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Rules of Logic

Investigation Holmes always wears one of the two vests he owns: one tweed and one mint green. He always wears either the green vest or red shoes.

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Investigation

Holmes always wears one of the two vests he owns: one tweed and one mint green. He always wears either the green vest or red shoes. Whenever he wears a purple shirt and the green vest, he chooses to not wear a bow tie. He never wears the green vest unless he is also wearing either a purple shirt or red shoes. Whenever he wears red shoes, he also wears a purple shirt. Today, Holmes wore a bow tie. What else did he wear?

Logic studies the ways statements can interact with each other. More precisely, we consider the way the logical form statements can interact. The study of logic does not care about the content of the atomic statements or the meaning of predicates. For example, the claims, If spiders have six legs, then Sam walks with a limp, and, If the moon is made of cheese, then cheddar is a type of cheese, are identical from a logical perspective. Logic doesn't care about whether Sam is a spider or the culinary makeup of the moon. Both statements have the same form: They are implications, \(P \imp Q\).

Of course, in mathematics we often do know some relationship between various atomic statements. For example, we know a relationship between being even and being a multiple of 10. That relationship allows us to make claims such as, If the number I'm thinking of is a multiple of 10, then it is even. Suppose I also told you that I am now thinking of a number that is not even. We can deduce that I am not thinking of a multiple of 10! Crucially, if we accept the truth of the statements here, we can make this deduction without thinking about the nature of numbers. It can feel very liberating and provide much-needed clarity when trying to understand complicated reasoning if we can separate the content from the logical form of arguments.

Our goal in this section is to establish some procedures for analyzing how the truth or falsity of statements interact, based on their logical form. We will see that some molecular statements must be true regardless of whether their atomic parts are true or false, while some statements must always be false. For other statements, it can be that two statements are always true or false together, or that whenever one statement is true, another statement must also be true.

The main method for establishing these relationships will be truth tables. There is a very clear procedure for constructing and analyzing truth tables, but for complicated arguments that contain many atomic statements, the truth tables become very large and unwieldy. We will therefore use truth tables to understand some basic equivalences and deductions that can be applied in a sequence of reasoning to construct larger arguments.

Truth Tables

Here's a question about playing Monopoly:

If you get more doubles than any other player, then you will lose, or if you lose, then you must have bought the most properties.

True or false? We will answer this question and won't need to know anything about Monopoly. Instead, we will look at the logical form of the statement.

We need to decide when the statement \((P \imp Q) \vee (Q \imp R)\) is true. Using the definitions of the connectives in , we see that for this to be true, either \(P \imp Q\) must be true or \(Q \imp R\) must be true (or both). Those are true if either \(P\) is false or \(Q\) is true (in the first case) and \(Q\) is false or \(R\) is true (in the second case). Soyeah, it gets a bit messy. Luckily, we can make a chart to keep track of all the possibilities with a truth table.

The idea is this: On each row, we list a possible combination of T's and F's (Trues and Falses) for each of the propositional variables, and then mark down whether the (molecular) statement in question is true or false in that case. We do this for every possible combination of T's and F's. Then we can clearly see the cases in which the statement is true or false. For complicated statements, we will first fill in values for each part of the statement, as a way of breaking up our task into smaller, more manageable pieces.

Since the truth value of a statement is completely determined by the truth values of its parts and how they are connected, all you need to know is the truth tables for each of the logical connectives, which we have already seen in

The truth tables we consider here all build off the basic ones, applying the basic rules multiple times.

Example

Make a truth table for the statement \(\neg P \vee Q\).

Solution

Note that this statement is not \(\neg(P \vee Q)\); the negation belongs to \(P\) alone. The main connective here is the \(\vee\), which means we will use that truth table last. First, we apply the truth table for \(\neg\), and then apply the truth table for \(\vee\) using inputs from \(\neg P\) and \(Q\).

Since there are two variables, there are four possible combinations of T's and F's. Putting this all together gives us the following truth table.

