In this example, we should use a truth table to verify each propositional equivalence. Because in this case the exercises only involve the use of conjunctions (∧) and disjunctions (∧), we start by having at hand (or in memory) the truth tables for those logical operators. Truth tables for conjunction and disjunction a) p∧T≡p In […]
Use truth tables to verify these equivalences Read More »
As usual, we should start with the appropriate definitions. Remember that in mathematics, we must know the definitions before we can start trying to solve a problem. Definitions Let p and q be propositions. The disjunction of p and q, denoted by p ∨ q, is the proposition “p or q.” The disjunction p ∨
Solve the following exercise. 11. Let p and q be the propositions p: It is below freezing. q: It is snowing. Write these propositions using p and q and logical connectives (including negations). a) It is below freezing and snowing.b) It is below freezing but not snowing.c) It is not below freezing and it is
Suppose that during the most recent fiscal year, the annual revenue of Acme Computer was 138 billion dollars and its net profit was 8 billion dollars, the annual revenue of Nadir Software was 87 billion dollars and its net profit was 5 billion dollars, and the annual revenue of Quixote Media was 111 billion dollars
Propositional Logic: Exercise 7 from the textbook Read More »
In this post, I’ll show some examples of how to negate propositions. As I always recommend, let’s start with the definitions. Definitions Let p be a proposition. The negation of p, denoted by ¬p, is the statement “It is not the case that p.” The proposition ¬p is read “not p.” The truth value of
What is the negation of each of these propositions? Read More »
Propositional logic is a very important topic in Discrete Mathematics. It is part of the foundations every student should know. In this post, I’ll show how I solve this specific type of exercise. In mathematics, definitions are very important. As I always recommend to my students, when you are starting a new topic, and you
We will follow a similar approach to previous exercises to solve this exercise. Let’s define 5 states as follows: S0, the initial state S1 is the state when the automata read exactly one 0 S2 is the state when the automata read exactly two 0s S3 is the state when the automata read exactly three
We will follow a similar approach to previous exercises to solve this exercise. Let’s define 4 states as follows: S0, the initial state S1 is the state when the automata read a 0 S2 is the state when the automata read a sequence of two 0s S3 is the state when the automata read a
We will follow a similar approach to previous exercises to solve this exercise. Let’s define 4 states as follows: S0, the initial state S1 is the state when the automata read a 1 S2 is the state when the automata read a 0 after a 1 S3 is the state when the automata read a
This exercise is a bit more difficult than the previous one. To solve this exercise, let’s use the following strategy: define states in which the automaton can be if it just reads a 1 or a 0. – We can have a state S1 which means the automaton read 0s. Therefore, from the state S0
