Now look at all the rows for which both \(P \imp Q\) and \(\neg P \imp Q\) are true. Simplify the statements below (so negation appears only directly next to predicates). So if \(P\imp Q\) and \(P\) are both true, we see that \(Q\) must be true as well. A proposition is simply a statement. In fact, it is equally true that “If the moon is made of cheese, then Elvis is still alive, or if Elvis is still alive, then unicorns have 5 legs.”, You might have noticed in Example 3.1.1 that the final column in the truth table for \(\neg P \vee Q\) is identical to the final column in the truth table for \(P \imp Q\text{:}\). \neg(\neg P \vee Q)\text{.} Don't just say, “it is false that …”. }\) The first is saying we can find one \(y\) that works for every \(x\text{. \neg \exists x \forall y (x \le y) Here they are: The truth table for negation looks like this: None of these truth tables should come as a surprise; they are all just restating the definitions of the connectives. Negation/ NOT (¬) 4. You are looking for a row in which \(P\) is true, and the whole statement is true. Note that this statement is not \(\neg(P \vee Q)\text{,}\) the negation belongs to \(P\) alone. To verify that two statements are logically equivalent, you can use truth tables or a sequence of logically equivalent replacements. Rather, we end with a two examples of logical equivalence and deduction, to pique your interest. \(\neg \exists x \forall y (\neg O(x) \vee E(y))\text{. There is a sequence that is both arithmetic and geometric. However, predicate logic allows us to analyze statements at a higher resolution, digging down into the individual propositions \(P\text{,}\) \(Q\text{,}\) etc. Hint: you should get three T's and one F. It's your birthday, but the cake is a lie. Suppose further that, is a valid deduction rule. Edith ate her vegetables. Propositional logic studies the ways statements can interact with each other. \neg \neg P \text{ is logically equivalent to } P\text{.} }\), \(\neg((P \imp \neg Q) \vee \neg (R \wedge \neg R))\text{. So—yeah, it gets kind of messy. That is, \(P\) and \(Q\) have the same truth value under any assignment of truth values to their atomic parts. \newcommand{\vl}[1]{\vtx{left}{#1}} Now let's answer our question about monopoly: Analyze the statement, “if you get more doubles than any other player you will lose, or that if you lose you must have bought the most properties,” using truth tables. To see this, we should provide an interpretation of the predicate \(P(x,y)\) which makes one of the statements true and the other false. Here we go through the first lecture of our curriculum, talking about Propositional Logic. We will answer this question, and won't need to know anything about Monopoly. Master Discrete Mathematics: Sets, Math Logic, and More. \exists y \forall x P(x,y) \imp \forall x \exists y P(x,y) It also happens that \(R\) is true in these rows as well. But this can be easily dealt with: Example: “It is not the case that \(c\) is not odd” means “\(c\) is odd.”. Prove that the statements \(\neg(P \imp Q)\) and \(P\wedge \neg Q\) are logically equivalent without using truth tables. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. So our statement about monopoly is true (regardless of how many properties you own, how many doubles you roll, or whether you win or lose). Yesterday, Holmes wore a bow tie. Here is the full truth table: The first three columns are simply a systematic listing of all possible combinations of T and F for the three statements (do you see how you would list the 16 possible combinations for four statements?). \neg\neg P \wedge \neg Q\text{.} Justify your answer by writing all of Tommy's statements using sentence variables (\(P, Q, R, S, T\)), taking their negations, and using these to deduce what Tommy actually ate. Could both trolls be knights? Are you convinced that it is a valid deduction rule? }\), \(\neg \forall x \neg \forall y \neg(x \lt y \wedge \exists z (x \lt z \vee y \lt z))\text{. We saw this before, in Section 0.2, but it is so important and useful, it warants a second blue box here: The negation of an implication is a conjuction: That is, the only way for an implication to be false is for the hypothesis to be true AND the conclusion to be false. We have a similar rule for distributing over conjunctions (“and”s): This suggests there might be a sort of “algebra” you could apply to statements (okay, there is: it is called Boolean algebra) to