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I started by making a truth table for A and B, with

$\begin{array}{c:c|c}A&B&A \wedge B\\\hline T&T& T\\ T&F& F\\ F&T& F\\ F&F& F\end{array}$

To see what would happen, I made a table for ~A and ~B:

$\begin{array}{c:c|c}\neg A&\neg B&\neg A \wedge\neg B\\\hline F&F& T\\ F&T& F\\ T&F& F\\ T&T& F\end{array}$

I am confident this is wrong, it doesn't make sense that the union of two statements is equal to the union of the opposite of those statements, but as far as I know, it also does make sense that the output for the first line is true if I say "neither statement A nor statement B is true" as it satisfies the condition of the union of the negation of the original statements.

Why is the table wrong? Why doesn't the union of those negations equal the preceding table?

GaneGoe
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  • For $A \land B$ the third line must be : F-T. – Mauro ALLEGRANZA Sep 13 '18 at 06:52
  • For the second formula, the case T-T must result into T. – Mauro ALLEGRANZA Sep 13 '18 at 06:53
  • But T-T means..."Yes the original statement A is true and yes the other original statement B is true" so how does that qualify as the union of negations? Because the statement is that "Not A as well as Not B is true", so if "A and B" is true then that seems like a contradiction. – GaneGoe Sep 13 '18 at 06:54
  • Your third column must be the AND of the first two columns. So if the first two columns have $F$ and $F$ in the first row, the corresponding third column's entry must be $F \land F = F$. – Jaap Scherphuis Sep 13 '18 at 08:32
  • You have to decide if you are reding the column $\lnot A \land \lnot B$ against the columns $A \mid B$ or against $\lnot A \mid \lnot B$. – Mauro ALLEGRANZA Sep 13 '18 at 08:38
  • In the first case, the fourth row is $A$ is F and $B$ is F and thus $\lnot A \land \lnot B$ is $T \land T$ i.e. T. In the second case the fourth row is $\lnot A$ is T and $\lnot B$ is T and thus $\lnot A \land \lnot B$ is again $T \land T$ i.e. T. – Mauro ALLEGRANZA Sep 13 '18 at 08:40

1 Answers1

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They are not equal. Note that the second table is upside down (relative to the first).

Put them together and you get:

$$\begin{array}{c:c|c:c|c:c} A & B&\neg A& \neg B& A\land B& \neg A\land\neg B\\\hline T & T &F&F& T & F\\ T & F&F&T & F & F\\F & T&T&F& F& F\\F & F& T&T&F & T\end{array}$$

Graham Kemp
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  • If I make a table for ~A as the first column, and ~B as the second column, then don't I arrive at the first two columns of the second table in the OP? Why not? I made the third column based on those two columns, and I see no reason why I wouldn't be able to. – GaneGoe Sep 13 '18 at 07:07
  • Upside down is upside down. – Graham Kemp Sep 13 '18 at 07:10
  • I guess that would explain tables. Why isn't it right to derive the results from the 3rd and 4th column in your table? What is the generalized fallacy in that? I do not understand why that won't work. – GaneGoe Sep 13 '18 at 07:26
  • $\neg (A \wedge B) \neq (\neg A \wedge \neg B).$ – user376343 Sep 13 '18 at 08:26
  • Right I've already known that since the OP, what I am asking is why the method I used in the table doesn't work even though it seems it should logically follow. – GaneGoe Sep 13 '18 at 12:14
  • If not-A is false, and not-B is false, why would you say not-A-and-not-B is true? false-and-false is false. – Graham Kemp Sep 13 '18 at 23:33