\(P\) \(Q\) \(\neg P\) \(\neg P \vee Q\)
TTFT
TFFF
FTTT
FFTT

We added a column for \(\neg P\) to make filling out the last column easier. The entries in the \(\neg P\) column were determined by the entries in the \(P\) column. Then to fill in the final column, look only at the column for \(Q\) and the column for \(\neg P\) and use the rule for \(\vee\).

Now let's answer our question about Monopoly.

Condensed — the full section is in Levin, Discrete Mathematics: An Open Introduction.

Logical Equivalence

You might have noticed in that the final column in the truth table for \(\neg P \vee Q\) is identical to the final column in the truth table for \(P \imp Q\):

\(P\) \(Q\) \(P \imp Q\) \(\neg P \vee Q\)
TTTT
T FFF
FTTT
F FTT

This says that no matter what \(P\) and \(Q\) are, the statements \(\neg P \vee Q\) and \(P \imp Q\) are either both true or both false. We therefore say these statements are logically equivalent.

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.

In we claimed that whenever an implication is true, so is its contrapositive. We can now make this claim as the following theorem.

Recognizing two statements as logically equivalent can be quite helpful. Rephrasing a mathematical statement can often lend insight into what it is saying, or how to prove or refute it. By using truth tables we can systematically verify that two statements are indeed logically equivalent.

Example

Are the statements, It will not rain or snow, and, It will not rain and it will not snow, logically equivalent?

Solution

We want to know whether \(\neg(P \vee Q)\) is logically equivalent to \(\neg P \wedge \neg Q\). Make a truth table which includes both statements:

\(P\) \(Q\) \(\neg(P \vee Q)\) \(\neg P \wedge \neg Q\)
TTF F
TFFF
F TFF
FFTT

Since the truth values for the two statements are equal in every row, the two statements are logically equivalent.

Notice that this example gives us a way to distribute a negation over a disjunction (an or). We have a similar rule for distributing over conjunctions (ands):

Condensed — the full section is in Levin, Discrete Mathematics: An Open Introduction.

Equivalence for Quantified Statements

All the examples we have looked at so far have only involved propositional logic, where the basic units of logic are statements that are either true or false. It is also possible to say that two statements involving quantifiers and predicates are logically equivalent.

Sometimes the quantifiers have nothing to do with the equivalence. For example, \[\forall x (P(x) \imp Q(x)) \equiv \forall x (\neg P(x) \vee Q(x))\]. As soon as we replace the \(x\) with a constant, we are left with two statements that are logically equivalent based on their propositional form.

Other times, the more interesting times, it is exactly the logic of the quantifiers that makes the statements logically equivalent. What is especially interesting here is that we cannot use truth tables to verify these equivalences!

Instead, we need to reason about the domain of discourse as a set. For example, let's consider how negation interacts with quantifiers.

Consider the claim that all odd numbers are prime. We might represent this symbolically as \(\forall x (O(x) \imp P(x))\). The statement clearly is not true, so what is true is that not all odd numbers are prime (i.e., \(\neg\forall x(O(x) \imp P(x))\)). How do we know? Easy: 9. Yes, 9 is odd but not prime. But is it enough that just one odd number isn't prime?

To dispute a universal claim, you just need one single counterexample. You just need to show there exists a number for which the claim is false. In our case, we have the equivalence, \[\neg\forall x (O(x) \imp P(x)) \equiv \exists x (O(x) \wedge \neg P(x))\]. If we ignore the quantifiers for a minute, we are left with \[\neg(O \imp P) \equiv O \wedge \neg P\] which is exactly an example of . The new, interesting part is that when we negated the universal quantifier, we got an existential quantifier.

Negating an existential quantifier results in a universal quantifier. This makes sense. If there does not exist something with a property, then everything does not have that property.

\(\neg \forall x P(x)\) is equivalent to \(\exists x \neg P(x)\).

\(\neg \exists x P(x)\) is equivalent to \(\forall x \neg P(x)\).

Symbolically, we can pass the negation symbol over a quantifier, but that causes the quantifier to switch type.

Condensed — the full section is in Levin, Discrete Mathematics: An Open Introduction.

