My question is regarding the inference of a truth table for an operator given how it behaves according to introduction and elimination. This follows from an exercise I read, and it got me thinking if it can be multiple solutions to it.

Consider for example the operator $\otimes$ and the following rules:

$(1)$ Introduction: Given $A$ and $\neg B$ follows conclude $A \otimes B$, given $\neg A$ and $B$ follows $A\otimes B$.

$(2)$ Elimination: Given $A\otimes B$ and $A$ follows $\neg B$; given $A\otimes B$ and $B$ follows $\neg A$.

Looks like is possible to determine uniquely two entries in the table for $\otimes$: $$\begin{array}{|c|c|c|c|} \hline A\otimes B& A &B \\ \hline &T & T\\ \hline T& T& F\\ \hline T& F&T \\ \hline &F &F \\ \hline \end{array}$$

But are the remaining values determined?. For example, for $A,B$ I can conclude $A\otimes\neg B$, but do I have enough information to say that $A\otimes \neg B=\neg (A\otimes B)$ ?.

The first elimination rule and true value for $A$ imply $\neg B$ so seems that is not possible to have $A\otimes B$ as truth in the first row. And considering the last row, if $A\otimes B$ were to have the value True then together with the False for $A$ imply True for $B$, then the complete table would be given by

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

If this is okay, here the truth table seems to be okay, but is this always possible?, introduction and elimination rules always determine uniquely the truth table for an operator?, does this depends of the logic system we're using?.


1 Answer 1


Your completion of the last row is not correct: from the given information, $A\otimes B$ could be either true or false. In the last row each of your four conditions (two introduction, two elimination) has at least one false premise (either $A$ or $B$), so no conclusion can be drawn: whatever value you give to $A\otimes B$, there will be no inconsistency.

So the answer to the question in your title is "no", and in fact your own example proves it.

  • $\begingroup$ Thanks a lot, but I have one question regarding your comment: By your argument I believe that not only the last row could have both true values, but also the first row, and there would be no inconsistency in defining $A\otimes B$ as $T$ or $F$ when $A$ and $B$ are both $T$, is that right?. $\endgroup$
    – Cure
    Jul 30, 2014 at 3:29
  • $\begingroup$ No, in the first row $A\otimes B$ has to be $F$. The argument you gave for this in your question is correct. $\endgroup$
    – David
    Jul 30, 2014 at 4:25

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