Does (If not P then Q) imply (If P then Q)? My truth table says yes but I want verification
As the title says, is this true?
$$(\lnot P \to \lnot Q) \to (P \to Q)$$
The truth table is
\begin{array}{rrrrrr} P & Q & \lnot P & \lnot Q & \lnot P \to \lnot Q & P \to Q & (\lnot P \to \lnot Q) \to (P \to Q) \\ \hline T & T & F & F & T & T & T \\ T & F & F & T & T & F & F \\ F & T & T & F & F & T & T \\ F & F & T & T & T & T & T \\ \end{array}
It seems like it's true from the table.
If it is true, is it true because $$(\lnot P \to \lnot Q) \to (P \to Q)$$ has the same truth table corresponding to the $\to$ connective which is false only when the antecedent is T but the consequent is F?
Or is it true because the statement is true when the premises of $\lnot P \to \lnot Q$ and $P \to Q$ are true?
If it's not true, why not?
$\endgroup$ 22 Answers
$\begingroup$$(\lnot P\to\lnot Q)\to(P\to Q)$ is not a tautology because it is not true when $P$ is true but $Q$ is false. That is shown in the second row of your truth table.
Likewise, it is not a contradiction. The statement is conditionally true.
The statement is logically equivalent to $\lnot(P\land\lnot Q)$, also to $(\lnot P\lor Q)$.
Now $(\lnot P\to\lnot Q)\to(Q\to P)$ is a tautology in classical logic. Notice the order of the terms.
Indeed $\lnot P\to \lnot Q$ is the contrapositive of $Q\to P$, and the two are logically equivalent.
$\endgroup$ 7 $\begingroup$No, if we have a statement "$P$ then $Q$", then "$\neg P$ then $\neg Q$" is the inverse of the statement. The inverse being true does not imply the statement is true.
For instance consider a class where the cutoff for an $A$ is $90\%$. Consider the statement $$ \text{"If you have above an }80\%\text{, then you will receive an }A\text{."} $$ This statement is not true. However its inverse is true.$$ \text{"If you do not have above an }80\%\text{, then you will not receive an }A\text{."} $$
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