll get," he added. The tenacious Jeev, who qualified and played in all four rounds in the 2002 US Open in Bethpage, also in New York, is riding on a crest of new-found confidence. Since April, he has notched one victory, his first since 1999, and posted three other top-10s. (PTI)
Topics : Golf
Get the latest Sri Lanka vs India 2017 news, check Sri Lanka vs India 2017 schedule and live score. Like us on Facebook or follow us on Twitter for more Sri Lanka vs India 2017 updates. You can also download the NDTV Cricket app for Android or iOS
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text: Prune your royal poinciana tree during its first three years to encourage strong branches. Remove all suckers--stems that sprout from the side of the tree--as well as limbs that are less than 8 to 12 feet from the ground. Remove branches that are dead, dying, broken or crossed, and prune to open up the center of the tree in order to increase sunlight and air circulation. Cut back growing tips to limit height and width, if desired.


text:
Last.fm Fridays | July 2017
last.fm college generator upload to imgur
[Stream Now] Meek Mill's Wins & Losses
Spotify https://open.spotify.com/album/2SCQAZZ5pc90l9SPBsWCIQ Youtube Playlist https://www.youtube.
**VIC MENSA ALBUM DEBUT ANNOUNCED : The Autobiography**
https://twitter.com/VicMensa/status/884871348241055744 Didn't I (Say I Didn't) Memories on 47th St.
Forum Guidelines
The Genius forums are a place where everyone should feel comfortable to contribute knowledge and take
Make Love is a better diss track than No Frauds
Agree or disagree? I was thinking about how Nicki had like 5 dis verses against Remy Ma. Even though "No
I found this youtube channel and its fucking retarded
https://www.youtube.com/channel/UCKOlU7SLdtqmOx1w0nyAPtw?&ab_channel=Hip-HopUniverse Literally every
https://soundcloud.com/ygtut/ragstobxxches https://soundcloud.com/ygtut/hotfire
Tyler, The Creator Is NOT Gay
https://www.youtube.com/watch?v=RHWmTWl5mks Ya'll mad or naw?
Rakim reacts to 4:44
Underrated And Overrated Rappers and Singers?
20 years ago today, this classic was released
Let's be completely honest
If Tupac and Biggie were still alive, they'd now be pop-rap artists like Pitbull and Flo Rida.
2017 AOTY (so far) elimination game
Rules; vote for a rapper to be eliminated, after 8 or so votes the rapper with the most votes will be
[New Freestyle] Meek Mill on Funk Flex
So this Tyler album....
might be aoty
Asher Roth has had a tumultuous experience in the Hip-Hop industry since he rose to mainstream prominence five years ago. Issues with major labels presented problems of their own, but perhaps even more detrimental to his career has been his reputation as a "frat-rapper", a label he earned after releasing his debut single "I Love College" in 2009. While the song remains his biggest commercial hit that helped him land a spot on the XXL Freshman cover that summer, it also turned away some listeners who were left with the misconception that his entire catalog consisted of similar juvenile, stereotypical "white rapper" songs. However, Roth's talent is not limited to boasting of his beer pong skills and chanting "Chug Freshman!" at frat parties. No, the truth remains that Asher Roth is actually capable of making music about more diverse and introspective subjects that most listeners can relate to, and his new record "RetroHash" is proof.
Although the core elements of Asher's style remains, "RetroHash" establishes a new sound in his music. Simply classifying the album as "Hip-Hop" doesn't really suffice; at the very least, it's some unique brand of Alternative Hip-Hop. The influence of California is very evident in the smooth, laid-back, sometimes hippy production, which is entirely handled by Blended Babies, and bears influences of several other genres, including blues, alternative rock and pop, among others. Singing plays a heavier role on "RetroHash" than it has on Roth's previous projects, an impressive skill he possesses in addition to his rhyming proficiency.
"RetroHash" begins with the reflective "Parties at the Disco", which also features vocals from songstress ZZ Ward. Asher opens his verse with a Wu-Tang reference, explaining how he copes when he feels disconnected with the world: "Cause the cash rules everything around / So, every now and then / Feeling like I'm out of town / Got a chill, sit down, remember how to be a child". Asher briefly switches back to a classic Hip-Hop sound up with the boom-bap sound of "Dude", a track in which he offers the listener a humorously candid glimpse of his life with full disclosure ("TED talks on my iPad / Old search says Bang Bros, my bad"). Fellow emcee Curren$y also drops a slick verse over the bouncy instrumental.
