In how many ways can a string of length 5 be formed using the letters ABCDEFG without repetitions. It should begin with the letter F and end with the letter A.
Let A= {1, 2, 3, 4}, and let R And S be the relations on A As follow: R = {(1,1),(1, 4),(3,2),(4,2),(4,3)} S = { (1,1),(1,2),(2,2),(3,3),(4,2),(4,4)) Then find out: 1. M𝑅𝑐 2. 𝑀𝑠̅ 3. M (𝑅∪ S)
Find these values ⌊−3.1⌋
b) Given
p="Front door is open
q="Back door is open"
r="Light is on"
s="The lift doors are open"
t="The lift is moving",
i)
write the following sentence in logic symbols: "When the front and back
doors are closed then the light is off"
write the following in words: SV (t As)
ii)
Find the simplest form of given Boolean expressions using algebraic methods.
i. A(A+B) + B(B+C) + C(C+A)
ii. (A+|B|)(B+C) + (A+B)(C+|A|)
iii. (A+B)(AC+A|C|)_+AB +B
iv. |A|(A+B) + (B+A)(A+|B|)
A says “If B is a knight, then I am a knave”, B says nothing.
(a) Find the inverse of 19 modulo 141, using the Extended Euclidean Algorithm.
Show your steps.
Translate these system specifications into English where the predicate
S(x, y) is “x is in state y” and where the domain for x and y consists of all systems
and all possible states, respectively.
(a) ∃S(x, open)
(b) ∀x(S(x, malfunctioning) ∨ S(x, diagnostic))
(c) ∃xS(x, open) ∨ ∃xS(x, diagnostic)
1
For each of the given statements:
1 - Express each of the statements using quantifiers and propositional functions.
2 - Form the negation of the statement so that no negation is to the left of the quantifier.
3 - Express the negation in simple English. (Do not simply use the words “it is not
the case that...”).
(a) Some drivers do not obey the speed limit.
(b) All Swedish movies are serious.
(c) No one can keep a secret.
(d) No monkey can speak French.
(e) There is someone in the class who does not have a good attitude.
For each of the following compound propositions give its truth table and derive an
equivalent compound proposition in disjunctive normal formal (DNF) and in conjunctive normal form (CNF).
(a) (p → q) → r
(b) (p ∧ ¬q) ∨ (p ↔ r)