PDF
... concrete system and a VLTL formula, it is possible to reduce the infinite domain to a finite one, and reduce the problem to LTL model checking. The reduction involves an LTL formula that is exponential in the VLTL formula, and we prove that the problem is indeed EXPSPACE-complete. Hardness in EXPSPA ...
... concrete system and a VLTL formula, it is possible to reduce the infinite domain to a finite one, and reduce the problem to LTL model checking. The reduction involves an LTL formula that is exponential in the VLTL formula, and we prove that the problem is indeed EXPSPACE-complete. Hardness in EXPSPA ...
Logic - UNL CSE
... 2 + 2 = 4 if and only if 2 < 2 true: for the same reason above. x2 ≥ 0 if and only if x ≥ 0. false: The converse holds. That is, “if x ≥ 0 then x2 ≥ 0”. However, the implication is false; consider x = −1. Then the hypothesis is true, 12 ≥ 0 but the conclusion fails. ...
... 2 + 2 = 4 if and only if 2 < 2 true: for the same reason above. x2 ≥ 0 if and only if x ≥ 0. false: The converse holds. That is, “if x ≥ 0 then x2 ≥ 0”. However, the implication is false; consider x = −1. Then the hypothesis is true, 12 ≥ 0 but the conclusion fails. ...
Conjunctive normal form - Computer Science and Engineering
... boolean formula expressed in Conjunctive Normal Form, such that the formula is true. The k-SAT problem is the problem of finding a satisfying assignment to a boolean formula expressed in CNF in which each disjunction contains at most k variables. 3-SAT is NP-complete (like any other k-SAT problem wi ...
... boolean formula expressed in Conjunctive Normal Form, such that the formula is true. The k-SAT problem is the problem of finding a satisfying assignment to a boolean formula expressed in CNF in which each disjunction contains at most k variables. 3-SAT is NP-complete (like any other k-SAT problem wi ...
Document
... a-Variable names must begin with a letter b-The rest of name can contain letters, digits and underscore characters c-MATLAB is case-sensitive ...
... a-Variable names must begin with a letter b-The rest of name can contain letters, digits and underscore characters c-MATLAB is case-sensitive ...
1 borrow A -1 - Help-A-Bull
... Binary Multiplication and Division Using Shifting • We can do binary multiplication and division by 2 very easily using an arithmetic shift operation • A left arithmetic shift inserts a 0 in for the rightmost bit and shifts everything else left one bit; in effect, it multiplies by 2 • A right arith ...
... Binary Multiplication and Division Using Shifting • We can do binary multiplication and division by 2 very easily using an arithmetic shift operation • A left arithmetic shift inserts a 0 in for the rightmost bit and shifts everything else left one bit; in effect, it multiplies by 2 • A right arith ...
2010 Exam
... There are three questions in this section, each with a number of parts. Write your answers in the space provided below each part. If you need more space there are extra pages at the end of the examination paper. 1. A shop sells pencils of five colours: red, orange, yellow, green and blue. Mrs Schwarz ...
... There are three questions in this section, each with a number of parts. Write your answers in the space provided below each part. If you need more space there are extra pages at the end of the examination paper. 1. A shop sells pencils of five colours: red, orange, yellow, green and blue. Mrs Schwarz ...
Boolean Functions
... © 2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved ...
... © 2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved ...
Presentation
... 1-1 Variables and Expressions Check It Out! Example 2b Miriam is 5 cm taller than her sister, than her sister who is m centimeters tall. Write an expression for Miriam’s height in centimeters. m represents Miriam’s sister's height in centimeters. m + 5 Think: Miriam's height is 5 added to her ...
... 1-1 Variables and Expressions Check It Out! Example 2b Miriam is 5 cm taller than her sister, than her sister who is m centimeters tall. Write an expression for Miriam’s height in centimeters. m represents Miriam’s sister's height in centimeters. m + 5 Think: Miriam's height is 5 added to her ...
