# Truth Table - implies false

I'm work with a task where I am not exactly sure if I proceed right. The task is saying: "We define the operation $\oplus$ by $a \oplus b = (a \wedge \neg b) \vee (\neg a \wedge b)$. Give the truth table of:"

$$(a \oplus b) \to \text{False}$$

So my question is, do I answer this with a truth table like this:

Or do I answer like this:

Or is it something I am not aware of? Thanks for any answer! :D

• $[{a}\oplus{b} \implies false] \iff [a=b]$ – barak manos Jun 9 '16 at 9:12
• And to answer your question, the truth table should consist of only two variables, so I would use the second picture, but not the first. – barak manos Jun 9 '16 at 9:13
• The existence of such a confusion as this one as if 'False' or 'f' meant a variable in this context might suggest a good case for using '0' for false and '1' for true. – Doug Spoonwood Jun 9 '16 at 23:03

"False" is not a variable, so you should not add it into the truth table. It makes no sense to say "what if false is true", which is exactly what the second row in your first table asks.

• okey, so how is the correct truth table gonna look like? – Christian Iversen Jun 9 '16 at 8:20
• @ChristianIversen I told you what the first line will look like. I suggest you now try to re-do your calculation, since I pointed out your error. – 5xum Jun 9 '16 at 8:20
• But I dont understand. Cause u said that if a is false and b is false, then a⊕b is false, which means that the statement (a⊕b)⟹⊥ should be true, right? But in my secound truth table, it is true... – Christian Iversen Jun 9 '16 at 8:25
• @ChristianIversen Whoops, my bad. Your second table is correct. – 5xum Jun 9 '16 at 8:27
• Nice! Thanks for the help man! Or girl... xD – Christian Iversen Jun 9 '16 at 8:37

Your second table is correct. "False" is not a variable, so you can't assign it two different truth values.

If the truth table was to be written for a proposition such as $a+b \implies c$, then you would have 8 rows as in the first table. But even then, the first table doesn't look correct. Consider the first row of the table. If $a=F, b=F, c=F$, then $a+b = T$ and $T \implies F$ is $F$. So you should have $F$ in the last column.

$G\equiv[(A\land\lnot B)\lor(\lnot A\land B)]$
$====================$
$C\equiv[\lnot B]$
$D\equiv[A\land C]$
$E\equiv[\lnot A]$
$F\equiv[E\land B]$
$G\equiv[D\lor F]$

$\begin{array}{cc|cccc|c} A&B&C&D&E&F&G\\ \hline 0&0&1&0&1&0&0\\ 0&1&0&0&1&1&1\\ 1&0&1&1&0&0&1\\ 1&1&0&0&0&0&0 \end{array}$