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I wanted to see how a natural deduction proof system works, so I started reading these notes. On page four we find the following rule:

$$\frac{ \begin{matrix} & & \left[\phi[t/v]\right]\\ & & \vdots\\ \exists v\phi & & \psi \end{matrix}} {\psi}$$ provided the constant $t$ does not occur in $∃vφ$, or in $ψ$, or in any undischarged assumption other than $φ [t/v]$ in the proof of $ψ$.

Here is something I ponder, but which the notes don't seem to address:

Why is $t$ not allowed to appear in in any undischarged assumption?

As I understand it, the sentence $\phi[t/v]$ is in square brackets to mean that it is being assumed (without us knowing if it's true). This is called a discharged assumption. A more intuitive example of this is the following:

$$\frac{ \begin{matrix} [\phi] & & [\phi]\\ \vdots & & \vdots\\ \psi & & \neg\psi \end{matrix}}{\neg \phi}.$$

Kijeong Lim
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Sam
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    The reason is: we know that there is something that is $\phi$ and we call it $t$, but we have to avoid that $t$ has already a meaning, in which case it can be incompatible with being $\phi$. – Mauro ALLEGRANZA Nov 09 '24 at 11:57
  • Silly example from arithmetic: we prove that there is a number x greater than 1. What happens if we state "let call it 0"? – Mauro ALLEGRANZA Nov 09 '24 at 12:02
  • @MauroALLEGRANZA I see. Though wouldn't the rule allow us to call it "$0$" as long as $0$ does not appear in (certain) other sentences of the proof of $\psi$? Is the idea here that if $0$ does not appear in any undischarged sentence, then the proof could be carried out with another arbitrary constant $c$, and thus it's immaterial whether we use $0$ or $c$? – Sam Nov 09 '24 at 12:13
  • @MauroALLEGRANZA could we modify the rule so that the condition becomes "for a new term $t$ just added to the language"? I think that would guarantee that $t$ does not appear in any undischarged assumption, nor in $\exists v\phi$, nor in $\psi$ (since these are sentences of the language). – Sam Nov 09 '24 at 12:15
  • The issue is that 0 already occurs elsewhere: onto the axioms. – Mauro ALLEGRANZA Nov 09 '24 at 13:04

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