Contradiction in Discrete Mathematics: Liars and De Morgan's Laws, Exercises of Mathematics

Two problems in discrete mathematics. The first problem deals with a group of people, one of whom is a knave who always lies. That a contradiction arises when we know one person is a knave but all people claim they are not. The second problem involves verifying de morgan's laws using a truth table. The document concludes that the first de morgan law has been verified.

Typology: Exercises

2019/2020

Uploaded on 03/31/2020

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Discrete Mathematics
1.2.30
We know one of these people is a knave, who always lies, so a knave
never says “I am not the spy” because it’s the truth. However, all the
people say the same words, which means there isn’t any knave. But it’s a
contradiction. Therefore, there isn’t any possible solution.
1.3.6
p q pq¬p ¬q ¬(pq) ¬p¬q
T T T F F F F
T F F F T T T
F T F T F T T
F F F T T T T
We have verified the first De Morgan law.

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Discrete Mathematics 1.2. We know one of these people is a knave, who always lies, so a knave never says “I am not the spy” because it’s the truth. However, all the people say the same words, which means there isn’t any knave. But it’s a contradiction. Therefore, there isn’t any possible solution. 1.3. p q p∧q ¬p ¬q ¬(p∧q) ¬p∨¬q T T T F F F F T F F F T T T F T F T F T T F F F T T T T We have verified the first De Morgan law.