lec26-first-order
... The language accepted by has 7 words: all assignments other than x1 = x2 = x3 = F. The first 2-CNF clause removes ¼ of the assignments, which leaves us with 6 accepted words. Additional clauses only remove more assignments. ...
... The language accepted by has 7 words: all assignments other than x1 = x2 = x3 = F. The first 2-CNF clause removes ¼ of the assignments, which leaves us with 6 accepted words. Additional clauses only remove more assignments. ...
Letter to the Editor
... In the February 1983 issue of this Journal, D. H. and Emma Lehmer introduced a set of polynomials and, among other things, derived a partial formula for the discriminant of those polynomials (Vol. 21, no. 1, p. 64). I am writing to send you the complete formula; namely, D(Pn(x)) ...
... In the February 1983 issue of this Journal, D. H. and Emma Lehmer introduced a set of polynomials and, among other things, derived a partial formula for the discriminant of those polynomials (Vol. 21, no. 1, p. 64). I am writing to send you the complete formula; namely, D(Pn(x)) ...
Working Out Formulae
... WORKING OUT FORMULAE A formula is a quick way of writing down a general method for solving a problem, which can be used whatever the numbers are. For example, to find the area of any rectangle we multiply the length by the width. The formula is ...
... WORKING OUT FORMULAE A formula is a quick way of writing down a general method for solving a problem, which can be used whatever the numbers are. For example, to find the area of any rectangle we multiply the length by the width. The formula is ...
PDF
... function symbols and equality are built into first-order logic. Since the formulation of firstorder logic in this course does not make this assumption we have to add the functionality axioms for + and *, and the equality axioms ref, sym, and trans. We also need a restricted substitution axiom that p ...
... function symbols and equality are built into first-order logic. Since the formulation of firstorder logic in this course does not make this assumption we have to add the functionality axioms for + and *, and the equality axioms ref, sym, and trans. We also need a restricted substitution axiom that p ...
MA 311: Exercises 1
... 1. Assume that there are nine billiard balls, four of them are black and the other five are colored blue, yellow, red, green, and orange. In how many ways can one choose five balls out of these nine (so that with respect to the black balls only their number is relevant)? 2. How many results are poss ...
... 1. Assume that there are nine billiard balls, four of them are black and the other five are colored blue, yellow, red, green, and orange. In how many ways can one choose five balls out of these nine (so that with respect to the black balls only their number is relevant)? 2. How many results are poss ...
Dalton`s Atomic Theory
... Earlier we used the Particle theory of Matter to explain observations of matter. However, this theory cannot explain everything we have just learned regarding chemical changes. For example it cannot explain the electrolysis of water. ...
... Earlier we used the Particle theory of Matter to explain observations of matter. However, this theory cannot explain everything we have just learned regarding chemical changes. For example it cannot explain the electrolysis of water. ...
pdf
... An interesting consequence of Church's Theorem is that rst-order logic is incomplete (as a theory), because it is obviously consistent and axiomatizable but not decidable. This, however, is not surprising. Since there is an unlimited number of models for rst-order logic, there are plenty of rst-o ...
... An interesting consequence of Church's Theorem is that rst-order logic is incomplete (as a theory), because it is obviously consistent and axiomatizable but not decidable. This, however, is not surprising. Since there is an unlimited number of models for rst-order logic, there are plenty of rst-o ...
Ch 11 Patterns (WP)
... 1. (a) Copy and complete this table, showing the number of legs . No. of spiders (S) ...
... 1. (a) Copy and complete this table, showing the number of legs . No. of spiders (S) ...
Empirical Formula/Molecular Formula
... 2. When a 14.2 gram sample of mercury(II) oxide is decomposed into its elements by heating, 13.2 grams of Hg is obtained. What is the percent composition of the compound? ...
... 2. When a 14.2 gram sample of mercury(II) oxide is decomposed into its elements by heating, 13.2 grams of Hg is obtained. What is the percent composition of the compound? ...
Percent composition
... tells actual # of each kind of atom in a molecule of the cmpd. either the same as its experimentally determined empirical formula, or a simple whole‐number multiple of empirical formula ...
... tells actual # of each kind of atom in a molecule of the cmpd. either the same as its experimentally determined empirical formula, or a simple whole‐number multiple of empirical formula ...
(Jed Liu's solutions)
... Without loss of generality, assume X = T (φ) for some formula φ. We proceed by induction on the degree of φ. Base case. If φ has degree 0, then φ is a propositional variable, and the branch already contains atomic conjugates. Inductive step. Let n ≥ 0 and assume for all formulæ φ of degree at most n ...
... Without loss of generality, assume X = T (φ) for some formula φ. We proceed by induction on the degree of φ. Base case. If φ has degree 0, then φ is a propositional variable, and the branch already contains atomic conjugates. Inductive step. Let n ≥ 0 and assume for all formulæ φ of degree at most n ...