Asher's experimentation with other genres is best exemplified with "Tangerine Girl", a dreamy, upbeat disco-esque ode dedicated to a female Asher likes. The entire song is sung in hypnotic echoes until the final verse which Asher quickly raps. Syrupy guitar chords and horns adorn the mellow instrumental of "Fast Life"—perhaps the album's best track— which touches on the chaos and suffering people subject themselves in their daily lives by trying to live their lives too fast. Guest rapper Vic Mensa lends the track an excellent verse while Asher sings a beautiful hook and tells the story of a young girl whose lack of guidance and shoddy home life cause her to grow up at an accelerated pace–she sleeps with guys much older and eventually turns to using drugs and alcohol as her vices:
Far too delicate, it's hard when a father is yelling and
Takes it out on her or the dog for the hell of it
Only thirteen when she started developing
By fifteen, attracting older gentlemen
She let them in, mom wasn't looking, she might have been
Turned a blind eye found it better than fighting with
Her little girl caught in a sinister world
The somber mood lightens and becomes more hopeful with two Major Myjah collaborations. The first, "Last of the Flohicans", features an inspirational hook from Major Myjah and outstanding rapped verses from a very self-aware Asher:
Need a new beginning, let me start again
Let me start again this time without an awkward ending
I walk pretending my thoughts are never in this planet
But I abandoned them not long ago and left them stranded
There aren't many missteps with "RetroHash". The only real grievance is the short length of the album, which clocks in at just over forty minutes across the span of ten tracks. And for some listeners, the multitude of different genres and alternative sound may be intimidating at first, but the level of originality Asher exhibits with this album outweighs any negative aspects of his decision to dabble further into other genres.
Asher Roth certainly covers new ground with the album, from a musical standpoint as well as furthering the progression of his career. He's endured several years of delays and label issues, but Asher Roth has finally broken free from the shackles of the majors to release this album on his own terms, which must feel like a huge accomplishment, and deservedly so. "RetroHash" is an impeccable demonstration of Roth's creativity and ability to create self-described "fun music" that simultaneously reaches a deeper, more insightful level. While he may never reach the commercial success he once enjoyed, Roth's music is continuing to develop, and regardless of Billboard charts and record sales, it's a safe bet that Asher Roth will be around for a while as he continues to improve. His "College" days are over.
Asher delivered
The summer jams are here, album has been on repeat since it dropped, I also copped the autographed vinyl.
nice review, i like the new sound of this one. its fresh and i think he pulled if off well where many other rappers would have made a disaster if they tried this style. vic mensa's verse was hot on fast life for sure
Also, Nice review, actually agree (even though album is like an 8.5 for me personally) nice breakdown of how it's actually different than the first, however long time asher fans who've kept up with him were ready for this, his sound hasn't really been the same for a long time now, but he's still delivered dope shit.
however long time asher fans who've kept up with him were ready for this, his sound hasn't really been the same for a long time now
That's a good point, "Pabst and Jazz" had some tracks that shared similarities with some songs on "RetroHash", and since Blended Babies produced the majority of both as well it makes sense.
Pabst and Jazz is a classic EP/mixtape fo sho
Yea, I enjoyed that release a lot. Not only was Asher dope lyrically but all of the guests and the production were nice.
Sounds like I'm gonna have to give this right here a chance, it seems like my kind of album. I'm into Alt Hip-Hop as much as you'd expect me to be (if you consider my taste in music).
It's worth checking out with an open mind. I suppose it can be described as more of a retro-pop kind of alternative sound, for the most part. A lot of singing and light guitar chords on quite a few songs.
Thanks, appreciate it man. Curious, what did you think of the album? Assuming you heard it.
@DeLillo I thought it was alright not bad just not album of the year
Yea, it's pretty good but I agree, there's already been several superior releases this year so far lol
Sorry for bumping this thread. But just first listened to this project now. Pretty dope project from Roth. agree with the score 7.7/10
I'd personally give it a 8.5/10 because of my personal taste. Loving the laid back vibes. Also excelent review, keep it up.