Deductions

Earlier, we claimed that the following was a valid argument:

If Edith eats her vegetables, then she can have a cookie. Edith ate her vegetables. Therefore Edith gets a cookie.

How do we know this is valid? Let's look at the form of the statements. Let \(P\) denote, Edith eats her vegetables and \(Q\) denote, Edith can have a cookie. The logical form of the argument is then:

\(P \imp Q\)
\(P\)
\(\therefore\) \(Q\)

This is an example of a deduction rule, an argument form that is always valid. This one is a particularly famous rule called modus ponens. Are you convinced that it is a valid deduction rule? If not, consider the following truth table:

\(P\) \(Q\) \(P\imp Q\)
TTT
TFF
FT T
FFT

This is just the truth table for \(P \imp Q\), but what matters here is that all the lines in the deduction rule have their own column in the truth table. Remember that an argument is valid provided the conclusion must be true given that the premises are true. The premises in this case are \(P \imp Q\) and \(P\). Which rows of the truth table correspond to both of these being true? \(P\) is true in the first two rows, and of those, only the first row has \(P \imp Q\) true as well. And lo-and-behold, in this one case, \(Q\) is also true. So if \(P\imp Q\) and \(P\) are both true, we see that \(Q\) must be true as well.

Think of deduction rules as a sort of one-way form of logical equivalence. Two statements are logically equivalent provided that in every row of the truth table in which the first statement is true, so is the second, and in every row in which the second statement is true, so is the first. A deduction only requires the first of these two parts.

Here are a few more examples.

Example

Show that the following is a valid deduction rule.

\(P \imp Q\)
\(\neg P \imp Q\)
\(\therefore\) \(Q\)
Solution

We make a truth table which contains all the lines of the argument form:

\(P\) \(Q\) \(P\imp Q\) \(\neg P\) \(\neg P \imp Q\)
TTTFT
TFFF T
FTTTT
FFTTF

(we include a column for \(\neg P\) just as a helping step to get the column for \(\neg P \imp Q\)).

Now look at all the rows for which both \(P \imp Q\) and \(\neg P \imp Q\) are true. This happens only in rows 1 and 3. Hey! In those rows \(Q\) is true as well, so the argument form is valid (it is a valid deduction rule).

Condensed — the full section is in Levin, Discrete Mathematics: An Open Introduction.

Practice (33)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Spend a few minutes thinking about the Investigate! question above. Of the six statements in the puzzle, only one is atomic. Use this atomic statement and one other statement to deduce a new statement about what Holmes might (or might not) be wearing. Explain why you think your new statement is true.

    జవాబు వెల్లడి చేయండి

    Hint:

    The atomic statement is, Holmes wore a bow tie. Only one of the molecular statements has this as one of its atoms.

  2. Consider the statement, Whenever Holmes wears a purple shirt and the green vest, he chooses to not wear a bow tie. Let \(P\) be the statement, Holmes wears a purple shirt, \(G\) be the statement, Holmes wears the green vest, and \(B\) be the statement, Holmes wears a bow tie. Which of the following is the best translation of the statement into propositional logic?

  3. Consider the statement, Holmes never wears the green vest unless he is also wearing either a purple shirt or red shoes. With \(P\) and \(G\) as in the previous question, and \(R\) being the statement, Holmes wears red shoes, which of the following is the best translation of the statement into propositional logic?

  4. Consider the statement, If you major in math, then you will get a high-paying job, and the statement, Either you don't major in math, or you will get a high-paying job. In which of the following cases are both statements true? Select all that apply.

  5. To check whether two statements are logically equivalent, you can use a truth table. Explain what you would look for in the truth table to conclude that the two statements are logically equivalent. What would tell you they are not logically equivalent?

  6. To check whether a deduction rule is valid, you can use a truth table. Explain what you would look for in the completed truth table to say that the deduction rule is valid, and what would tell you the deduction rule is not valid.

  7. What questions do you have after reading this section? Write at least one question about the content of this section that you are curious about.

