As in previous cases, we start by expressing the conjecture as a conditional statement. If m is irrational and n is rational, then m+n is irrational. In this case, we must use proof by contradiction. Remember that a contradiction is a compound proposition that is always false. To give a proof by contradiction of a […]
To use this method, we must prove that a theorem, stated as a conditional statement p-> q is true. In this case, we can restate the conjecture as follows: If m and n are even integers, then m+n is even. If we assume that p is true (m and n are evens), then we have
Use a direct proof to show that the sum of two even integers is even Read More »
Direct proof is one the easiest method to construct proofs. To use this method, we must prove that a theorem, stated as a conditional statement p-> q is true. To this, we assume that p is true, and we apply rules of inference, axioms, definitions and previously proved theorems to prove that q is also
Use a direct proof to show that the sum of two odd integers is even Read More »
As with the previous examples, there are two main things we need to solve this exercise. First, we need to know the rules of inference so we can apply them. Second, we need to identify whether we must use propositional variables or predicates. The first step will always be to translate the argument using propositional
As in the previous examples, first, we need to know the rules of inferences. In this example, you need to realize that you must use predicates instead of propositional variables. Because of that, we also need the rules of inference for quantified statements. Let’s use predicates to represent the argument, then we will apply the
To solve exercises where you have to apply rules of inference, the first step is to know the rules of inference. Once we know the rules, we can then start to apply them. So, let’s start solving the exercise. “Randy works hard,” “If Randy works hard, then he is a dull boy,” and “If Randy
To answer this question, we should substitute the values of x in the predicate. Once we do that, the predicate is transformed into a proposition, and we can state the truth value. So, let’s start answering the questions. a) P(0) P(x): “x=x2” P(0): “0=02=0” Answer: true. b) P(1) P(x): “x=x2” P(1): “1=12=1” Answer: true. c)
To prove the first De Morgan law, we need to use the truth table for conjunctions, disjunctions, and negation of propositions. Truth tables for conjunction, disjunctions and negation Now we can start solving the exercise. First De Morgan Law The following propositional equivalence is the First De Morgan Law. ¬(p ∧ q) ≡ ¬p ∨
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To prove the associative laws, we need to use the truth table for conjunctions and disjunctions. Truth tables for conjunction and disjunctions Now we can start solving the exercise. a) (p∨q)∨r ≡p∨(q∨r) As in previous examples, let’s examine what will be the structure of the truth table we must create. There are three propositional variables,
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As usual, to solve this type of exercise we should look at the logical operators. In this case, we only have negation (¬), so let’s start by having the truth table for that specific operator. Truth table for the negation (¬) of a proposition Solution to the exercise If we analyze what we have to
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