text:
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text:
Hello, i have a couple of computers that it wont scan. There is no firewall on them and the remote access on them is open. I am using the domain admin account to do scanning and it keeps telling me that the logon is wrong. Can anyone please help
6 Replies
Jan 3, 2007 at 9:31 UTC
1st Post
I presume its a windows based machines on a domain
if thats the case then you should try the domain admin account & pass
Username: DOMAIN\administrator
check for any sort of port blocking such as firewalls and other security software.
if this doesnt work the the local admin account of the machine to audit it, this would help you narrow down the problem.
Jan 3, 2007 at 1:31 UTC
I've tried DOMAIN\administrator account and the local admin account for authentication and still nothing. There are a couple workstations on the same subnet as computers that are scanning properly that are getting auth errors. Open ports and installed software are the same for all of them too.
Jan 3, 2007 at 1:55 UTC
yes this is a windows domain and i am using the domain admin account. Its also in the format that you provided/
There is no other security software that is preventing the scan. no windows firewall or 3 party firewall. All remote ports are open.
i tired the local admin account and it still failed. Im not sure what is going on here
Aug 25, 2008 at 8:20 UTC
1st Post
I am having the same issue . . . any one know how to fix this?
Feb 5, 2009 at 4:19 UTC
Any resolution to this problem?
Apr 9, 2010 at 3:48 UTC
Niagara Technology Group (NTG) is an IT service provider.
And another follow up one year later :)
If anyone has an issue today, please post a NEW question. Tacking onto old, zombie threads doesn't help anyone.
This discussion has been inactive for over a year.
You may get a better answer to your question by starting a new discussion.


text:
The roots of this label lie with the traditional Italian fashion house Trussadi who are renowned for their elegant aesthetics within the fashion industry. The story of this timeless brand began in 1910 in Bergamo, in a small glove factory which is where the fashion designer Dante Trussardi crafted fine leather goods. Over the years, Trussadi has expanded to become a global phenomenon who has taken the world by storm. In 1978, the fashion empire extended its family to include a junior line entitled Trussardi Junior which takes inspiration from the mainline collection.
New products from the Spring/Summer 2017 season will be arriving shortly. Meanwhile have a look at our New Arrivals.
The Trussardi Junior collection incorporates a noble sense of style and makes class and refined elegance contemporary to satisfy the trends and styles that characterise today. With a timeless colour palette, simple subtle details and a natural fabric selection which is sumptuous and comfortable.


text:
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text:
My Notes:
This home plan could reduce your energy costs »
From I-440: take US 1 South, take exit #95 Hwy 55 east towards Holly Springs. Stay on 55/55-By Pass for approximately 4 miles. Turn right onto Green Oaks Parkway. Go approximately 1 mile until you see the entrance to the 12 Oaks Community in front of you. Continue straight into 12 Oaks. Take your 1st left onto Ribbon Walk Lane. Our Model will be the second on the left. The address is 305 Ribbon Walk Lane.
If you are using GPS, just enter 12 Oaks, Holly Springs, NC.
Welcome home to the Villa Camellia II! This inviting front or side porch makes this home unique. This ranch floorplan has a spacious owner's suite, guest room, study and 2 full baths - all on first floor. The open and bright great room is adjacent to a large dining area with 12 ft. ceilings and opens to a large patio or optional screened in porch. Upstairs there is a 2nd floor loft, bedroom and optional full bath, or opt for a ton of 2nd floor storage, you take your pick!


text:
This content has been marked as final. Show 2 replies
1. Re: informatica 9.0.1 hf2 and Dac 10.1.3.4.1 only Platform Softwaresuser10675696 Oct 15, 2012 9:42 AM (in response to user10675696)HI,
Can any one provide me DAC Platfform softwre
2. Re: informatica 9.0.1 hf2 and Dac 10.1.3.4.1 only Platform SoftwaresAndy Coates Oct 15, 2012 10:15 AM (in response to user10675696)Hi,
Presuming client machine is a Windows desktop.
Sadly Oracle do not currently provide a download for a standalone DAC client; you need to install the client/server and then (as it is a Java application) copy the DAC install location to the client machine and edit config files as appropriate. On the client machine you will obviously not be running the DAC server.
The Informatica client tools are bundled with the Windows server installation downloads on eDelivery.
OBIEE 11g client tools can be installed from the Windows download on eDelivery (it is a setup option).
Please mark if helpful / correct,