  8. Make a truth table for the statement \((P \wedge Q) \rightarrow (P \vee Q)\).

    జవాబు వెల్లడి చేయండి

    If you think about what this statement is saying, it makes sense that it is a tautology (that it is true in every case). The complete truth table is:

    \(P\)\(Q\)\(P \wedge Q\)\(P \vee Q\)\((P \wedge Q) \rightarrow (P \vee Q))\)
    TTTTT
    TFFTT
    FTFTT
    FFFFT

  9. Complete a truth table for the statement \(\neg Q \vee (Q \rightarrow P))\). What can you conclude about \(P\) and \(Q\) if you knew the statement above was false?

    జవాబు వెల్లడి చేయండి

    The truth table is:

    \(P\)\(Q\)\(\neg Q\)\(Q \rightarrow P\)\(\neg Q \vee (Q \rightarrow P))\)
    TTFTT
    TFTTT
    FTFFF
    FFTTT

    If this statement is false, we must be in the third row, making \(P\) false and \(Q\) true.

  10. Construct a truth table for the statement \(Q \rightarrow (\neg P \vee R)\).

    జవాబు వెల్లడి చేయండి

    The complete truth table is:

    \(P\)\(Q\)\(R\)\(\neg P\)\(\neg P \vee R\)\(Q \rightarrow (\neg P \vee R)\)
    TTTFTT
    TTFFFF
    TFTFTT
    TFFFFT
    FTTTTT
    FTFTTT
    FFTTTT
    FFFTTT

  11. Determine whether the statements \(P \rightarrow (Q \vee R)\) and \((P \rightarrow Q) \vee (P\rightarrow R)\) are logically equivalent by completing a truth table for both statements.

    జవాబు వెల్లడి చేయండి

    The complete truth table is:

    \(P\)\(Q\)\(R\)\(P \rightarrow (Q \vee R)\)\((P \rightarrow Q) \vee (P\rightarrow R)\)
    TTTTT
    TTFTT
    TFTTT
    TFFFF
    FTTTT
    FTFTT
    FFTTT
    FFFTT

    Since the two columns are identical, the statements are logically equivalent.

  12. Determine if the following is a valid deduction rule:

    \(P \rightarrow Q\)
    \(\neg Q\)
    \(\therefore\)\(\neg P\)
    జవాబు వెల్లడి చేయండి

    The complete truth table is:

    \(P\)\(Q\)\(P \rightarrow Q\)\(\neg Q\)\(\neg P\)
    TTTFF
    TFFTF
    FTTFT
    FFTTT

    There is only one row in which both premises are true (row 4). In this row, the conclusion is also true. Thus the deduction rule is valid.

  13. Determine if the following is a valid deduction rule:

    \(P \rightarrow (Q \vee R)\)
    \(\neg(P \rightarrow Q)\)
    \(\therefore\)\(R\)
    జవాబు వెల్లడి చేయండి

    The complete truth table is:

    \(P\)\(Q\)\(R\)\(P \rightarrow (Q \vee R)\)\(\neg(P \rightarrow Q)\)
    TTTTF
    TTFTF
    TFTTT
    TFFFT
    FTTTF
    FTFTF
    FFTTF
    FFFTF

    There is only one row in which both premises are true (row 3). In this row, the conclusion is also true, so the deduction rule is valid.

  14. Determine if the following is a valid deduction rule:

    \((P \wedge Q) \rightarrow R\)
    \(\neg P \vee \neg Q\)
    \(\therefore\)\(\neg R\)
    జవాబు వెల్లడి చేయండి

    The complete truth table is:

    \(P\)\(Q\)\(R\)\((P \wedge Q) \rightarrow R\)\(\neg P \vee \neg Q\)\(\neg R\)
    TTTTFF
    TTFFFT
    TFTTTF
    TFFTTT
    FTTTTF
    FTFTTT
    FFTTTF
    FFFTTT

    In rows 3, 5 and 7 both of the premises are true, but the conclusion is false. Thus the deduction rule is not valid.