Math 101 Lecture Notes Ch. 2.1 Page 1 of 4 2.1 Simplifying Algebraic
... Suppose we want to simplify 2x + 7x. First, notice that by the distributive property (2 + 7)x = 2x + 7x By reversing the equation above and simplifying 2x + 7x = (2 + 7)x = 9x ...
... Suppose we want to simplify 2x + 7x. First, notice that by the distributive property (2 + 7)x = 2x + 7x By reversing the equation above and simplifying 2x + 7x = (2 + 7)x = 9x ...
Solutions
... There are 18 sums. Consider trying to prove the opposite of what is required, i.e. that there are not more than 2 of each sum. Since there are nine sums lying between 0 and 8, each sum would have to appear twice. (A) A sum of 8 can not appear in a diagonal otherwise it is impossible to obtain sum 0. ...
... There are 18 sums. Consider trying to prove the opposite of what is required, i.e. that there are not more than 2 of each sum. Since there are nine sums lying between 0 and 8, each sum would have to appear twice. (A) A sum of 8 can not appear in a diagonal otherwise it is impossible to obtain sum 0. ...
Consider an ideal J of A and an A-module M . Define the product JM
... Suppose now that A is the internal direct sum of ideals B1 , . . . , Bn regarded as A-modules. Then 1 = e1 + . . . + en for unique ei ∈ Bi . If b ∈ Bi then b = 1 b = (e1 + . . . + en) b = e1 b + . . . + ei b + . . . + en b , so, by uniqueness of linear combinations, ei b = b . ...
... Suppose now that A is the internal direct sum of ideals B1 , . . . , Bn regarded as A-modules. Then 1 = e1 + . . . + en for unique ei ∈ Bi . If b ∈ Bi then b = 1 b = (e1 + . . . + en) b = e1 b + . . . + ei b + . . . + en b , so, by uniqueness of linear combinations, ei b = b . ...
pdf
... the formula φ.” Let’s compute the truth-value of the sentence ∃xP x in the structure M . In other words, let’s compute the set [∃xP x]M . Recall that, in general, [~x.∃vφ(~x, v)]M results from projecting out the last coordinate of [~x, v.φ(~x, v)]M . How do we apply this rule to the special case of ...
... the formula φ.” Let’s compute the truth-value of the sentence ∃xP x in the structure M . In other words, let’s compute the set [∃xP x]M . Recall that, in general, [~x.∃vφ(~x, v)]M results from projecting out the last coordinate of [~x, v.φ(~x, v)]M . How do we apply this rule to the special case of ...
Complexity of Regular Functions
... another restriction: the register update functions are all of the form r ← r0 + c for some register r0 and some semiring element c. Thus assume that the function f is computed by a CRA M of this form. Let M have k registers r1 , . . . , rk . It is straightforward to see that the following functions ...
... another restriction: the register update functions are all of the form r ← r0 + c for some register r0 and some semiring element c. Thus assume that the function f is computed by a CRA M of this form. Let M have k registers r1 , . . . , rk . It is straightforward to see that the following functions ...
doc - Brown CS
... which assigns an interpretation to a given vocabulary V is said to satisfy P if some relation can be assigned to P so that M assigns an interpretation to V + P (V with relation symbol P added) which satisfies . Essentially this is the same as existentially quantifying over a variable in expressi ...
... which assigns an interpretation to a given vocabulary V is said to satisfy P if some relation can be assigned to P so that M assigns an interpretation to V + P (V with relation symbol P added) which satisfies . Essentially this is the same as existentially quantifying over a variable in expressi ...
Slides_04 - Canvas
... question. Based on the answer, they choose the next statement or group of statements to execute. In a robot program, the question asked might be, “Is the robot’s front blocked by a wall?” or “Is there something on this intersection the robot can pick up?” All of these questions have “yes” or “no” an ...
... question. Based on the answer, they choose the next statement or group of statements to execute. In a robot program, the question asked might be, “Is the robot’s front blocked by a wall?” or “Is there something on this intersection the robot can pick up?” All of these questions have “yes” or “no” an ...
