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Where I can find formal proof of there exists an equivalent parse tree for each derivation? There is a lot of informal proof of equivalency on the internet but I need formal proof to reference it in a paper.

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    $\begingroup$ There is no need to reference such a trivial fact in a paper. $\endgroup$ Sep 12 at 8:53
  • $\begingroup$ @YuvalFilmus can you please direct me to formal proof? $\endgroup$
    – Node.JS
    Sep 12 at 16:47
  • $\begingroup$ You can take look at any decent textbook. $\endgroup$ Sep 12 at 17:09
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Proposition 6.11

Full manuscript: https://www.cis.upenn.edu/~jean/gbooks/tocnotes.html

Definition 3.11.2: Given a context-free grammar $G = (V, \Sigma, P, > S)$, for any $A \in N$, if $\pi: A \stackrel{n}{\implies} \alpha$ is a derivation in $G$, we construct an A-derivation tree $t_\pi$ with yield $\alpha$ as follows.

  1. if $n = 0$, then $t_\pi$ is the one-node tree such that $dom(t_{\pi}) = \{ \epsilon \}$ and $t_{\pi}(\epsilon) = A$
  2. if $A \stackrel{n-1}{\implies} \lambda B \rho$ and if $t_1$ is the A-derivation tree with yield $\lambda B \rho$ associated with the derivation $A \stackrel{n-1}{\implies} \lambda B \rho$, and if $t_2$ is the tree associated with the production $B \rightarrow \gamma$ that is, if $\gamma = X_1 \dots X_n$,

then $dom(t_2) = \{ \epsilon, 1, \dots, k \}$, $t_2(\epsilon) = B$, and $t_2(i) = X_i$ or all $i$, $1 \leq i \leq k$, of if $\gamma = \epsilon$

then $dom(t_2) = \{ \epsilon, 1 \}$, $t_2(\epsilon) = B$, and $t_2(1) = \epsilon$,

then $t_{\pi} = t_1[u \gets t_2]$, where $u$ is the address of the leaf labeled $B$ in $t_1$.

The tree $t_{\pi}$ is the A-derivation tree associated with the derivation $A \stackrel{n-1}{\implies} \alpha$.

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