  15. Determine if the following is a valid deduction rule:

    \(P \rightarrow Q\)
    \(P \wedge \neg Q\)
    \(\therefore\)\(R\)
    జవాబు వెల్లడి చేయండి

    The complete truth table is:

    \(P\)\(Q\)\(R\)\(P \rightarrow Q\)\(P \wedge \neg Q\)
    TTTTF
    TTFTF
    TFTFT
    TFFFT
    FTTTF
    FTFTF
    FFTTF
    FFFTF

    There is no row in which both premises are true (indeed, these are contradictory premises; the second is the negation of the first). Thus every row in which both premises are true (i.e., no row), the conclusion is also true. Therefore the deduction rule is valid. (This is an example of how everything follows from a contradiction.)

  16. Which of the following statements is a law of logic? That is, which of the following are true no matter what your domain of discourse is and no matter what you interpret the predicates as meaning? Select all that apply.

  17. You stumble upon two trolls playing Stratego. They tell you:

    Troll 1: If we are cousins, then we are both knaves.

    Troll 2: We are cousins, or we are both knaves.

    Could both trolls be knights? Recall that all trolls are either always-truth-telling knights or always-lying knaves. Explain your answer and how you can use truth tables to find it.

    జవాబు వెల్లడి చేయండి

    Hint:

    You could probably reason through the cases by hand, but try making a truth table. Use two statements, \(P\) being we are cousins and \(Q\) being we are both knaves.

  18. Next you come upon three trolls, helpfully wearing name tags. They say:

    • Pat

      If either Quinn or I are knights, then so is Ryan.

    • Quinn

      Ryan is a knight, and if Pat is a knight, then so am I.

    • Ryan

      Quinn is a knave, but Pat and I share the same persuasion.

    Create a truth table that includes all three statements. Then use the truth table to determine the persuasion of each troll.

  19. Consider the statement about a party, If it's your birthday or there will be cake, then there will be cake.

    1. Translate the above statement into symbols. Clearly state which statement is \(P\) and which is \(Q\).

    2. Make a truth table for the statement.

    3. Assuming the statement is true, what (if anything) can you conclude if you know there will be cake?

    4. Assuming the statement is true, what (if anything) can you conclude if you know there will not be cake?

    5. Suppose you found out that the statement was a lie. What can you conclude?

    జవాబు వెల్లడి చేయండి

    1. \(P\)\(Q\)\((P \vee Q) \imp Q\)

    2. Hint: You should get three T's and one F.

    3. Only that there will be cake.

    4. It's NOT your birthday!

    5. It's your birthday, but the cake is a lie.

  20. Geoff Poshingten is out at a fancy pizza joint and decides to order a calzone. When the waiter asks what he would like in it, he replies, I want either pepperoni or sausage. Also, if I have sausage, then I must also include quail. Oh, and if I have pepperoni or quail, then I must also have ricotta cheese.

    1. Translate Geoff's order into logical symbols.

    2. The waiter knows that Geoff is either a liar or a truth-teller (so either everything he says is false, or everything is true). Which is it?

    3. What, if anything, can the waiter conclude about the ingredients in Geoff's desired calzone?

    జవాబు వెల్లడి చేయండి

    Hint:

    You should write down three statements using the symbols \(P, Q, R, S\). If Geoff is a truth-teller, then all three statements would be true. If he was a liar, then all three statements would be false. But in either case, we don't yet know whether the four atomic statements are true or false, since he hasn't said them by themselves.

    A truth table might help, although it is probably not entirely necessary.

  21. Determine whether the following two statements are logically equivalent: \(\neg(P \imp Q)\) and \(P \wedge \neg Q\). Explain how you know you are correct.

    జవాబు వెల్లడి చేయండి

    Make a truth table for each and compare. The statements are logically equivalent.

  22. Simplify the following statements (so that negation only appears right before variables).

    1. \(\neg(P \imp \neg Q)\)
    2. \((\neg P \vee \neg Q) \imp \neg (\neg Q \wedge R)\)
    3. \(\neg((P \imp \neg Q) \vee \neg (R \wedge \neg R))\)

    4. It is false that if Sam is not a man then Chris is a woman, and that Chris is not a woman.

    జవాబు వెల్లడి చేయండి

    1. \(P \wedge Q\)
    2. \((\neg P \vee \neg R) \imp (Q \vee \neg R)\)\((P \wedge Q) \vee (Q \vee \neg R)\)

    3. \((P \wedge Q) \wedge (R \wedge \neg R)\). This is necessarily false, so it is also equivalent to \(P \wedge \neg P\).

