Given the known theorems $P \Rightarrow Q$ and $\lnot P$, prove that we cannot deduce $\lnot Q.$
I made the truth table for $P \Rightarrow Q:$
Now, if $P \Rightarrow Q$ is a theorem, that means that we must exclude the second row of the truth table.
If $\lnot P$ is also a theorem, that means that $\lnot P$ is true, so $P$ must be false. Therefore, we must look in the last 2 rows of the truth table, since they're the only ones where $\lnot P$ and $P \Rightarrow Q$ are true at the same time.
I don't understand why we can't deduce $\lnot Q$ from $P \Rightarrow Q$ and $\lnot P.$ After all, if we are looking in the fourth row of the truth table where $P \Rightarrow Q$ is true and $\lnot P$ is true, then$\lnot Q$ is also true.
As you have explained, the truth table shows that when $(P⇒Q)$ and $(¬P)$ are both true, there are two possibilities: $(¬Q)$ is either true (row 4) or false (row 3).
So, $(P⇒Q)$ and $(¬P)$ both being true does not necessitate/imply that $(¬Q)$ is also true.
In other words, the result $(¬Q)$ cannot be deduced from the theorems $(P⇒Q)$ and $(¬P).$
Concretely: given that (if today is Friday, then today is a weekday) and that (today is not Friday), it is invalid to deduce that (today is a weekend).
The point is that there are infinitely many ways to give meanings to $P$ and $Q$ (like how I have just done), and whether row 3 or 4 applies depends on the assigned meanings. That is, the infinitely many interpretations of $P$ and $Q$ fall into two disjoint sets: those that correspond to row 3 and those that correspond to row 4.
To deduce a conclusion using a truth table requires that within the rows where the premises are all true, the conclusion is invariably true.