5.1: Basic
Studying the logic of individual statements is an important first step, but ultimately we will need to analyze how statements can be combined into an argument (mathematical, philosophical, political, or otherwise) that tries to convince others that some particular conclusion is true.
a finite collection of statements, called premises or hypotheses , along with a final statement, called the conclusion
\(A_1, A_2, \ldots, A_m \therefore C\) an argument with premises \(A_1, A_2, \ldots, A_m\) and conclusion \(C\)
\(\begin{array}{c} A_1 &\qquad \text{an argument with premises } A_1, A_2, \ldots, A_m \text{ and conclusion } C\\ A_2& \\ \vdots& \\ \dfrac{A_m }{C}& \end{array} \)
| (premise) | If the world is flat, it has an edge. |
| (premise) | The world does not have an edge. |
| (conclusion) | Therefore, the world is not flat. |
| (premise) | If the world is round, there exists a Titan named Atlas who holds it aloft in the heavens. |
| (premise) | The world is round. |
| (conclusion) | Therefore, the Titan Atlas exists. |
| (premise) | Rectangles are geometric objects that have four sides. |
| (premise) | Parallelograms have four sides. |
| (premise) | Tetrahedrons have four sides. |
| (conclusion) | Therefore, parallelograms and tetrahedrons are rectangles. |
When we analyse an argument, one component of the analysis should be to check whether or not its logical structure is valid , regardless of the content and truth/falsity of the individual statements making up the argument.
Of the three provided English-language example arguments above, which are “true”? Which are “logically correct”? Is there a difference?
whenever the premises are all true, the conclusion must be true as well
Whether the conclusion of an argument is actually true is irrelevant to the validity of the argument! It is the combination of possibilities of truth and falsity of the premises and conclusion together that determine whether an argument is valid.
If there is no choice of truth values for the statement variables that simultaneously make the premises all true but the conclusion false, then the argument is valid.
Test the validity of the following argument.
\(\begin{aligned} &p \rightarrow q \\ &q \rightarrow r \\ &\hline p \rightarrow r\end{aligned}\)
Solution 1
Let's write out the truth tables for the statements in the argument. However, we are only concerned with truth table rows where every premise is true , so we won't bother completing any rows where a premise ends up being false.
| (pr) | (pr) | (c) | ||||
| \(p\) | \(q\) | \(r\) | \(p \rightarrow q\) | \(q \rightarrow r\) | \(p \rightarrow r\) | |
| \(T\) | \(T\) | \(T\) | \(T\) | \(T\) | \(T\) | \(\checkmark\) |
| \(T\) | \(T\) | \(F\) | \(T\) | \(F\) | \(\ast\) | |
| \(T\) | \(F\) | \(T\) | \(F\) | \(\ast\) | \(\ast\) | |
| \(T\) | \(F\) | \(F\) | \(F\) | \(\ast\) | \(\ast\) | |
| \(F\) | \(T\) | \(T\) | \(T\) | \(T\) | \(T\) | \(\checkmark\) |
| \(F\) | \(T\) | \(F\) | \(T\) | \(F\) | \(\ast\) | |
| \(F\) | \(F\) | \(T\) | \(T\) | \(T\) | \(T\) | \(\checkmark\) |
| \(F\) | \(F\) | \(F\) | \(T\) | \(T\) | \(T\) | \(\checkmark\) |
As every row that resulted in both premises true also resulted in the conclusion true (as indicated by \(\checkmark\)), the argument is valid. (The \(\ast\) symbol indicates a truth value that we don't care about, since it is in a row with at least one premise false.)
Solution 2 Alternative solution
Rather than work out the whole truth table, let us consider the question: is there any possible way for the conclusion to be false but all the premises true? Start with the following partial truth table row.
| (pr) | (pr) | (c) | |||
| \(p\) | \(q\) | \(r\) | \(p \rightarrow q\) | \(q \rightarrow r\) | \(p \rightarrow r\) |
| \(T\) | \(T\) | \(F\) |
The conclusion is only false when \(p = T\) and \(r = F\text{;}\) fill these into the row. Now since \(p = T\text{,}\) the first premise can only be true if \(q = T\text{;}\) fill this into the row.
| (pr) | (pr) | (c) | ||||
| \(p\) | \(q\) | \(r\) | \(p \rightarrow q\) | \(q \rightarrow r\) | \(p \rightarrow r\) | |
| \(T\) | \(T\) | \(F\) | \(T\) | \(T\) | \(F\) | \(\times\) |
We have marked this row as “incorrect,” because \(q=T\) and \(r=F\) should make the second premise false! So the above truth table row is inconsistent, and therefore there is no way for conclusion to be false and all the premises true. Conclude that the argument is valid.