PATTERNS, CONTINUED: EXPLICIT FORMULAS
... An explicit formula is a formula that enables you to compute any given term in a pattern without knowing the previous term; that is, a formula giving the nth term in terms of n. In the example of {4, 11, 18, 25, 32…}, we might use the formula 4 + 7(n -1) to find the nth term. 1. Recall the sequence ...
... An explicit formula is a formula that enables you to compute any given term in a pattern without knowing the previous term; that is, a formula giving the nth term in terms of n. In the example of {4, 11, 18, 25, 32…}, we might use the formula 4 + 7(n -1) to find the nth term. 1. Recall the sequence ...
Normal Forms
... The Skolem form of a formula F in RPF is the result of applying the following algorithm to F : while F contains an existential quantifier do Let F = ∀y1 ∀y2 . . . ∀yn ∃z G (the block of universal quantifiers may be empty) Let f be a fresh function symbol of arity n that does not occur in F . F := ∀y ...
... The Skolem form of a formula F in RPF is the result of applying the following algorithm to F : while F contains an existential quantifier do Let F = ∀y1 ∀y2 . . . ∀yn ∃z G (the block of universal quantifiers may be empty) Let f be a fresh function symbol of arity n that does not occur in F . F := ∀y ...
IUMA Máster MTT, Métodos, 2015-2016 Examen 22 febrero 2016
... 3. Comentar: A1, . . . ,An ⊨ B means that formula B is valid: it is true in all situations in which A1, . . . ,An are true. It implies the use of semantics. However a theorem is a formula that can be established (‘proved’) by a given proof system. We write A is a theorem as ⊢ A, A is a theorem. A pr ...
... 3. Comentar: A1, . . . ,An ⊨ B means that formula B is valid: it is true in all situations in which A1, . . . ,An are true. It implies the use of semantics. However a theorem is a formula that can be established (‘proved’) by a given proof system. We write A is a theorem as ⊢ A, A is a theorem. A pr ...
.pdf
... of those are there? If there are N distinct propositional variables occuring in ϕ, then there are 2N different interpretations to consider (since each variable can get one of two truth values). That’s a lot, but at least, it’s a finite number, and it is possible to simply enumerate all 2N interpreta ...
... of those are there? If there are N distinct propositional variables occuring in ϕ, then there are 2N different interpretations to consider (since each variable can get one of two truth values). That’s a lot, but at least, it’s a finite number, and it is possible to simply enumerate all 2N interpreta ...
Logic: Propositional Tree Rules (Handout)
... (Step 1) List the elements of the initial list, adding a line number for each. ...
... (Step 1) List the elements of the initial list, adding a line number for each. ...
INF3170 Logikk Spring 2011 Homework #8 Problems 2–6
... d. Use the definition of [[−]]v to compute [[φ ∨ ¬φ]]v . Conclude that RAA does not follow from the other deduction rules. e. Is this semantics complete? That is, is it the case that Γ I φ ⇒ Γ ` φ for Γ a finite set of formulas? Justify your answer. 8. Do problem 1 on page 60. ? 9. Do problem 4 on ...
... d. Use the definition of [[−]]v to compute [[φ ∨ ¬φ]]v . Conclude that RAA does not follow from the other deduction rules. e. Is this semantics complete? That is, is it the case that Γ I φ ⇒ Γ ` φ for Γ a finite set of formulas? Justify your answer. 8. Do problem 1 on page 60. ? 9. Do problem 4 on ...
Class notes - Nayland Maths
... eg V = 3 πr3 In this formula ‘V’ is the subject (the formula has V=) eg E = ...
... eg V = 3 πr3 In this formula ‘V’ is the subject (the formula has V=) eg E = ...
1 Quantifier Complexity and Bounded Quantifiers
... So far we have used ordinary quantifiers ∀ and ∃. In order to study quantifier complexity, we now introduce bounded versions, defined here: (∀y ≤ t)A(y) ↔ (∀y)(y ≤ t → A(y)) (∃y ≤ t)A(y) ↔ (∃y)(y ≤ t ∧ A(y)) where t is a term not involving y. Define a formula to be ∆0 if all of its quantifiers are bounde ...
... So far we have used ordinary quantifiers ∀ and ∃. In order to study quantifier complexity, we now introduce bounded versions, defined here: (∀y ≤ t)A(y) ↔ (∀y)(y ≤ t → A(y)) (∃y ≤ t)A(y) ↔ (∃y)(y ≤ t ∧ A(y)) where t is a term not involving y. Define a formula to be ∆0 if all of its quantifiers are bounde ...
Full text
... Equating the values in the upper left-hand corners of the two expressions in (4) gives formula (1) when n > 2. A simple check shows that (1) holds in fact, for all n > 0. The above method for ohtainingthe explicit formula (1) is quite general; it can be used to obtain explicit formulae for terms in ...
... Equating the values in the upper left-hand corners of the two expressions in (4) gives formula (1) when n > 2. A simple check shows that (1) holds in fact, for all n > 0. The above method for ohtainingthe explicit formula (1) is quite general; it can be used to obtain explicit formulae for terms in ...