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Usually WolframAlpha understands what logical expression I have entered but sometimes I can't find a way to do it especially when you are using equivalent sign.

For example if I have $$ p \land q \Leftrightarrow \lnot q$$ I type (p AND q) equivalent (NOT q) and it doesn't work, I have tried other various methods it didn't work either.

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  • $\begingroup$ This is very strange, perhaps a bug, since "a equivalent b" is interpreted correctly. $\endgroup$
    – magma
    Jan 24, 2012 at 20:01

2 Answers 2

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Use XNOR instead of equivalent.

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  • $\begingroup$ yeah it's same and it works, thanks! $\endgroup$
    – Templar
    Jan 23, 2012 at 12:58
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Many people (including mathematicians) often mistakenly use $ P \Leftrightarrow Q \Leftrightarrow R $ to indicate equivalence, that is $ (P \Leftrightarrow Q) \wedge (Q \Leftrightarrow R) $, which is a way of saying $ P $ is equivalent to $ Q $ which is also equivalent to $ R $. (See goblin's answer to the question “IFF” (if and only if) vs. “TFAE” (the following are equivalent).)

I think it is confusing to think of if-and-only-if as equivalence relationship, since being an associative operator(1), we are allowed to write expressions such as $ P \Leftrightarrow Q \Leftrightarrow R $, which are not ambiguous at all (compare with $ P \Rightarrow Q \Rightarrow R $, which is only equivalent to $ P \Rightarrow (Q \Rightarrow R) $).

lhf's solution works fine if you are trying to establish an equivalence relationship between only two sub-expressions, but XNOR is not an equivalent of iff operator! So, for instance, compare the following two expressions: ($ \bar\veebar $ is used for XNOR)

$$ \begin{array}{|c|c|c|c|c|} \hline P & Q & R & P \Leftrightarrow Q \Leftrightarrow R & P\ \bar\veebar\ Q\ \bar\veebar\ R \\ \hline \top & \top & \top & \top & \bot \\ \hline \top & \top & \bot & \bot & \top \\ \hline \top & \bot & \top & \bot & \top \\ \hline \top & \bot & \bot & \top & \bot \\ \hline \bot & \top & \top & \bot & \top \\ \hline \bot & \top & \bot & \top & \bot \\ \hline \bot & \bot & \top & \top & \bot \\ \hline \bot & \bot & \bot & \bot & \top \\ \hline \end{array} $$

As can be seen in the truth table, XNOR and iff operators are not equivalent; furthermore, you might have noticed that they are negation of each other!

I am going to omit the proof, simply because I don't have one, but a given

$$ P_1 \Leftrightarrow P_2 \Leftrightarrow \ldots \Leftrightarrow P_n $$

is equivalent to

$$ P_1\ \bar\veebar\ P_2\ \bar\veebar\ \ldots\ \bar\veebar\ P_n $$

if and only if $ n $ is an even number; else is equivalent to

$$ P_1 \veebar P_2 \veebar \ldots \veebar P_n $$

where $ \veebar $ is used for XOR.

Both XNOR() and XOR() operators are available on WolframAlpha.


(1): $ P \Leftrightarrow Q \Leftrightarrow R $ is same as $ (P \Leftrightarrow Q) \Leftrightarrow R $ which is also same as $ P \Leftrightarrow (Q \Leftrightarrow R) $.

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