# Meaning of a Logical Operator

Is it possible to know what those operator mean if they must be involved in this logicical condition? What is all the possible meaning of those two symbol if you don't know the symbol's meaning beforehand: (¬A) ⊕ A is always true, A ⊕ A is always false and with just those definition given, others unknown

If it is not possible to get the outcome, how to prove it is insuffient?

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Insufficient data for meaningful answer. – Pedro Tamaroff Mar 21 '12 at 0:00
How to prove this is insufficient? – Victor Mar 21 '12 at 0:01
Are you talking about $¬$ and $⊕$? You can look up logic symbols in Wikipedia. – Pedro Tamaroff Mar 21 '12 at 0:03
No, i think it could be of great useful example if you could prove it insufficient to get the outcome – Victor Mar 21 '12 at 0:04
@PeterT.off I don't think it is insufficient. "Given two rules specifying the operation of ⊕, find ⊕": that's a legitimate question. See my answer. – user2468 Mar 21 '12 at 0:10

Based on the description given in the question, we can build the following truth table: $$\begin{matrix} A & A & | & ⊕ \\ \hline F & F & | & F & \color{red}{\text{A ⊕ A}}\\ F & T & | & T & \color{blue}{\text{¬A ⊕ A}}\\ T & F & | & T & \color{blue}{\text{A ⊕ ¬A}}\\ T & T & | & F & \color{red}{\text{A ⊕ A}} \\ \end{matrix}$$ Now, we can deduce $⊕$ is the exclusive or operation. Deduced directly from the rules stated in the question.

Addednum: We can easily deduce the meaning of $\neg.$ Given the set $\mathbb{B} = \{ T, F \},$ any unary operator $\neg : \mathbb{B} \to \mathbb{B}$ will either operate as identity or as negation. Since the 2 equations in the given problem differ only by $\neg,$ $$A \oplus \neg A = T, A \oplus A = F,$$ we can deduce that $\neg$ can not be identity (assume $\neg A \equiv A$ and you'll get contradiction in the given system of formulas). Hence $\neg$ is negation. QED. Now we proceed as above.

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i mean you don't even know the other symbol ¬ mean beforehand... – Victor Mar 21 '12 at 0:12
@Victor I can easily deduce the meaning of $\neg.$ Given the set $\mathbb{B} = \{ T, F \},$ any unary operator $\neg$ will either operate as identity or as negation. Since the 2 equations in the given problem differ only by $\neg,$ I can deduce that $\neg$ can not be identity, i.e., it is negation. – user2468 Mar 21 '12 at 0:19
@PeterT.off I've just posted a comment above this one answering this very same question. – user2468 Mar 21 '12 at 0:19