In a conditional statement: if a then b is not necessarily equivalent to ~a then ~b. But in a truth table, when a is false and b is false, the statement is said to be "true". For example, in the conditional statement: "If you pass the exam, I will buy you dinner," the truth table says that F, F = T: if you don't pass the exam, I will not buy you dinner - would be considered a true statement. But this translation is also exactly the same as the logical inverse of the statement, "~a then ~b," which is considered to be "not necessarily true", a seeming contradiction. My question is three fold: (1) how to reconcile the contradiction between results of the truth table (Fales, False = true, and results of the inverse statement ~A then ~ B, not necessarily true. (2) Is there a difference between the idea of "false" from truth table and the "not" from the inverse statement? If so, what is it, and why is it that the translation to an actual example results in the same sentence (as indicated by example above, "if you do not pass the exam, I will not buy you dinner") (3) for the truth table, why would a false condition and a false conclusion result in a true statement (nothing is said about the implication when the condition is false, so shouldn't this make the (false, false) or (false, true) statements "undetermined" rather than "true")? Thanks.
-
$\begingroup$ I"m going to make a guess as to what your problem is. You wrote that "if a then b" is not necessarily equivalent to "if ~a then ~b", and this is correct. But "not necessarily equivalent" doesn't mean "never equivalent". If a and b are both false, then "if a then b" is true, and so is "if ~a then ~b". The situation where "if a then b" is not equivalent to "if ~a then ~b" is when a and b have different truth values. $\endgroup$– Andreas BlassCommented Sep 23, 2015 at 16:06
2 Answers
An implication is equivalent to its contrapositive, not its inverse. So ""If you passed the exam, then I bought you dinner" is equivalent to "If I did not buy you dinner, then you did not pass the exam". In terms of the truth table:
$$\boxed{\begin{array}{cc|ccc} A & B & A \implies B & \neg A \implies \neg B & \neg B \implies \neg A \\[1ex]\hline T & T & T & T & T \\[1ex] T & F & F & T & F \\[1ex] F & T & T & F & T \\[1ex] F & F & T & T & T \end{array}}$$
-
$\begingroup$ Thanks, my question is not about contrapositive equivalence. I am confused by values of the truth table in the "false" sections (F,F = T and F,T = T), and how those truth values of the statement seems to lead to inconsistent conclusions when considering the inverse of the conditional statement. $\endgroup$ Commented Sep 23, 2015 at 7:12
-
1$\begingroup$ $A \implies B$ is the same as $\neg (A \wedge \neg B)$, which may help you understand the truth table. A falsehood implies everything. $\endgroup$– HenryCommented Sep 23, 2015 at 7:18
-
$\begingroup$ I'm new to this and need a bit more help. Is there an intuitive explanation for why F,F = T, and not merely "undetermined"? Furthermore, the truth value of the statement seems to lead to inconsistent conclusions when considering the inverse of the conditional statement. When using a venn diagram, it is clear that if a then b is not the same as ~a then ~b. But using a truth table, the row for F, F, seems to cause the conditional sentence to read the same as the inverse which is not necessarily "true" $\endgroup$ Commented Sep 23, 2015 at 7:29
-
$\begingroup$ @user52438: "If the moon is made of cheese, then pigs can fly" is a true statement. $\endgroup$ Commented Sep 23, 2015 at 16:17
-
$\begingroup$ @llmari, thanks. it seems the second part the conditional statement doesn't matter, when the first part is false, which also explains f,t=t. But, how is it that a false hypothesis always results in true statements? Can you share a bit more insight? $\endgroup$ Commented Sep 25, 2015 at 17:45
Beginners often mistakenly attribute some kind of causal aspect to the logical relation $A\implies B$.
Consider for example, the statement: If it is raining, then it is cloudy.
$Raining \implies Cloudy$
This does not say that rain causes cloudiness or that cloudiness causes rain. It just says that it is not the case that it is both raining and not cloudy.
$\neg [Raining \land \neg Cloudy]$
This would always be true when it is not raining, whether cloudy or not.
In your example, you have: "If you pass the exam, then I will buy you dinner."
$Pass\implies FreeDinner$
Or equivalently:
$\neg [Pass \land \neg FreeDinner]$
So, it is not the case that you both pass and do not get a free dinner.
This would always be true if you did not pass, whether you got a free dinner or not.