text:
A FOOTY doctor linked to the Western Bulldogs has been prescribing drugs to Hells Angels, police say.
Dr Richard Vucinic allegedly issued hundreds of medical scripts to Hells Angels associates and members of the outlaw gang's feeder club, the Red Devils.
His former VFL club, Williamstown, told the Herald Sun Dr Vucinic hadn't been employed there since September last year.
Williamstown Football Club chief executive Brendan Curry said Dr Vucinic had been recruited by Footscray Bulldogs last year and had worked with the Western Bulldogs.
Bulldogs chief executive Simon Garlick said last on Friday the club had now terminated Dr Vucinic's services.
Police say the 37-year-old GP issued up to 20 prescriptions a day to bikies.
In covert phone recordings picked up during an investigation into Hells Angels' boss Mohammed Khodr, Red Devils associate James Turner allegedly boasted he had a doctor "on board" who could provide scripts for any medications.
The Melbourne Magistrates' Court heard Dr Vucinic prescribed Xanax, Valium and other drugs.
Police said Dr Vucinic "was linked to the Western Bulldogs or its affiliate the Williamstown Football Club".
Williamstown is not an affiliate club to the Western Bulldogs this year, with Footscray instead representing the AFL club in the VFL.
Dr Vucinic told Magistrate Simon Garnett: "I'm not guilty. I didn't know he was a bikie".
He said his arrest had the potential to destroy his career.
Dr Vucinic is under investigation by the Medical Board of Victoria. He has been granted bail with strict conditions.
Originally published as Bulldogs doctor linked to bikies, police say


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text:
|Episode 1458
|Production code
|ITV transmission date
|8th January 1975 (Wednesday)
|Stories by
| Esther Rose
Peter Tonkinson
|Script editor
|Leslie Duxbury
|Brian Finch
|Brian Mills
|Susi Hush
|Previous episode
|6th January 1975
|Next episode
|13th January 1975
Albert adopts a pigeon. He calls it Gilbert. Ray tells Jerry that Mavis is getting married. Betty and Maggie realise they're going to have to tell Gordon. Battered wife Lynn Johnson asks Len for help but refuses to go to the police. He tells her that he's willing to talk to her husband but, afraid, she rejects the offer. Having heard from Alf that racing pigeons can be worth up to one thousand pounds, Stan tries to get Albert interested in racing Gilbert and goes in 50-50 with him. Maggie tells Gordon that Betty is his mother.
Regular castEdit
- Len Fairclough - Peter Adamson
- Albert Tatlock - Jack Howarth
- Stan Ogden - Bernard Youens
- Ken Barlow - William Roache
- Ray Langton - Neville Buswell
- Jerry Booth - Graham Haberfield
- Betty Turpin - Betty Driver
- Granny Hopkins - Jessie Evans
- Vera Hopkins - Kathy Staff
- Deirdre Hunt - Anne Kirkbride
- Alf Roberts - Bryan Mosley
- Maggie Cooke - Irene Sutcliffe
- Gordon Clegg - Bill Kenwright
Guest castEdit
- TV Times synopsis: Will Gordon be let into an old secret - and will Albert keep his new "lodger"?
- Viewing Figures: First UK broadcast - 7,850,000 homes (6th place).
|January 1975 episodes
|Wed 1st • Mon 6th • Wed 8th • Mon 13th • Wed 15th • Mon 20th • Wed 22nd • Mon 27th • Wed 29th


text:
SANTA MONICA, Calif. (AP) Former first lady Nancy Reagan was hospitalized Sunday after falling in her home in Bel-Air but is doing well, her spokeswoman said.
Reagan, 86, was taken to St. John's Health Center, where doctors determined she did not break a hip as initially feared, spokeswoman Joanne Drake said.
Drake said Reagan was doing well and would be staying the night in the same room where former President Ronald Reagan stayed after he broke his hip at home in 2001. He died June 5, 2004, after a 10-year battle with Alzheimer's disease. The two were married for 52 years.
Reagan's family physician recommended the overnight stay "as a precaution," Drake said.
"She's joking and visiting in her room," Drake said.
A message left for a hospital spokesman Sunday night was not immediately returned.
Nancy Reagan's last major public appearance was at the Jan. 30 Republican presidential debate at the Reagan Library in Simi Valley, Calif., where she sat with Gov. Arnold Schwarzenegger.
She did not attend a 97th birthday celebration held for her late husband at the library on Feb. 6.
Copyright 2008 The Associated Press. All rights reserved. This material may not be published, broadcast, rewritten or redistributed.