    4. Either Sam is a woman and Chris is a man, or Chris is a woman.

  23. Use De Morgan's Laws and any other logical equivalence facts you know to simplify the following statements. Show all your steps. Your final statements should have negations only appear directly next to the sentence variables or predicates (\(P\), \(Q\), \(E(x)\), etc.), and no double negations. It would be a good idea to use only conjunctions, disjunctions, and negations.

    1. \(\neg((\neg P \wedge Q) \vee \neg(R \vee \neg S))\)
    2. \(\neg((\neg P \imp \neg Q) \wedge (\neg Q \imp R))\)

    3. For both parts above, verify your answers are correct using truth tables. That is, use a truth table to check that the given statement and your proposed simplification are actually logically equivalent.

  24. Consider the statement, If a number is triangular or square, then it is not prime

    1. Make a truth table for the statement \((T \vee S) \imp \neg P\).

    2. If you believed the statement was false, what properties would a counterexample need to possess? Explain by referencing your truth table.

    3. If the statement were true, what could you conclude about the number 5657, which is definitely prime? Again, explain using the truth table.

    జవాబు వెల్లడి చేయండి

    Hint:

    1. There will be three rows in which the statement is false.

    2. Consider the three rows that evaluate to false, and say what the truth values of \(T\), \(S\), and \(P\) are there.

    3. You are looking for a row in which \(P\) is true and the whole statement is true.

  25. Tommy Flanagan was telling you what he ate yesterday afternoon. He tells you, I had either popcorn or raisins. Also, if I had cucumber sandwiches, then I had soda. But I didn't drink soda or tea. Of course, you know that Tommy is the world's worst liar, and everything he says is false. What did Tommy eat?

    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.

    జవాబు వెల్లడి చేయండి

    Hint:

    Write down three statements, and then take the negation of each (since he is a liar). You should find that Tommy ate one item and drank one item. (\(Q\) is for cucumber sandwiches.)

  26. Can you chain implications together? That is, if \(P \imp Q\) and \(Q \imp R\), does that means the \(P \imp R\)? Prove that the following is a valid deduction rule:

    \(P \imp Q\)
    \(Q \imp R\)
    \(\therefore\) \(P \imp R\)

  27. Suppose \(P\) and \(Q\) are (possibly molecular) propositional statements. Prove that \(P\) and \(Q\) are logically equivalent if and only if \(P \iff Q\) is a tautology.

    జవాబు వెల్లడి చేయండి

    Hint:

    What do these concepts mean in terms of truth tables?

  28. Suppose \(P_1, P_2, \ldots, P_n\) and \(Q\) are (possibly molecular) propositional statements. Suppose further that

    \(P_1\)
    \(P_2\)
    \(\vdots\)
    \(P_n\)
    \(\therefore\) \(Q\)

    is a valid deduction rule. Prove that the statement \[(P_1 \wedge P_2 \wedge \cdots \wedge P_n) \imp Q\] is a tautology.

  29. Consider the statements below. Translate each into symbols, using the predicate \(F(x,y)\) for person \(x\) can be fooled at time \(y\). Decide whether any of the statements are equivalent to each other, or whether any imply any others, in this context or in general.

    1. You can fool some people all of the time.

    2. You can fool everyone some of the time.

    3. You can always fool some people.

    4. Sometimes you can fool everyone.

  30. Suppose \(P(x)\) is some predicate for which the statement \(\forall x P(x)\) is true. Is it also the case that \(\exists x P(x)\) is true? In other words, is the statement \(\forall x P(x) \imp \exists x P(x)\) always true? Is the converse always true? Assume the domain of discourse is non-empty.

    జవాబు వెల్లడి చేయండి

    Hint:

    Try an example. What if \(P(x)\) was the predicate, \(x\) is prime? What if it was, If \(x\) is divisible by 4, then it is even? Of course examples are not enough to prove something in general, but that is entirely the point of this question.