transform one statement into another. What, if anything, can the waiter conclude about the ingredients in Geoff's desired calzone? Simplify the following statements (so that negation only appears right before variables). \newcommand{\twoline}[2]{\begin{pmatrix}#1 \\ #2 \end{pmatrix}} Propositional Logic – Wikipedia Principle of Explosion – Wikipedia Discrete Mathematics and its Applications, by Kenneth H Rosen. \newcommand{\N}{\mathbb N} \newcommand{\st}{:} Are the statements, “it will not rain or snow” and “it will not rain and it will not snow” logically equivalent? What else did he wear? Propositional Logic – Wikipedia Principle of Explosion – Wikipedia Discrete Mathematics and its Applications, by Kenneth H Rosen. Suppose \(P\) and \(Q\) are (possibly molecular) propositional statements. The technical term for these is predicates and when we study them in logic, we need to use predicate logic. Oh, and if I have pepperoni or quail then I must also have ricotta cheese.”. Using the definitions of the connectives in Section 0.2, we see that for this to be true, either \(P \imp Q\) must be true or \(Q \imp R\) must be true (or both). \((P \wedge Q) \wedge (R \wedge \neg R)\text{. Let's look at the form of the statements. But I didn't drink soda or tea.” Of course you know that Tommy is the worlds worst liar, and everything he says is false. For all numbers \(n\text{,}\) if \(n\) is prime, then \(n+3\) is not prime. Here all three premises of the argument are true, but the conclusion is false. Let's try another one. Then translate this back into English. And lo-and-behold, in this one case, \(Q\) is also true. }\) The second allows different \(y\)'s to work for different \(x\)'s, but there is nothing preventing us from using the same \(y\) that work for every \(x\text{. Consider the statement, “If a number is triangular or square, then it is not prime”, Make a truth table for the statement \((T \vee S) \imp \neg P\text{.}\). }\) This is necessarily false, so it is also equivalent to \(P \wedge \neg P\text{.}\). Like above, only now you will need 8 rows instead of just 4. They are both implications: statements of the form, \(P \imp Q\text{.}\). \newcommand{\gt}{>} After simplifying, you should get \(\forall x(\neg E(x) \wedge \neg O(x))\text{,}\) for the first one, for example. The applications of propositional logic today in computer science is countless. }\) Can you chain more implications together? It is false that if Sam is not a man then Chris is a woman, and that Chris is not a woman. So instead, let's make a truth table: Look at the fourth (or sixth) row. }\) Then you are back in the case in part (a) again. \newcommand{\vtx}[2]{node[fill,circle,inner sep=0pt, minimum size=4pt,label=#1:#2]{}} Recall that all trolls are either always-truth-telling knights or always-lying knaves. Let's see how we can apply the equivalences we have encountered so far. De Morgan's laws do not do not directly help us with implications, but as we saw above, every implication can be written as a disjunction: Example: “If a number is a multiple of 4, then it is even” is equivalent to, “a number is not a multiple of 4 or (else) it is even.”. If you believed the statement was false, what properties would a counterexample need to possess? What can you conclude? We do this for every possible combination of T's and F's. Tautologies are always true but they don't tell us much about the world. To verify that two statements are logically equivalent, you can make a truth table for each and check whether the columns for the two statements are identical. \), \begin{equation*} }\) What can you conclude about \(P\) and \(Q\) if you know the statement is true? \end{equation*}, \begin{equation*} \renewcommand{\iff}{\leftrightarrow} Simplify the statements below to the point that negation symbols occur only directly next to predicates. A proposition is simply a statement. Let's find out: Prove that the following is a valid deduction rule: Prove that the following is a valid deduction rule for any \(n \ge 2\text{:}\). OR (∨) 2. Since the truth value of a statement is completely determined by the truth values of its parts and how they are connected, all you really need to know is the truth tables for each of the logical connectives. \end{equation*}, \begin{equation*} \newcommand{\Imp}{\Rightarrow} Mean in terms of truth tables or a sequence that is, use a truth method... 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