A conditional statement, as we saw in SectionΒ 1.1, has the form βif \(p\) then \(q\text{.}\)β We use the connective \(p\rightarrow q\) for conditional statements.
Suppose I say to you βIf it rains tomorrow, then class is cancelled.β When would you accuse me of lying to you? For example, if it doesnβt rain and we have class, would you have thought my statement false? No. Now, you might not think my statement is true, either. But remember, statements must be either true or false. So if it is not false, then it is true.
Now, what if it doesnβt rain, but I cancel class anyway? Would I have lied? No, I didnβt tell you I wouldnβt cancel class. So, again, this would be a true statement.
The only time you could accuse me of having made a false statement is if it rains and we donβt cancel class. If we think of βit rains tomorrowβ as \(p\) and βclass is cancelledβ as \(q\text{,}\) then the only time the statement is false is when \(p\) is true and \(q\) is false.
Give an example of a conditional statement, not one of the examples from above. Write in your own words why \(p\rightarrow q\) must be true whenever \(p\) is false.
Use a truth-table to show \(\sim(p \rightarrow q)\) is logically equivalent to \(p\ \wedge\sim q\text{.}\) This is the rule for negating an if...then. Like DeMorganβs Law, it is worth committing to memory.
The second equivalence is the negation of the conditional. It is going to be really important that we understand that the negation of a conditional is not a conditional itself.
Give the truth-table for \(q\rightarrow p\text{.}\) Is it equivalent to \(p\rightarrow q\text{?}\) The statement \(q\rightarrow p\) is the converse of \(p\rightarrow q\text{.}\)
Give the truth-table for \(\sim q\rightarrow \sim p\text{.}\) Is it equivalent to \(p\rightarrow q\text{?}\) The statement \(\sim q\rightarrow \sim p\) is the contrapositive of \(p\rightarrow q\text{.}\)
The statement \((p \rightarrow q) \wedge (q\rightarrow p)\) is useful in mathematics, but it is not very concise. We can simplify this statement by introducing an additional logical connective.
The biconditional is really just the combination of two conditionals: \(p\rightarrow q\) and \(q\rightarrow p\text{.}\) In particular, \(p\leftrightarrow q\equiv (p\rightarrow q)\wedge (q\rightarrow p)\text{.}\)
We read the biconditional \(p \leftrightarrow q\) as β\(p\) if and only if \(q\text{.}\)β Determine which of the two statements \(p \rightarrow q\) and \(q\rightarrow p\) is equivalent to β\(p\) if \(q\)β and which is β\(p\) only if \(q\text{.}\)β Think about when each statement should be true (or false).
In mathematics we might see statements such as β\(p\) is necessary for \(q\)β or β\(p\) is sufficient for \(q\)β. These are really conditional statements.
\(p\) is sufficient for \(q\) means \(p\rightarrow q\text{.}\)
Using the truth-table for \(\sim p\rightarrow \sim q\text{,}\) explain why \(\sim p\rightarrow \sim q\) is not the negation of \(p\rightarrow q\text{.}\)
Write each of the following statements in symbolic form and determine whether they are logically equivalent. Include a truth-table and clearly state your conclusion.
Write each of the following statements in symbolic form and determine whether they are logically equivalent. Include a truth-table and clearly state your conclusion.
If statements \(P\) and \(Q\) are logically equivalent, then the statement \(P\leftrightarrow Q\) is a tautology. Conversely, if \(P\leftrightarrow Q\) is a tautology, then \(P\) and \(Q\) are logically equivalent. Convert the logical equivalence \(p\rightarrow (q\rightarrow r)\equiv (p\ \wedge q)\rightarrow r\) to a statement using \(\leftrightarrow\) and use a truth-table to verify the statement is a tautology.