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Discrete Math for Shockers

John Hammond

Section 2.4 Logical Arguments

“Vote for me, and your taxes will go down.” “If \(x^2\) is divisible by 4, then \(x\) is even.” “If you sneeze with your eyes open, your eye will burst.” These statements are all logical arguments. In this section we’ll discuss how to determine if such a statement is valid or invalid.

Subsection

Definition 2.4.1. Logical Arguments.

An argument is a sequence of statements (premises) that ends with a conclusion.
A valid argument is one where the conclusion follows from the truth of the premises.
For the sequence of premises \(p_1, p_2, \dots, p_n\) and conclusion \(q\text{,}\) an argument is valid if:
\begin{equation*} p_1 \wedge p_2 \wedge \dots \wedge p_n \to q \end{equation*}
is a tautology.
A fallacy is a form of incorrect reasoning that leads to an invalid argument.

Example 2.4.2.

Consider the following argument:
  • If you give a mouse a cookie, then he’ll ask for a glass of milk.
  • You gave a mouse a cookie.
  • Therefore he asked for a glass of milk.
This argument is of the form:
\(p \to q\)
\(p\)
\(\therefore q\)
and is valid since \(((p \to q) \wedge p) \to q\) is a tautology.
Below is a table of common argument forms, also known as rules of inference. Observe that we’re using \(\Rightarrow\) to show “if, then”. This symbolic choice is to highlight the difference between the hypotheses and conclusion of the argument, and carries no additional meaning for the purposes of this course.
List 2.4.3. Basic argument forms - rule of inference
Let \(p, q \and r\) be logical propositions. The following are tautologies:
Modus Ponens/Direct reasoning
\(\displaystyle (p\to q) \land p \Rightarrow q\)
Modus Tollens / Indirect Reasoning
\(\displaystyle (p \to q) \land \neg q \Rightarrow \neg p\)
Disjunctive Addition
\(\displaystyle p\Rightarrow (p\lor q)\)
Conjunctive Simplification
\((p \land q) \Rightarrow p\)
\((p \land q) \Rightarrow q\)
Disjunctive Simplification
\((p \lor q) \land \neg p \Rightarrow q\)
\((p \lor q) \land \neg q\Rightarrow p\)
Chain Argument / Syllogism
\(\displaystyle (p \to q) \land ( q \rightarrow r) \Rightarrow (p\to r)\)
Conjunction
\(\displaystyle (p) \land (q) \Rightarrow (p \land q)\)
Resolution
\(\displaystyle (p \lor q) \land (\neg p \lor r) \Rightarrow (q \lor r)\)
It would be good practice to verify that each of the arguments in table List 2.4.3 is a tautology using either truth tables or logical equivalences.

Example 2.4.4.

What rules of inference are being used here?
  • Kangaroos live in Australia and are marsupials. Therefore, kangaroos are marsupials.
  • If I go swimming, then I will stay in the sun too long. If I stay in the sun too long, then I will sunburn. Therefore, if I go swimming, then I will sunburn.
  • I read a horror story. When I read horror stories, I have weird dreams. Therefore I had weird dreams.
Video / Answer.

Definition 2.4.5. Quantified Arguments.

  • Existential instantiation describes the argument:
    \begin{equation*} \exists x P(x) \implies P(c) \end{equation*}
    for some \(c\) in the domain.
  • Existential generalization describes the argument:
    \begin{equation*} P(c) \implies \exists x P(x) \text{.} \end{equation*}
  • Universal modus ponens describes the argument:
    \begin{equation*} \forall x (P(x) \to Q(x)) \wedge P(a) \implies Q(a) \end{equation*}
    where \(a\) is a particular element in the domain.
  • Universal modus tollens describes the argument:
    \begin{equation*} \forall x (P(x) \to Q(x)) \wedge \neg Q(a) \implies \neg P(a) \end{equation*}
    where \(a\) is a particular element in the domain.

Example 2.4.6.

Show that the premises, “A student in this class has not read the book,” and “Everyone in this class passed the first exam” imply the conclusion “Someone who passed the first exam has not read the book.” Let \(C(x)\) be “\(x\) is in this class,” \(B(x)\) “x has read the book”, and \(P(x)\) “\(x\) passed the first exam.”
Video / Answer.
Note that the number of the example is wrong, the video says 1.4.7, Sorry!
Solution.
The statement, “someone who passed the first exam has not read the book” means there is someone in class did those things. \(\exists x (P(x) \wedge \neg B(x))\)
Table 2.4.7. Argument step-by-step
Step Reason
1. \(\exists x(C(x) \wedge \neg B(x))\) Premise
2. \(C(a) \wedge \neg B(a) \) Existential instantiation
3. \(C(a)\) Simplification of 2
4. \(\forall x (C(x) \to P(x))\) Premise
5. \(C(a) \to P(a)\) Universal instantiation
6. \(P(a)\) Modus ponens of 3 and 5
7. \(\neg B(a)\) simplification of 2 a different way
8. \(P(a) \wedge \neg B(a)\) Conjunction of 6 and 7
9 . \(\exists x (P(x) \wedge \neg B(x))\) Existential generalization from 8.

