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I am struggling with a particular case in the (inductive) proof of Theorem 2.8.3 (i) of Logic and Structure by Dirk Van Dalen ($c \neq x$ in the Theorem statement is a variable)

Theorem 2.8.3

The cases when we consider proof trees for $\Gamma \vdash \phi$ for all rules but and-elimination/if-elimination I don't encounter any difficulty with as the inductive hypothesis (on the weight of proof tree) can be straightforwardly applied, but when the proof tree is that of and-elimination (say), the parent of the consequent may have occurrences of the variable $x$. To make matters worse I couldn't eliminate the problem by attempting to use the induction hypothesis with a 'fresh' variable $m$ replacing all occurrences of $x$ in the parent of the consequent since all such occurrences may be bound.

Any help with this matter would be much appreciated.

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A priori, you are right, there might be a problem in the rules $\land_E$ and $\to_E$, because the . But in fact, the problem is easily solved because there is another nice property for natural deduction:

Lemma: If $\Gamma \vdash \varphi$ and $x \in FV(\varphi)$ then $x \in FV(\psi)$ for some $\psi \in \Gamma$.

This lemma can be easily proved by induction on the derivation of $\Gamma \vdash \varphi$ (if you prefer, you can prove it simultaneously to the proof of van Dalen's Theorem 2.8.3.(i)). Note that you are in language where the only connectives are $\land, \to, ⊥$ and $\forall$ (p. 91).

Thanks to the lemma above, you do not have any problem in the proof of Theorem 2.8.3.(i) with the cases $\land_E$ and $\to_E$. For instance, for $\land_{E_i}$ (with $i \in \{1,2\}$), you have that \begin{align} \dfrac{\quad\Gamma\\\quad \ \vdots\\\varphi_1 \land \varphi_2}{\varphi_i}\land_{E_i} \end{align} According to your hypothesis, $x$ does not occur in $\Gamma$ or $\varphi_i$, but what about $\varphi_j$ with $j\neq i$? By the lemma above, if $x$ occurred free in $\varphi_j$ then $x$ would occur free in $\Gamma$, which contradicts the hypothesis. So, $x \notin FV(\varphi_j)$. Moreover, if $x$ occurred bound in $\varphi_j$ then you could rename bound variables in $\varphi_j$ so that $x$ does not occur bound in $\varphi_j$. Therefore, $x$ do not occur in $\Gamma$ or $\varphi_1$ or $\varphi_2$. Hence, you can apply the inductive hypothesis to the derivation of $\Gamma \vdash \varphi \land \varphi_2$. Then, you can easily conclude by yourself.

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  • $\begingroup$ Thank you kindly for your reply, if I am understanding correctly, this helper lemma eliminates some difficulty but if $\phi_j = \phi_2 \neq \phi_1$ contains bound occurrences of $x$, would we not need another helper lemma giving us a proof of $\phi_1 \wedge \phi_2^{*}$ where $\phi_2^{*}$ is $\phi_2$ with all bound occurrences of $x$ replaced by a new variable, of the same weight as the and elimination tree $\wedge_{E_i}$ in your answer, so that we can proceed and apply the inductive hypothesis? $\endgroup$ – porridgemathematics Jun 4 '20 at 11:32
  • $\begingroup$ Basically its the '" rename all bound variables in $\phi_j$ " that is troubling me. $\endgroup$ – porridgemathematics Jun 4 '20 at 11:41
  • $\begingroup$ Also isn't the lemma false as stated? Consider $ \{ \forall x. (x=x)\} \vdash (v=v)$ where $v \neq x$, clearly $v$ is free in the consequent, but not free in any hypothesis? $\endgroup$ – porridgemathematics Jun 4 '20 at 11:46
  • $\begingroup$ (answer to the first two comments) - Yes, but actually this further lemma is implicit, because formulas in predicate logic can be identified up to renaming of bound variables. In other words, for instance, $\forall x \, x = x$ and $\forall y \, y =y$ are the same formula. See Theorem 2.5.6 in van Dalen's textbook. $\endgroup$ – Taroccoesbrocco Jun 4 '20 at 11:50
  • $\begingroup$ Oh I see! I overlooked this completely as I hadn't looked over that theorem (I am actually using notes that are only loosely based on Van Dalen's text), will do now $\endgroup$ – porridgemathematics Jun 4 '20 at 11:54

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