Remark \(\PageIndex{1}\)
The reasoning in the Alternative solution above is an example of a proof by contradiction — see Section 6.9 .
Demonstrate that the following argument is in valid.
| If \(n\) is even, then \(n\) is divisible by \(2\text{.}\) |
| If \(n\) is odd, then \(n+1\) is even. |
| If \(n\) is not divisible by \(2\text{,}\) then \(n\) is not divisible by \(4\text{.}\) |
| Therefore, if \(n\) is not divisible by \(4\text{,}\) then \(n+1\) is even. |
Solution
Introduce statement variables and write the argument in symbolic form.
\begin{align*} p & = \text{ “} n \text{ is even”} \\ q & = \text{ “} n + 1 \text{ is even”} \\ r & = \text{ “} n \text{ is divisble by } 2 \text{”} \\ s & = \text{ “} n \text{ is divisble by } 4 \text{”} \end{align*}
\(\begin{aligned} &\phantom{\neg }p \rightarrow r \\ &\neg p \rightarrow q \\&\neg r \rightarrow \neg s \\ &\hline \neg s \rightarrow q\end{aligned}\)
Try to construct a truth table row in which all the premises are true but the conclusion is false.
| (pr) | (pr) | (pr) | (c) | |||||||
| \(p\) | \(q\) | \(r\) | \(s\) | \(\neg p\) | \(\neg r\) | \(\neg s\) | \(p \rightarrow r\) | \(\neg p \rightarrow q\) | \(\neg r \rightarrow \neg s\) | \(\neg s \rightarrow q\) |
| \(T\) | \(T\) | \(T\) | \(F\) |
Start with the above partial truth table row. The conditional in the conclusion can only be false when \(\neg s = T\) and \(q = F\text{;}\) fill these into the row, also entering \(s = F\text{.}\) Now since \(q = F\text{,}\) the second premise will only be true when \(\neg p = F\text{;}\) fill this and \(p = T\) into the row. Now since \(p = T\text{,}\) the first premise will only be true when \(r = T\text{;}\) fill this and \(\neg r = F\) into the row. Finally, check that our choices of truth values for \(p,q,r,s\) are consistent with the imposed truth value for the third premise.
| (pr) | (pr) | (pr) | (c) | |||||||
| \(p\) | \(q\) | \(r\) | \(s\) | \(\neg p\) | \(\neg r\) | \(\neg s\) | \(p \rightarrow r\) | \(\neg p \rightarrow q\) | \(\neg r \rightarrow \neg s\) | \(\neg s \rightarrow q\) |
| \(T\) | \(F\) | \(T\) | \(F\) | \(F\) | \(F\) | \(T\) | \(T\) | \(T\) | \(T\) | \(F\) |
Since there exists a choice of truth values for the statement variables which makes all premises true but the conclusion false, the argument is invalid.
- If the conclusion is a tautology, the argument is automatically valid.
- If the premises are all contradictions (i.e. logically false), the argument is automatically valid.
- If the argument is valid and the premises are all tautologies, then the conclusion must also be a tautology.
- If the argument is valid and the conclusion is a contradiction, then the premises can't all be true at the same time. (That is, in this situation the conjunction of all the premises must be a contradiction.)
Remark \(\PageIndex{2}\)
- In the logical analysis of an argument, we don't care if the premises are actually true. We only care that the conclusion follows from the premises.
- The order of the premises is irrelevant to the validity of the argument. For arguments written in English language, there may be a preferred order that best illuminates validity or invalidity, but this is essentially only aesthetic from a logical-analysis point of view.
Check your understanding.
Verify that an argument \(A_1, A_2, \ldots, A_m \therefore C\) is valid if and only if \(A_1 \land A_2 \land \cdots \land A_m \Rightarrow C\text{.}\)