We have the following argument:
$P \implies Q$ (Premise 1)
$P$ (Premise 2)
$—$
$∴Q$ (Conclusion)
The accompanying truth table is:
I don't understand this section of my textbook:
The premises are both true only in line four of the table, and in this
line the conclusion is true as well.You can also see from the truth table that both premises are needed to
make this argument valid. But if we were to change the truth table for
the conditional statement to make P → Q false in the first line of the
table, then the second premise of this argument would no longer be
needed.We would end up with the conclusion that, just from the single
premise P → Q, we could infer that Q must be true, since in the two
lines of the truth table in which the premise P → Q would still be
true, lines two and four, the conclusion Q is true too.
My thinking:
If we would change the $P → Q$ to be false in the first line of the truth table. Then why couldn't the conclusion be false as well? I don't understand what line $2$ and $4$ have anything to do with line $1$?
Best Answer
The discussion is about why the statement $\bot \to \bot$ is considered "true" rather than "false". That is, why the truth table of the conditional connective is defined as it is.
An argument is considered valid if, it guarantees the conclusion is true when all the premises are true.
So if $\to$ is defined as it is, then the truth of both premises, $\{P\to Q, P\}$ are required to guarantee the truth of the conclusion $Q$. (Knowing only that one is true is not enough to ensure the conclusion is true.)
$$\begin{array}{cc|cc|cc}P & Q & P\to Q & P & Q \\ \hline \bot & \bot & \top & \bot & \bot \\ \bot & \top & \top & \bot & \top \\ \top & \bot & \bot & \top & \bot \\ \top & \top & \top & \top & \top & \star \end{array}$$
However if we defined $\dot\to$ so that $\bot \dot\to \bot = \bot$ we only need the one premise $P\dot\to Q$ to guarantee the conclusion. The conclusion is always true when that premise is, whatever $P$ may be.
$$\begin{array}{cc|c|cc}P & Q & P\dot\to Q & Q \\ \hline \bot & \bot & \bot & \bot \\ \bot & \top & \top & \top & \star \\ \top & \bot & \bot & \bot \\ \top & \top & \top & \top & \star \end{array}$$