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I want to prove that $\{\leftrightarrow,+\}$ is not a complete/adequate set of connectives.(+ is $xor$)

I define a set $\Phi$ as the smallest set of all wffs that use only connectives from this set $\{\leftrightarrow,+\}$ and :

  1. If A is a propositional variable then $A\in\Phi$.
  2. If $\phi_1,\phi_2\in\Phi$ then $\phi_1 \leftrightarrow \phi_2\in \Phi$ and $\phi_1+\phi_2\in\Phi$.

Now I have to find a property that holds for every $\phi\in\Phi$ but not in general. After some experiments I found that every $\phi\in\Phi$ has an even number of True values in its truth table. Then if I take the disjunction of $\phi_1,\phi_2\in\Phi$, we prove that the property doesn't hold. So, we have proved that the set $\{\leftrightarrow,+\}$ is not adequate. How can I prove this property? I tried with induction but I got stuck.

Thanks!

Mauro ALLEGRANZA
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Failousa
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2 Answers2

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What you’re suggesting doesn’t quite work as you’ve stated it: a propositional variable has only one True value in its truth table. However, it is true that every $\varphi\in\Phi$ either is a propositional variable or has an even number of True values in its truth table.

The standard approach is to prove it by induction on the complexity of the wff. The basis step is to check that every propositional variable has it; this is clearly true. The induction step is to show that if $\varphi,\psi\in\Phi$, and $\varphi$ and $\psi$ both have the property, then $\varphi\leftrightarrow\psi$ and $\varphi+\psi$ both have the property.

You can see an example of such an argument in my answer to this question.

Brian M. Scott
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  • Great!Thank you for your answer!! – Failousa Apr 03 '20 at 18:50
  • @Failousa: You’re welcome! – Brian M. Scott Apr 03 '20 at 18:50
  • I have a question. Even though the property holds for $\phi$ and $\psi$ how do I prove that there will be an even number of pairs, for example ${T,T}$ and ${F,F}$ in the table in order to give us an even number of True values in $\phi \leftrightarrow \psi$. – Failousa Apr 03 '20 at 22:23
  • @Failousa: Let $n_{TT}$ be the number of rows in which $\varphi$ is true and $\psi$ is true, $n_{TF}$ the number in which $\varphi$ is true and $\psi$ is false, and so on. Suppose that $n_{TT}$ is odd; $\varphi$ is true in an even number of rows, so $n_{TT}+n_{TF}$ is even, and $n_{TF}$ must be odd. Similarly, $\psi$ is true in an even number of rows, so $n_{FT}$ must be odd. But then $n_{TT}+n_{TF}+n_{FT}$ is odd, and since the total number of rows is even, $n_{FF}$ must be odd. Then $n_{TT}+n_{FF}$ is even, so $\varphi+\psi$ are true in an even number of rows. Can you finish it now? – Brian M. Scott Apr 03 '20 at 22:49
  • Yes of course! Thank you very much! – Failousa Apr 04 '20 at 10:32
  • @Failousa: You're welcome! – Brian M. Scott Apr 04 '20 at 10:35
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We have $(T \leftrightarrow T)=(F \leftrightarrow F)=T$ and $(T \oplus T)=(F \oplus F)=F$. It follows that $\{\leftrightarrow, \oplus\}$ cannot be complete, for only connectives $\varphi$ for which $\varphi(T,T)=\varphi(F,F)$ can be generated from $\leftrightarrow$ and $\oplus$.

Geoffrey Trang
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