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.
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.
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.
Test5.1.6.For validity of an argument in symbolic language.
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.
Letβs 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 wonβt bother completing any rows where a premise ends up being false.
As every row that resulted in both premises true also resulted in the conclusion true (as indicated by \(\correct \)), the argument is valid. (The \(\ast \) symbol indicates a truth value that we donβt care about, since it is in a row with at least one premise false.)
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.
The conclusion is only false when \(p = \lgctrue \) and \(r = \lgcfalse \text{;}\) fill these into the row. Now since \(p = \lgctrue \text{,}\) the first premise can only be true if \(q = \lgctrue \text{;}\) fill this into the row.
We have marked this row as βincorrect,β because \(q=\lgctrue \) and \(r=\lgcfalse \) 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.
\begin{align*}
p \amp = \text{ β} n \text{ is evenβ} \\
q \amp = \text{ β} n + 1 \text{ is evenβ} \\
r \amp = \text{ β} n \text{ is divisble by } 2 \text{β} \\
s \amp = \text{ β} n \text{ is divisble by } 4 \text{β}
\end{align*}
Start with the above partial truth table row. The conditional in the conclusion can only be false when \(\lgcnot s = \lgctrue \) and \(q = \lgcfalse \text{;}\) fill these into the row, also entering \(s = \lgcfalse \text{.}\) Now since \(q = \lgcfalse \text{,}\) the second premise will only be true when \(\lgcnot p = \lgcfalse \text{;}\) fill this and \(p = \lgctrue \) into the row. Now since \(p = \lgctrue \text{,}\) the first premise will only be true when \(r = \lgctrue \text{;}\) fill this and \(\lgcnot r = \lgcfalse \) 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.
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 argument is valid and the conclusion is a contradiction, then the premises canβt all be true at the same time. (That is, in this situation the conjunction of all the premises must be a contradiction.)
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.
Verify that an argument \(A_1, A_2, \dotsc, A_m \therefore C \) is valid if and only if \(A_1 \lgcand A_2 \lgcand \dotsb \lgcand A_m \lgcimplies C \text{.}\)