  31. Simplifying negations will be especially useful when we try to prove a statement by considering what would happen if it were false. For each statement below, write the negation of the statement as simply as possible. Don't just say, It is false that

    1. Every number is either even or odd.

    2. There is a sequence that is both arithmetic and geometric.

    3. For all numbers \(n\), if \(n\) is prime, then \(n+3\) is not prime.

    జవాబు వెల్లడి చేయండి

    Hint:

    It might help to translate the statements into symbols and then use the formulaic rules to simplify negations (i.e., rules for quantifiers and De Morgan's laws). After simplifying, you should get \(\forall x(\neg E(x) \wedge \neg O(x))\) for the first one, for example. Then translate this back into English.

  32. We can simplify statements in predicate logic using our rules for passing negations over quantifiers before applying logical equivalence to the inside propositional part. Simplify the statements below (so negation appears only directly next to predicates).

    1. \(\neg \exists x \forall y (\neg O(x) \vee E(y))\)
    2. \(\neg \forall x \neg \forall y \neg(x \lt y \wedge \exists z (x \lt z \vee y \lt z))\)

    3. There is a number \(n\) for which no other number is less than or equal to \(n\).

    4. It is false that for every number \(n\) there are two other numbers which \(n\) is between.

    జవాబు వెల్లడి చేయండి

    1. \(\forall x \exists y (O(x) \wedge \neg E(y))\)
    2. \(\exists x \forall y (x \ge y \vee \forall z (x \ge z \wedge y \ge z))\)

    3. There is a number \(n\) for which every other number is strictly greater than \(n\).

    4. There is a number \(n\) which is not between any other two numbers.

  33. Simplify the statements below to the point that negation symbols occur only directly next to predicates.

    1. \(\neg \forall x \forall y (x \lt y \vee y \lt x)\).

    2. \(\neg(\exists x P(x) \imp \forall y P(y))\).

Symbols used here

n!
factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.
a \equiv b \pmod n
congruent modulo n
n divides a − b; a and b have the same remainder.
\forall,\ \exists
for all, there exists
Quantifiers: every x; at least one x.
\neg,\ \wedge,\ \vee,\ \Rightarrow,\ \Leftrightarrow
not, and, or, implies, iff
Logical connectives.
P(A),\ P(A \mid B)
probability, conditional probability
Chance of A; chance of A given that B happened.
i
imaginary unit
i² = −1.
\neq
not equal
The two sides are different.
O(n^2),\ \Theta,\ \Omega
big-O notation
Grows no faster than n² (up to a constant), for large n.
\binom{n}{k}
binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).
\sum_{k=1}^{n} a_k
summation
Add a_k for k = 1 up to n.
x \in A,\ A \subseteq B
element of, subset
x belongs to A; every element of A is in B.
A \cup B,\ A \cap B,\ A \setminus B
union, intersection, difference
In either; in both; in A but not B.
\emptyset,\ |A|
empty set, cardinality
The set with no elements; the number of elements of A.
\blacksquare\ \text{or}\ \square
end of proof (halmos)
Marks the point where the statement has been established.
a \bmod n
remainder
What is left after dividing a by n.

How to: Rules of Logic

  1. Use truth tables to determine whether two statements are logically equivalent.
  2. Use truth tables to determine whether a deduction rule is valid.
  3. Use logical equivalence and deduction rules to simplify statements and make deductions.

Questions people ask

What makes mathematics "discrete"?

It deals with separate, countable objects — integers, graphs, statements — rather than continuous quantities. No limits, no infinitesimals; instead induction, counting and logic.

How does a proof by induction work?

Show the statement for the first case, then show that whenever it holds for n it holds for n + 1. Like dominoes: the first falls, and each knocks over the next.

మీ సొంత ప్రయత్నించండి

Parts of this page are adapted from Levin, Discrete Mathematics: An Open Introduction (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

ఇంకా Discrete Math & Logic

Truth tablesSums and inductionProof by inductionAlgorithms and growth of functions