Example 2.4.8.

For the following argument, explain the rules of inference for each step leading the the conclusion:
Linda, a student in this class owns a red convertible. Everyone who owns a red convertible has gotten at least one speeding ticket. Therefore, someone in this class has gotten at least one speeding ticket.
Video / Answer.
Here are some common fallacies. They’re not typeset in a table because although you should be aware of what they are, they are not valid arguments!
  • Fallacy of affirming the conclusion:
    • “If you get an A, you pass this class. You passed the class, therefore you got an A.” (not valid!)
    • “If an animal is a type of cat, it has hair. I have hair. Therefore, I must be some kind of cat.” (not valid!)
    • \(((p \to q) \wedge q \to p)\) is not a tautology!
  • Fallacy of denying the hypothesis
    • “If you get an A, you pass this class. You didn’t get an A. Oh no! You must not have passed this class!” (still not valid!)
    • “If an animal is a type of cat, it has hair. My dog Ruby isn’t a cat. Therefore she must not have hair.” (nope)
    • \(((p\to q) \wedge \neg p \to q)\) is not a tautology!

Exercises Exercises

1.

Determine if the following deduction rule is valid:
\(p \vee q\)
\(\neg p\)
\(\therefore q\)
Solution.
The deduction rule is valid. To see this, make a truth table which contains \(p \vee q\) and \(\neg p\) (and \(p\) and \(q\) of course). Look at the truth value of \(q\) in each of the rows that have \(p \vee q\) and \(\neg p\) true.

2.

Determine if the following is a valid deduction rule:
\(p \to (q \vee r)\)
\(\neg(p \to q)\)
\(\therefore r\)
Solution.
This is valid. Make a truth table.

3.

Determine if the following is a valid deduction rule:
\((p \wedge q) \to r\)
\(\neg p \vee \neg q\)
\(\therefore \neg r\)
Solution.
It isn’t; make a truth table and see \((((p \land q) \to r) \land (\neg p \lor \neg q)) \to \neg r\) isn’t a tautology.

4.

Show that the following is a valid deduction rule by building a truth table.
\(p \to q\)
\(q \to r\)
\(\therefore (p \to r)\)

5.

Come up with all valid conclusions for this set of premises: “If you get out the leash, the dog wants to go for a walk,” “the dog wants to go for a walk if you put on shoes,” “the dog wanting to go for a walk is sufficient for me to want a cat.” “The dog doesn’t want to go for a walk.” Explain your answer noting each step of the argument.

6.

Consider the statement “Superman saves the day or Lois Lane dies. Superman saved the day, therefore Lois didn’t die.” Is the argument valid? Explain your answer.
Solution.
This is an argument of the form:
\(\text{Superman saves} \lor \text{Lois dies}\)
\(\text{Superman saved}\)
\(\therefore \neg \text{Lois didn't die}\)
This is a fallacy. In fact, it’s the fallacy of denying the hypothesis in disguise. Rewrite the disjunction as an implication to see this:
\(\neg \text{Superman saves} \to \text{Lois dies}\)
\(\text{Superman saved}\)
\(\therefore \neg \text{Lois didn't die}\)

7.

Prove that the folowing is a valid argument, explaining each rule of inference used to arrive at the conclusion.
  1. “I take the bus or I walk. If I walk I get tired. I do not get tired. Therefore I take the bus.”
  2. “All lions are fierce. Some lions do not drink coffee. Therefore, some fierce creatures do not drink coffee.”
Hint.
  1. Write each in terms of logical symbols.
  2. Let the universe of discourse be all creatures, \(L(x)\) be the statement “\(x\) is a lion,” \(C(x)\) be “\(x\) drinks coffee,” and \(F(x)\) is “\(x\) is fierce.”

8.

Show that the following argument is invalid.
If wages increase, then there will be inflation. The cost of living will not increase if there is no inflation. Wages will increase. Therefore, the cost of living will increase.
Hint.
Make a truth table.
Solution.
The argument has three propositional variables and is of the form:
\(\text{wages increase} \to \text{inflation}\)
\(\neg\text{inflation} \to \neg \text{cost of living increase}\)
\(\text{wages increase}\)
\(\therefore \text{cost of living increases}\)
... now make a truth table.
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