3.1.3 Subformulas
... The formula G = ((p ∧ (p → q)) → q) is valid. To prove this, we would have to consider all interpretations. Because there are infinitely many propositional variables in the alphabet of propositional logic, and because an interpretation is uniquely characterized by the assignment of the propositional ...
... The formula G = ((p ∧ (p → q)) → q) is valid. To prove this, we would have to consider all interpretations. Because there are infinitely many propositional variables in the alphabet of propositional logic, and because an interpretation is uniquely characterized by the assignment of the propositional ...
File
... Terms – Are the parts of an expression that are added or subtracted together. For example, the terms of 3x + 10 are 3x and 10. Coefficient – Is the number that is multiplied by a variable in an algebraic expression. For example, the coefficient of 3x is 3. ...
... Terms – Are the parts of an expression that are added or subtracted together. For example, the terms of 3x + 10 are 3x and 10. Coefficient – Is the number that is multiplied by a variable in an algebraic expression. For example, the coefficient of 3x is 3. ...
PS12
... 4. Variables in rules are (consecutively) renamed before applying the unification algorithm. If the unification succeeds an edge in the proof tree is created. 5. A node in the proof tree is a leaf if: the goal list is empty (success), or the goal list is not empty but no rule can be applied to the s ...
... 4. Variables in rules are (consecutively) renamed before applying the unification algorithm. If the unification succeeds an edge in the proof tree is created. 5. A node in the proof tree is a leaf if: the goal list is empty (success), or the goal list is not empty but no rule can be applied to the s ...
EXAMPLE: /* A Program to add N even positive numbers: Sum= 2+4
... int ADD_Even(int, int); // Function's prototype /* notice that the function declaration: int ADD_Even(int M, int SUM); */ /* notice that you write function prototype without parameters if you want */ /* notice that the parameters in calling statement are dummy (as you notice the declaration statemen ...
... int ADD_Even(int, int); // Function's prototype /* notice that the function declaration: int ADD_Even(int M, int SUM); */ /* notice that you write function prototype without parameters if you want */ /* notice that the parameters in calling statement are dummy (as you notice the declaration statemen ...
Chapter 4, Propositional Calculus
... 3.1. Conjunction = = “and.” If both p and q are true, then p q is true, otherwise p q is false. 3.2. Disjunction = = “or.” If both p and q are false, then p q is false, otherwise p q is true. 3.3. Negation = = “not.” If p is true, then p is false. If p is false, then p is true. 4. Pro ...
... 3.1. Conjunction = = “and.” If both p and q are true, then p q is true, otherwise p q is false. 3.2. Disjunction = = “or.” If both p and q are false, then p q is false, otherwise p q is true. 3.3. Negation = = “not.” If p is true, then p is false. If p is false, then p is true. 4. Pro ...
Propositional Logic: Normal Forms
... SAT: if false is NOT marked, let ν be an interpretation that assign true to all marked atoms, and false to the others. If φ is not true under ν, it means that there exists a conjunct P1 ∧ . . . ∧ Pki → P 0 of φ that is false. By the semantics, this can only mean that P1 ∧ . . . ∧ Pki is true but P 0 ...
... SAT: if false is NOT marked, let ν be an interpretation that assign true to all marked atoms, and false to the others. If φ is not true under ν, it means that there exists a conjunct P1 ∧ . . . ∧ Pki → P 0 of φ that is false. By the semantics, this can only mean that P1 ∧ . . . ∧ Pki is true but P 0 ...
Discrete Mathematics Chapter 1 The Foundations: Logic and Proof
... • Quantifiers provide a notation that allows us to quantify (count) how many objects in the univ. of disc. satisfy a given predicate. • “∀” is the FOR∀LL or universal quantifier. ∀x P(x) means for all x in the u.d., P holds. • “∃” is the ∃XISTS or existential quantifier. ∃x P(x) means there exists a ...
... • Quantifiers provide a notation that allows us to quantify (count) how many objects in the univ. of disc. satisfy a given predicate. • “∀” is the FOR∀LL or universal quantifier. ∀x P(x) means for all x in the u.d., P holds. • “∃” is the ∃XISTS or existential quantifier. ∃x P(x) means there exists a ...
