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 \title{Theory I, Sheet 8 Solution}

 \author{Eugen Sawin}

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 \section*{Exercise 8.1}

 Let $s, t \in T(\Sigma, X)$ be terms over $X$ and $p, q$ be strings over the natural numbers. We provide proofs by induction.

 \subsection*{1. $pq \in \Pos(s) \implies p \in \Pos(s)$ }

 $(IH): pq \in \Pos(t_i) \implies p \in \Pos(t_i)$ for all $0 < i \leq n$\\\\

 \emph{Base case:}\\

 From $s \in X$ or $s \in \Sigma^{(0)}$ follows $\Pos(s) = \{\epsilon\}$.\\

 Trivially it follows that $pq \in \Pos(s) \iff pq = \epsilon$ and $pq \in \Pos(s) \implies p \in \Pos(s)$ with $pq \notin \Pos(s)$ or $pq = p = \epsilon$. \\ 

 \noindent \emph{Inductive case:}\\

 Let $s = f(t_1, ...,t_n) \in T(\Sigma, X)$ it follows $\Pos(s) = \{\epsilon\} \cup \bigcup_{i = 1}^{n}\{ip \mid p \in \Pos(t_i)\}$.\\

 \begin{align*}

 pq \in \Pos(s) &\implies p \in \Pos(s)\\ 

 \iff pqr \in \Pos(f(t_1,...,t_n)) &\implies pr \in \Pos(f(t_1,...,t_n))\\

 \iff pqr \in \{\epsilon\} \cup \bigcup_{i = 1}^{n}\{im \mid m \in \Pos(t_i)\} &\implies pq \in \{\epsilon\} \cup \bigcup_{i = 1}^{n}\{im \mid m \in \Pos(t_i)\}\\

 \end{align*}

 By (IH) we know that $qr \in \Pos(t_i) \implies r \in \Pos(t_i)$ for all $i \leq n$ and by the left concatenation of $p$ for $0 < p \leq n$ it holds that $pqr \in \Pos(f(t_1,...,t_n)) \implies pq \in \Pos(f(t_1,...,t_n))$. \\ \qed

 \subsection*{2. $pq \in \Pos(s) \implies s|_{pq} = (s|_p)|_q$ }

 $(IH): pq \in \Pos(t_i) \implies t_i|_{pq} = (t_i|_p)|_q$ for all $0 < i \leq n$\\\\

 \emph{Base case:}\\

 From $s \in X$ or $s \in \Sigma^{(0)}$ follows $\Pos(s) = \{\epsilon\}$.\\

 Trivially it follows that $pq \in \Pos(s) \iff pq = q = \epsilon$ and $pq \in \Pos(s) \implies s|_{pq} = (s|_p)|_q$ with $pq \notin \Pos(s)$ or $pq = p = \epsilon$ with $s_{pq} = s = s|_p = (s|_p)|_q$.\\

 \noindent \emph{Inductive case:}\\

 Let $s = f(t_1, ...,t_n) \in T(\Sigma, X)$ it follows $\Pos(s) = \{\epsilon\} \cup \bigcup_{i = 1}^{n}\{ip \mid p \in \Pos(t_i)\}$.\\

 By definition we know $f(t_1, ...,t_n)|_{pqr} = t_p|_{qr}$ and from $(IH)$ it follows that $t_p|_{qr} = (t_p|_q)|_r$. \\ \qed

 \subsection*{3. $p \in \Pos(s) \land q \in \Pos(t) \implies (s[t]_p)|_{pq} = t|_q$ }

 $(IH): p \in \Pos(t_i) \land q \in \Pos(k_j) \implies (t_i[k_j]_p)|_{pq} = k_j|_q$ for all $0 < i \leq n, 0 < j \leq m$\\\\

 \emph{Base case:}\\

 From $s,t \in X$ or $s,t \in \Sigma^{(0)}$ follows $\Pos(s) = \Pos(t) = \{\epsilon\}$.\\

 Trivially it follows that $p \in \Pos(s) \land q \in \Pos(t) \iff p = q = \epsilon$ and $p \in \Pos(s) \land q \in \Pos(t) \implies (s[t]_p)|_{pq} = t|_q$ with $p \notin \Pos(s) \lor q \notin \Pos(t)$ or $p = q = \epsilon$ with $s[t]_p = (s[t]_p)|_{pq} = t|_q = \epsilon$.\\

 \noindent \emph{Inductive case:}\\

 Let $s = f(t_1, ...,t_n), t = g(k_1, ...,k_m) \in T(\Sigma, X)$.\\

 By definition we know that $f(t_1, ...,t_n)[g(k_1, ...,k_m)]_{p} = f(t_1, ...t_p[g(k_1, ...,k_m)]_\epsilon, ..t_n) = f(t_1, ...g(k_1, ...,k_m), ..t_n)$ and from $(IH)$ follows that $f(t_1, ...g(k_1, ...,k_m), ..t_n)|_{pq} = g(k_1, ...,k_m)|_q = k_q = t|_q$. \\ \qed

 \section*{Exercise 8.2}

 We use symbols $\top, \bot$ for $T, F$ respectively.

 \subsection*{1. $t = \neg (x \land (\top \lor y)), \sigma = \{x \mapsto \bot, y \mapsto \top \land x\}, \tau = \{z \mapsto \top, x \mapsto \top\}$}

 \begin{align*}

 \sigma(t) &= \neg (\bot \land (\top \lor ( \top \land x)))\\

 \tau \sigma(t) &= \tau(\neg (\bot \land (\top \lor ( \top \land x)))) = \neg (\bot \land (\top \lor (\top \land \top)))\\

 \end{align*}

 \subsection*{2. $t = \neg(\top \land (\bot \lor y)), s = \neg(x \land (\bot \lor (x \land \bot)))$}

 \[\sigma(t) = \sigma(s) \text{ with } \sigma = \{x \mapsto \top, y \mapsto \top \land \bot \} \]

 \subsection*{3.}

 Substitution composition is \emph{not} commutative, we give a counterexample.

 Let $t = x$, $\sigma = \{x \mapsto \bot\}$ and $\tau = \{x \mapsto \top\}$, it follows:

 \[\tau\sigma(t) = \bot\]

 \[\sigma\tau(t) = \top\]

 We see that $\tau\sigma(t) \neq \sigma\tau(t)$.

 \end{document}